Thermodynamics

Given,

• Rate of flow of water = 3.0 L per minute = 3.0 × 10^-3 m³ min^-1
• Initial temperature of water = 27°C
• Final temperature of water = 77°C
• Heat of combustion of the fuel = 4.0 × 10⁴ J g^-1
• Specific heat of water, c = 4.18 × 10³ J kg^-1 °C^-1
• Density of water = 1.0 × 10³ kg m^-3

The mass of water flowing through the geyser per minute is

$$\begin{aligned}\text m = \text{volume} \times \text{density} \\[1em] = (3.0 \times 10^{-3}\ \text{m}^3\ \text{min}^{-1}) \times (1.0 \times 10^{3}\ \text{kg m}^{-3}) \\[1em] = 3.0\ \text{kg min}^{-1}\end{aligned}$$

The rise in the temperature of water is

$\Delta \text T = 77°\text C - 27°\text C = 50°\text C$

Therefore the rate at which heat is taken by the water is

$$\begin{aligned}\text{mc}\Delta \text T = (3.0\ \text{kg min}^{-1}) \times (4.18 \times 10^{3}\ \text{J kg}^{-1}\ °\text C^{-1}) \times (50°\text C) \\[1em] = 6.27 \times 10^{5}\ \text{J min}^{-1}\end{aligned}$$

This heat is supplied by the combustion of the fuel. Since the heat of combustion of the gas is 4.0 × 10⁴ J g^-1, the rate of consumption of the fuel is

$$\begin{aligned}= \frac{6.27 \times 10^{5}\ \text{J min}^{-1}}{4.0 \times 10^{4}\ \text{J g}^{-1}} \\[1em] = 15.7\ \text{g min}^{-1}\end{aligned}$$

Hence, the rate of consumption of the fuel is 15.7 g min^-1.

Given,

• Mass of nitrogen, m = 2.0 × 10^-2 kg = 20 g
• Rise in temperature, ΔT = 45°C = 45 K
• Molecular mass of N₂, M = 28
• R = 8.31 J mol^-1 K^-1
• The gas is heated at constant pressure

The number of moles of nitrogen present is

$\mu = \frac{\text m}{\text M} = \frac{20\ \text g}{28\ \text{g mol}^{-1}} = \frac{5}{7}\ \text{mol}$

The amount of heat required to raise the temperature of μ moles of a gas at constant pressure by ΔT is

$\text Q = \mu\text C_p\Delta \text T$

where C_p is the molar specific heat at constant pressure. Nitrogen (N₂) is a diatomic gas, for which

$\text C_p = \frac{7}{2}\text R$

Substituting the values,

$\text Q = \left(\frac{5}{7}\ \text{mol}\right) \times \left(\frac{7}{2} \times 8.31\ \text{J mol}^{-1}\ \text K^{-1}\right) \times (45\ \text K)$

$$\begin{aligned}= \frac{5}{7} \times 29.085 \times 45 \\[1em] = 935\ \text J\end{aligned}$$

Hence, the heat that should be given to the nitrogen is 935 J.

(a) Let m₁ and c₁ be the mass and the specific heat of the body at temperature T₁, and m₂ and c₂ those of the body at temperature T₂, with T₁ > T₂. If T is the common temperature reached, then by the principle of calorimetry,

$$\begin{aligned}\text{heat lost} = \text{heat gained} \\[1em] \text m_1\text c_1(\text T_1 - \text T) = \text m_2\text c_2(\text T - \text T_2)\end{aligned}$$

Solving for T,

$\text T = \frac{\text m_1\text c_1\text T_1 + \text m_2\text c_2\text T_2}{\text m_1\text c_1 + \text m_2\text c_2} \neq \frac{\text T_1 + \text T_2}{2}$

The final temperature equals the mean temperature $\frac{\text T_1 + \text T_2}{2}$ only when m₁c₁ = m₂c₂, that is, only when the two bodies have the same thermal capacity (mass × specific heat). In general their thermal capacities are different, and so the two bodies do not necessarily settle to the mean temperature.

(b) The coolant is used to prevent the different parts of the plant from getting too hot. The heat absorbed by a substance is proportional to its specific heat. Hence a coolant having a high specific heat will remove a large amount of heat from the plant for a small rise in its own temperature, and so it cools the plant more effectively.

(c) During driving, the friction between the tyre and the road raises the temperature of the tyre and hence of the air enclosed in it. The volume of the tyre remains practically constant, so the change is isochoric and $\frac{\text P}{\text T}$ = constant. As T increases, the air pressure in the tyre also increases.

(d) The specific heat of water is about 5 times higher than that of sand. Therefore, when the same quantity of heat is given to (or taken from) the same mass of water and of sand, the water undergoes a much smaller rise (or fall) in temperature than the sand. During the day the temperature of the sea water rises much more slowly than that of the sand of the desert, so in summer the harbour town remains cooler. At night the sea water cools much more slowly than the sand, so the harbour town remains warmer. Hence the climate of a harbour town is more temperate.

Given,

• Number of moles of hydrogen, μ = 3
• The gas is initially at standard temperature and pressure
• Final volume, $\text V_2 = \frac{\text V_1}{2}$
• The walls of the cylinder are made of a heat insulator and the piston is insulated

Since the cylinder is insulated on all sides, no heat can enter or leave the gas. Hence the compression is an adiabatic process, which obeys Poisson's law,

$\text{PV}^{\gamma} = \text{constant} \quad \Rightarrow \quad \text P_1\text V_1^{\gamma} = \text P_2\text V_2^{\gamma}$

Therefore

$\frac{\text P_2}{\text P_1} = \left(\frac{\text V_1}{\text V_2}\right)^{\gamma}$

Hydrogen is a diatomic gas, for which (the vibrational modes being inactive at room temperature)

$\gamma = \frac{\text C_p}{\text C_v} = \frac{7}{5} = 1.4$

Substituting $\text V_2 = \frac{\text V_1}{2}$,

$\frac{\text P_2}{\text P_1} = \left(\frac{\text V_1}{\text V_1/2}\right)^{1.4} = (2)^{1.4} = 2.64$

Hence, the pressure of the gas increases by a factor of 2.64.

Given,

• Work done on the gas in the adiabatic change A → B = 22.3 J, so W = − 22.3 J
• Heat absorbed by the gas in the second process, Q = 9.35 cal
• 1 cal = 4.19 J

The first law of thermodynamics applied to a system in equilibrium is

$\text Q = \Delta \text U + \text W$

where Q is the heat absorbed by the system, W is the work done by the system and ΔU is the change in the internal energy of the system.

For the adiabatic change from state A to state B : No heat enters or leaves the gas, so Q = 0, and the work is done on the gas, so W = − 22.3 J. Therefore

$\Delta \text U = \text Q - \text W = 0 - (-22.3) = 22.3\ \text J$

Since the internal energy U is a state function, the change ΔU for any process that takes the gas from state A to state B will be the same, that is, 22.3 J.

For the second process from A to B : The heat absorbed is

$\text Q = 9.35\ \text{cal} = 9.35 \times 4.19 = 39.2\ \text J$

Therefore the work done by the gas is

$$\begin{aligned}\text W = \text Q - \Delta \text U = 39.2 - 22.3 \\[1em] = 16.9\ \text J\end{aligned}$$

Hence, the work done by the gas in the second process is 16.9 J.

(a) When the stopcock is suddenly opened, the gas of cylinder A rushes into the evacuated cylinder B. The two cylinders are of equal capacity, so the volume available to the gas is doubled. At constant temperature, by Boyle's law, the pressure is halved. Since the gas was initially at standard temperature and pressure (1 atm), the final pressure of the gas in A and in B is 0.5 atm.

(b) By the first law of thermodynamics, the change in the internal energy of the gas is

$\Delta \text U = \text Q - \text W$

The entire system is thermally insulated, so there is no transference of heat with the surroundings, that is, Q = 0. Again, the gas expands into the vacuum of cylinder B, so it has nothing to push against and no external work is done, that is, W = 0. Consequently

$\Delta \text U = 0 - 0 = 0$

Hence the internal energy of the gas does not change.

(c) The internal energy of an ideal gas depends only upon its temperature. Since the internal energy does not change, the temperature of the gas also does not change.

(d) No, the intermediate states do not lie on the P-V-T surface. The process is a 'free expansion' and is too rapid. The intermediate states are non-equilibrium states, for which the pressure, volume and temperature of the gas are not defined for the system as a whole. Only the initial and the final equilibrium states lie on the P-V-T surface of the system.

Given,

• Rate at which heat is supplied to the system, Q = 100 W
• Rate at which the system performs work, W = 75 J s^-1 = 75 W

By the first law of thermodynamics, the change in the internal energy of the system is

$\Delta \text U = \text Q - \text W$

where Q is the heat taken by the system and W is the work done by the system. Since both the quantities are given as rates, the equation gives directly the rate of change of the internal energy,

$$\begin{aligned}\Delta \text U = 100\ \text W - 75\ \text W \\[1em] = 25\ \text W\end{aligned}$$

which is positive.

Hence, the internal energy of the system is increasing at the rate of 25 W (that is, 25 J per second).

Given, from the graph,

• At A : P = 300 N m^-2, V = 2.0 m³
• At B : P = 600 N m^-2, V = 2.0 m³
• At C : P = 300 N m^-2, V = 5.0 m³

The gas is taken from A to C through B and is then brought back from C to A by an isobaric process. Thus the gas undergoes a cyclic process A → B → C → A, which is traced clockwise on the P-V diagram.

The work done by a system in a cyclic process is equal to the area enclosed by the closed curve, and since the cycle is traced clockwise, the net work is done by the gas. Here the closed curve is the triangle ABC, so

$\text W = \text{area } \text{ABC} = \frac{1}{2} \times \text{AC} \times \text{AB}$

The base AC is the change in volume and the height AB is the change in pressure,

$$\begin{aligned}\text{AC} = (5.0 - 2.0)\ \text m^3 = 3.0\ \text m^3 \\[1em] \text{AB} = (600 - 300)\ \text{N m}^{-2} = 300\ \text{N m}^{-2}\end{aligned}$$

Therefore

$$\begin{aligned}\text W = \frac{1}{2} \times 3.0 \times 300 \\[1em] = 450\ \text{N m} = 450\ \text J\end{aligned}$$

Hence, the total work done by the gas from B to C to A is 450 J.

22.5 g

Reason — Given,

• Mass of cold water, m_w = 20 g at 10°C
• Steam at 100°C is passed into it
• Final temperature of the mixture = 80°C
• Specific heat of water, c = 1 cal g^-1 °C^-1
• Latent heat of steam, L = 540 cal g^-1

Heat taken by the cold water in rising from 10°C to 80°C is

$$\begin{aligned}\text Q_{\text{gained}} = \text m_w\text c\Delta \text T = 20 \times 1 \times (80 - 10) \\[1em] = 1400\ \text{cal}\end{aligned}$$

Heat given out by the steam : Let m gram of steam be condensed. The steam first condenses into water at 100°C, giving out the latent heat mL, and this water then cools from 100°C to 80°C, giving out mcΔT. Hence

$$\begin{aligned}\text Q_{\text{lost}} = \text{mL} + \text{mc}\Delta\theta = \text m \times 540 + \text m \times 1 \times (100 - 80) \\[1em] = \text m \times 540 + \text m \times 20 = \text m \times 560\ \text{cal}\end{aligned}$$

By the principle of calorimetry, heat lost = heat gained,

$\text m \times 560 = 1400 \quad \Rightarrow \quad \text m = \frac{1400}{560} = 2.5\ \text g$

The condensed steam remains in the vessel as water, so the total mass of water present is

$= 20\ \text g + 2.5\ \text g = 22.5\ \text g$

6 cal

Reason — Given,

• Along the path iaf : Q₁ = 50 cal and W₁ = 20 cal
• Along the path ibf : Q₂ = 36 cal, W₂ = ?

The internal energy is a state function, so the change in the internal energy in going from the state i to the state f is the same along both the paths,

$\Delta \text U_{iaf} = \Delta \text U_{ibf}$

By the first law of thermodynamics, ΔU = Q − W. Therefore

$\text Q_1 - \text W_1 = \text Q_2 - \text W_2$

Substituting the values,

$$\begin{aligned}50 - 20 = 36 - \text W_2 \\[1em] 30 = 36 - \text W_2\end{aligned}$$

$\text W_2 = 36 - 30 = 6\ \text{cal}$

zero

Reason —

From the graph, the four states of the gas are A (V₀, P₀), B (2V₀, 3P₀), C (V₀, 3P₀) and D (2V₀, P₀). The two straight lines AB and CD cross each other at the point O.

The work done in a thermodynamic process is equal to the area enclosed between the P-V curve and the volume-axis, and the closed figure here is made up of two triangles which meet at O.

Work done in the process O → D → A : This lower triangle is traced clockwise, so the work done is positive,

$\text W_1 = \frac{1}{2} \times \text P_0 \times \text V_0$

Work done in the process O → B → C : This upper triangle is traced anticlockwise, so the work done is negative and equal in magnitude,

$\text W_2 = -\frac{1}{2} \times \text P_0 \times \text V_0$

The two triangles are of equal area but are described in opposite senses. Hence the total work done by the system in the cycle is

$$\begin{aligned}\text W = \text W_1 + \text W_2 = \frac{1}{2}\text P_0\text V_0 - \frac{1}{2}\text P_0\text V_0 \\[1em] = \text{zero}\end{aligned}$$

460 J

Reason — Given,

• Heat added in the process AB, Q_AB = 400 J
• Heat added in the process BC, Q_BC = 100 J

From the graph, the states are A (2 × 10^-3 m³, 2 × 10⁴ Pa) and C (4 × 10^-3 m³, 6 × 10⁴ Pa), and the path AC is the straight line joining them.

Since the internal energy is a state function, the change in internal energy is the same along the path ABC and along the path AC.

For the path ABC : The work done is zero along AB (isochoric) and along BC the pressure is constant, but the total heat given is

$\text Q_{\text{ABC}} = 400 + 100 = 500\ \text J$

For the closed cycle A → B → C → A : Applying the first law to the complete cycle, in which ΔU = 0, the net heat absorbed equals the net work done, which is the area of the triangle ABC,

$400\ \text J + 100\ \text J + \text Q_{\text C \to \text A} = \frac{1}{2} \times (4 \times 10^{4}) \times (2 \times 10^{-3})$

$$\begin{aligned}500 + \text Q_{\text C \to \text A} = 40 \\[1em] \text Q_{\text C \to \text A} = -460\ \text J\end{aligned}$$

Since the heat absorbed in going from C to A is − 460 J, the heat absorbed by the system in the reverse process, that is, in the process A → C, is

$\text Q_{\text A \to \text C} = 460\ \text J$

$\left(\frac{13}{2}\right)$ P₀V₀

Reason —

From the graph the cycle consists of two isochoric and two isobaric processes, the states being (V₀, P₀) → (V₀, 2P₀) → (2V₀, 2P₀) → (2V₀, P₀) → (V₀, P₀). The gas is monoatomic, so

$\text C_v = \frac{3}{2}\text R \quad \text{and} \quad \text C_p = \frac{5}{2}\text R$

Heat is extracted from the source only in those parts of the cycle in which the heat is absorbed by the gas, that is, when the pressure grows at constant volume and when the volume grows at constant pressure.

Heat absorbed when the pressure grows from P₀ to 2P₀ at the constant volume V₀ :

$\text Q_1 = \text n\text C_v\Delta \text T = \frac{3}{2}\text n\text R(\text T_2 - \text T_1) = \frac{3}{2}(\text P_2\text V_2 - \text P_1\text V_1)$

$= \frac{3}{2}(2\text P_0\text V_0 - \text P_0\text V_0) = \frac{3}{2}\text P_0\text V_0 \qquad \text{...(i)}$

Heat absorbed when the volume grows from V₀ to 2V₀ at the constant pressure 2P₀ :

$\text Q_2 = \text n\text C_p\Delta \text T = \frac{5}{2}(\text P_2\text V_2 - \text P_1\text V_1)$

$= \frac{5}{2}(2\text P_0 \times 2\text V_0 - 2\text P_0 \times \text V_0) = \frac{5}{2}(2\text P_0\text V_0) = 5\text P_0\text V_0 \qquad \text{...(ii)}$

In the remaining two processes the pressure and the volume decrease, so heat is rejected by the gas.

Adding (i) and (ii), the total heat extracted from the source in a single cycle is

$$\begin{aligned}\text Q = \text Q_1 + \text Q_2 = \frac{3}{2}\text P_0\text V_0 + 5\text P_0\text V_0 \\[1em] = \left(\frac{13}{2}\right)\text P_0\text V_0\end{aligned}$$

8.4 kJ

Reason — Given,

• Mass of water, m = 100 g = 100 × 10^-3 kg
• Initial temperature = 30°C, final temperature = 50°C, so ΔT = 20°C = 20 K
• Specific heat of water, c = 4184 J kg^-1 K^-1
• The slight expansion of the water is to be ignored

Since the expansion of the water is ignored, the change in volume is zero and hence no external work is done, W = 0. By the first law of thermodynamics,

$\Delta \text U = \text Q - \text W = \text Q - 0 = \text Q$

The heat taken by the water is

$\text Q = \text{mc}\Delta \text T = (100 \times 10^{-3}) \times 4184 \times (50 - 30)$

$$\begin{aligned}= 0.1 \times 4184 \times 20 \\[1em] = 8368\ \text J = 8.368\ \text{kJ} \approx 8.4\ \text{kJ}\end{aligned}$$

$\frac{(γ - 1)}{2\text R} \text M v^2$ K

Reason — Given,

• Molecular mass of the gas = M
• Ratio of specific heats = γ
• Speed of the vessel = v, and the vessel is suddenly brought to rest
• The vessel is thermally insulated, so no heat is lost to the surroundings

Let the vessel contain μ mole of the gas. The mass of the gas is then μM, and the kinetic energy of the gas on account of the motion of the vessel is

$\text K = \frac{1}{2}(\mu \text M)v^2$

When the vessel is suddenly brought to rest, this ordered kinetic energy of the gas is converted into the internal energy of the gas, and hence the temperature of the gas rises. Since no heat is lost, the whole of it appears as the increase in internal energy.

If ΔT be the rise in temperature of μ mole of the gas, then

$\frac{1}{2}(\mu \text M)v^2 = \mu\text C_v\Delta \text T$

For an ideal gas the molar specific heat at constant volume is

$\text C_v = \frac{\text R}{\gamma - 1}$

Substituting this value,

$\frac{1}{2}\mu \text M v^2 = \mu \times \frac{\text R}{\gamma - 1} \times \Delta \text T$

Cancelling μ and solving for ΔT,

$\Delta \text T = \frac{\text M v^2(\gamma - 1)}{2\text R}\ \text K$

15.4%

Reason —

From the graph the cycle ABCDA consists of two isochoric and two isobaric lines, the states being A (V₀, P₀), B (V₀, 2P₀), C (2V₀, 2P₀) and D (2V₀, P₀). Helium is a monoatomic gas, so

$\text C_v = \frac{3}{2}\text R \quad \text{and} \quad \text C_p = \frac{5}{2}\text R$

Work done in the complete cycle : No work is done along AB and CD, since the volume does not change there. Therefore

$\text W = 0 + 2\text P_0(2\text V_0 - \text V_0) + 0 - \text P_0(2\text V_0 - \text V_0)$

$= 2\text P_0\text V_0 - \text P_0\text V_0 = \text P_0\text V_0$

Heat given to the gas from A → B (isochoric) :

$\text Q_1 = \text n\text C_v\Delta \text T = \frac{3}{2}\text n\text R\Delta \text T$

But for this path n R ΔT = V₀ ΔP = V₀(2P₀ − P₀) = P₀V₀, so

$\text Q_1 = \frac{3}{2}\text P_0\text V_0$

Heat given to the gas from B → C (isobaric) :

$\text Q_2 = \text n\text C_p\Delta \text T = \frac{5}{2}\text n\text R\Delta \text T$

But for this path n R ΔT = 2P₀ ΔV = 2P₀V₀, so

$\text Q_2 = \frac{5}{2} \times 2\text P_0\text V_0 = 5\text P_0\text V_0$

In the processes C → D and D → A heat is rejected by the gas. Hence the efficiency of the cycle is

$\eta = \frac{\text{work done by the gas}}{\text{heat absorbed by the gas}} \times 100 = \frac{\text P_0\text V_0}{\text Q_1 + \text Q_2} \times 100$

$= \frac{\text P_0\text V_0}{\frac{3}{2}\text P_0\text V_0 + 5\text P_0\text V_0} \times 100 = \frac{\text P_0\text V_0}{\frac{13}{2}\text P_0\text V_0} \times 100$

$= \frac{2}{13} \times 100 = 15.4%$

the temperature will fall

Reason — Given, for the process W = 0 and Q < 0.

By the first law of thermodynamics,

$\Delta \text U = \text Q - \text W$

Since W = 0, we get ΔU = Q. As Q is negative, ΔU is also negative, that is, the internal energy of the gas decreases.

For an ideal gas the change in internal energy is related to the change in temperature by

$\Delta \text U = \mu\text C_v\Delta \text T$

Since ΔU < 0 and both μ and C_v are positive quantities, ΔT must be negative. Hence the temperature of the gas will fall.

The volume cannot increase, because W = P ΔV = 0 means the volume stays constant.

The change in internal energy in the process BC is − 500 R.

Reason — Given,

• One mole of a diatomic ideal gas, so the degrees of freedom f = 5 and $\text C_v = \frac{5}{2}\text R$
• T_A = 400 K, T_B = 800 K, T_C = 600 K
• The process BC is adiabatic

For one mole of a gas the change in internal energy is

$\Delta \text U = \text C_v\Delta \text T = \frac{1}{2}f\text R\Delta \text T = \frac{5}{2}\text R\Delta \text T$

For the complete cyclic process : The gas returns to its initial state, so

$\Delta \text U = 0$

Hence option 1 is incorrect.

For the process CA :

$\Delta \text U = \frac{5}{2}\text R(400 - 600) = \frac{5}{2}\text R \times (-200) = -500\ \text R$

Hence option 2 is incorrect.

For the process AB :

$\Delta \text U = \frac{5}{2}\text R(800 - 400) = \frac{5}{2}\text R \times 400 = 1000\ \text R$

Hence option 3 is incorrect.

For the process BC :

$\Delta \text U = \frac{5}{2}\text R(600 - 800) = \frac{5}{2}\text R \times (-200) = -500\ \text R$

Hence the correct statement is that the change in internal energy in the process BC is − 500 R.

208 J

Reason — Given,

• Number of moles of helium, μ = 2
• Initial temperature = 30°C, final temperature = 35°C, so ΔT = 5°C = 5 K
• R = 8.31 J mol^-1 K^-1
• The balloon is fully expandable and requires no energy in its expansion

Since the balloon is fully expandable and needs no energy to expand, the gas inside always stays at the constant atmospheric pressure. The heating is therefore an isobaric process, for which the heat required is

$\text Q = \mu\text C_p\Delta \text T$

Helium is a monoatomic gas, for which

$\text C_p = \frac{5}{2}\text R$

Substituting the values,

$\text Q = 2 \times \frac{5}{2}\text R \times 5 = 25\ \text R$

$$\begin{aligned}= 25 \times 8.31 \\[1em] = 207.75\ \text J \approx 208\ \text J\end{aligned}$$

1. If V₂ = 2 V₁ and T₂ = 3 T₁, then the energy stored in the spring is $\frac{1}{4}$ P₁V₁.

2. If V₂ = 2 V₁ and T₂ = 3 T₁, then the change in internal energy is 3 P₁V₁.

3. If V₂ = 3 V₁ and T₂ = 4 T₁, then the work done by the gas is $\frac{7}{3}$ P₁V₁.

Reason —

The gas is monoatomic, so $\text C_v = \frac{3}{2}\text R$ and the total internal energy of the gas is $\text U = \frac{3}{2}\text{nRT} = \frac{3}{2}\text{PV}$.

Force on the piston : Initially the spring is in its relaxed state, so the initial gas pressure P₁ is balanced by the external pressure alone. When the piston moves out by x, the spring is deformed and exerts a force kx on it. The new force balance on the piston therefore gives

$\text P_2\text A = \text P_1\text A + \text{kx} \qquad \text{or} \qquad \text{kx} = (\text P_2 - \text P_1)\text A$

Since the heating is carried out very slowly, the pressure rises in step with the displacement, so P varies linearly with V between the initial and the final states.

Checking option 1 : For V₂ = 2V₁ and T₂ = 3T₁, the gas equation gives

$\frac{\text P_1\text V_1}{\text T_1} = \frac{\text P_2\text V_2}{\text T_2} = \frac{\text P_2(2\text V_1)}{3\text T_1} \quad \Rightarrow \quad \text P_2 = \frac{3\text P_1}{2}$

The energy stored in the spring is

$\frac{1}{2}\text{kx}^2 = \frac{1}{2}(\text{kx})\text x = \frac{1}{2}(\text P_2 - \text P_1)\text A\text x = \frac{1}{2}(\text P_2 - \text P_1)(\text V_2 - \text V_1)$

since A x is the change in volume. Substituting the values,

$= \frac{1}{2}\left(\frac{3}{2}\text P_1 - \text P_1\right)(2\text V_1 - \text V_1) = \frac{1}{2} \times \frac{1}{2}\text P_1 \times \text V_1 = \frac{1}{4}\text P_1\text V_1$

Hence option 1 is correct.

Checking option 2 : With P₂ = $\frac{3}{2}$P₁ and V₂ = 2V₁,

$\Delta \text U = \frac{3}{2}(\text P_2\text V_2 - \text P_1\text V_1) = \frac{3}{2}\left(\frac{3}{2}\text P_1 \times 2\text V_1 - \text P_1\text V_1\right)$

$= \frac{3}{2}(3\text P_1\text V_1 - \text P_1\text V_1) = \frac{3}{2} \times 2\text P_1\text V_1 = 3\text P_1\text V_1$

Hence option 2 is correct.

Checking option 3 : For V₂ = 3V₁ and T₂ = 4T₁,

$\frac{\text P_1\text V_1}{\text T_1} = \frac{\text P_2(3\text V_1)}{4\text T_1} \quad \Rightarrow \quad \text P_2 = \frac{4\text P_1}{3}$

Since the pressure varies linearly with the volume, the work done by the gas is the area under the straight line joining the initial and the final states,

$\text W = \frac{1}{2}(\text P_1 + \text P_2)(\text V_2 - \text V_1) = \frac{1}{2}\left(\text P_1 + \frac{4}{3}\text P_1\right)(3\text V_1 - \text V_1)$

$= \frac{1}{2} \times \frac{7}{3}\text P_1 \times 2\text V_1 = \frac{7}{3}\text P_1\text V_1$

Hence option 3 is correct.

Checking option 4 : For the same case the change in internal energy is

$\Delta \text U = \frac{3}{2}(\text P_2\text V_2 - \text P_1\text V_1) = \frac{3}{2}\left(\frac{4}{3}\text P_1 \times 3\text V_1 - \text P_1\text V_1\right)$

$= \frac{3}{2}(4\text P_1\text V_1 - \text P_1\text V_1) = \frac{9}{2}\text P_1\text V_1$

By the first law of thermodynamics, the heat supplied is

$\Delta \text Q = \Delta \text U + \text W = \frac{9}{2}\text P_1\text V_1 + \frac{7}{3}\text P_1\text V_1 = \frac{27 + 14}{6}\text P_1\text V_1 = \frac{41}{6}\text P_1\text V_1$

This is not $\frac{17}{6}\text P_1\text V_1$, so option 4 is incorrect.

Hence the correct statements are 1, 2 and 3.

Note: The printed answer key of the textbook gives option (b) alone, its hint having taken the spring force as P₂A = kx. Since the spring is relaxed in the initial state, the spring force accounts only for the rise in pressure, so kx = (P₂ − P₁)A. With this relation options 1, 2 and 3 are all correct, and the heat supplied in option 4 works out to $\frac{41}{6}$ P₁V₁.

diatomic

Reason — Given,

• Work done on the gas, W = − 146 kJ = − 146 × 10³ J
• Number of moles, μ = 1 kilo mole = 1 × 10³ mol
• Rise in temperature, ΔT = 7°C = 7 K
• R = 8.3 J mol^-1 K^-1

According to the first law of thermodynamics,

$\text Q = \Delta \text U + \text W$

For an adiabatic process Q = 0, so

$\Delta \text U = -\text W$

But the change in internal energy is also ΔU = μ C_v ΔT. Therefore

$\text C_v = \frac{-\text W}{\mu\Delta \text T} = \frac{-(-146 \times 10^{3})}{(1 \times 10^{3}) \times 7}$

$= \frac{146 \times 10^{3}}{7 \times 10^{3}} = 20.8\ \text{J mol}^{-1}\ \text K^{-1}$

For a diatomic gas the molar specific heat at constant volume is

$\text C_v = \frac{5}{2}\text R = \frac{5}{2} \times 8.3 = 20.75\ \text{J mol}^{-1}\ \text K^{-1}$

Since the calculated value agrees with this, the gas is diatomic.

T_f = $\frac{3}{2}$ T₀

Reason — Given,

• Box A : 1 mole of nitrogen (diatomic) at temperature T₀, so $\text C_{v_1} = \frac{5}{2}\text R$
• Box B : 1 mole of helium (monoatomic) at temperature $\frac{7}{3}$T₀, so $\text C_{v_2} = \frac{3}{2}\text R$
• The heat capacity of the boxes is to be ignored

The boxes are rigid, so the volume of each gas remains constant and no work is done. The heat lost by the hotter gas is therefore gained by the cooler gas, and the total change in internal energy is zero,

$\mu_1\text C_{v_1}\Delta \text T_1 + \mu_2\text C_{v_2}\Delta \text T_2 = 0$

Substituting the values,

$1 \times \left(\frac{5}{2}\text R\right) \times (\text T_f - \text T_0) + 1 \times \left(\frac{3}{2}\text R\right) \times \left(\text T_f - \frac{7}{3}\text T_0\right) = 0$

Multiplying throughout by $\frac{2}{\text R}$,

$5(\text T_f - \text T_0) + 3\left(\text T_f - \frac{7}{3}\text T_0\right) = 0$

$$\begin{aligned}5\text T_f - 5\text T_0 + 3\text T_f - 7\text T_0 = 0 \\[1em] 8\text T_f = 12\text T_0\end{aligned}$$

$\text T_f = \frac{12}{8}\text T_0 = \frac{3}{2}\text T_0$

2. heat flows out the gas during the path B → C → D.

4. positive work is done by the gas in the cycle ABCDA.

Reason —

By the first law of thermodynamics, ΔQ = ΔU + W.

Checking option 1 : For the path A → B the P-V curve is a part of the semi-circle ABC, and not a rectangular hyperbola. An isothermal curve for an ideal gas obeys PV = constant, which is a rectangular hyperbola. Hence the process A → B is not isothermal and option 1 is incorrect.

Checking option 2 : The states are B (3, 2), C (2, 1) and D (1, 2). Therefore

$\text P_{\text B}\text V_{\text B} = 3 \times 2 = 6 \qquad \text{and} \qquad \text P_{\text D}\text V_{\text D} = 1 \times 2 = 2$

Since P_DV_D < P_BV_B, the change in internal energy

$\Delta \text U = \frac{\text P_{\text D}\text V_{\text D} - \text P_{\text B}\text V_{\text B}}{\gamma - 1}$

is negative. Again, along the whole path B → C → D the volume falls steadily from 3 to 1, so the work done is also negative. Hence ΔQ is negative, that is, heat flows out of the gas along B → C → D, and option 2 is correct.

Checking option 3 : Along the path A → B → C the volume of the gas changes, so the work done, being the area under the curve, is not zero. Hence option 3 is incorrect.

Checking option 4 : The closed curve ABCDA is traced clockwise on the P-V diagram. For a clockwise cycle the net work is done by the gas and is therefore positive. Hence option 4 is correct.

2. C_p + C_v is larger for a diatomic ideal gas than for a monoatomic ideal gas.

4. C_p · C_v is larger for a diatomic ideal gas than for a monoatomic ideal gas.

Reason — For a diatomic ideal gas :

$\text C_v = \frac{5}{2}\text R \quad \text{and} \quad \text C_p = \frac{7}{2}\text R$

$$\begin{aligned}\text C_p - \text C_v = \frac{7}{2}\text R - \frac{5}{2}\text R = \text R \\[1em] \text C_p + \text C_v = \frac{7}{2}\text R + \frac{5}{2}\text R = 6\text R\end{aligned}$$

$$\begin{aligned}\frac{\text C_p}{\text C_v} = \frac{7/2\ \text R}{5/2\ \text R} = \frac{7}{5} \\[1em] \text C_p \cdot \text C_v = \frac{7}{2}\text R \times \frac{5}{2}\text R = \frac{35}{4}\text R^2\end{aligned}$$

For a monoatomic ideal gas :

$\text C_v = \frac{3}{2}\text R \quad \text{and} \quad \text C_p = \frac{5}{2}\text R$

$$\begin{aligned}\text C_p - \text C_v = \frac{5}{2}\text R - \frac{3}{2}\text R = \text R \\[1em] \text C_p + \text C_v = \frac{5}{2}\text R + \frac{3}{2}\text R = 4\text R\end{aligned}$$

$$\begin{aligned}\frac{\text C_p}{\text C_v} = \frac{5/2\ \text R}{3/2\ \text R} = \frac{5}{3} \\[1em] \text C_p \cdot \text C_v = \frac{5}{2}\text R \times \frac{3}{2}\text R = \frac{15}{4}\text R^2\end{aligned}$$

Comparing the two :

C_p − C_v = R for both the gases, so option 1 is incorrect.

C_p + C_v is 6R for the diatomic gas and 4R for the monoatomic gas, so option 2 is correct.

$\frac{\text C_p}{\text C_v}$ is $\frac{7}{5}$ = 1.4 for the diatomic gas and $\frac{5}{3}$ = 1.67 for the monoatomic gas, which is smaller for the diatomic gas, so option 3 is incorrect.

C_p · C_v is $\frac{35}{4}\text R^2$ for the diatomic gas and $\frac{15}{4}\text R^2$ for the monoatomic gas, so option 4 is correct.

1. The rate at which heat is absorbed in the range 0-100 K varies linearly with temperature T.

2. Heat absorbed in increasing the temperature from 0-100 K is less than the heat required for increasing the temperature from 400-500 K.

3. There is no change in the rate of heat absorption in range 400-500 K.

4. The rate of heat absorption increases in the range 200-300 K.

Reason —

The heat absorbed in raising the temperature of a solid of mass m is

$\Delta \text Q = \int \text{mc}\text{dT} = \text m \times (\text{area under the c-T graph})$

Differentiating with respect to time,

$\frac{\text{dQ}}{\text{dt}} = \text{mc}\frac{\text{dT}}{\text{dt}}$

The temperature is increased at a constant rate, so $\frac{\text{dT}}{\text{dt}}$ is constant, and the mass m is also constant. Therefore

$\frac{\text{dQ}}{\text{dt}} \propto \text c$

that is, the rate of heat absorption at any stage is proportional to the specific heat capacity c at that temperature.

Option 1 is correct, because in the range 0-100 K the graph of c against T is very nearly a straight line, so the rate of heat absorption varies linearly with T.

Option 2 is correct, because the area under the curve in the range 0-100 K is less than the area under the curve in the range 400-500 K, and the heat absorbed is proportional to this area.

Option 3 is correct, because in the range 400-500 K the graph of c against T is parallel to the temperature axis, so c is constant and the rate of heat absorption does not change.

Option 4 is correct, because in the range 200-300 K the specific heat capacity c increases with temperature, and hence the rate of heat absorption also increases.

Hence all the four statements are correct.

8.4 × 10⁶ J

Reason — Given,

• Heat taken in from the reservoir, Q₁ = 3000 kcal
• Temperature of the source, T₁ = 627°C = 627 + 273 = 900 K
• Temperature of the sink, T₂ = 27°C = 27 + 273 = 300 K
• 1 kcal = 4.2 × 10³ J

The efficiency of a Carnot engine is

$\eta = 1 - \frac{\text T_2}{\text T_1} = 1 - \frac{300}{900} = 1 - \frac{1}{3} = \frac{2}{3}$

The heat energy taken in from the source, expressed in joule, is

$\text Q_1 = 3000\ \text{kcal} = 3000 \times 10^{3}\ \text{cal} = 3 \times 10^{6}\ \text{cal}$

$= 3 \times 10^{6} \times 4.2\ \text J$

The work done by the engine is

$\text W = \text Q_1 \times \eta = 3 \times 10^{6} \times 4.2 \times \frac{2}{3}$

$= 8.4 \times 10^{6}\ \text J$

750 K

Reason — Given,

• Efficiency in the first case, η₁ = 40%
• Temperature of the source in the first case, T₁ = 500 K
• Desired efficiency, η₂ = 60%, the sink temperature remaining the same

For the first case : The efficiency of a Carnot engine is

$\eta = \left(1 - \frac{\text T_2}{\text T_1}\right) \times 100$

Substituting the values,

$$\begin{aligned}40 = \left(1 - \frac{\text T_2}{500}\right) \times 100 \\[1em] 0.4 = 1 - \frac{\text T_2}{500}\end{aligned}$$

$\frac{\text T_2}{500} = 0.6 \quad \Rightarrow \quad \text T_2 = 300\ \text K$

For the second case : The sink temperature stays at 300 K and the efficiency is to be 60%. If T₁′ be the new intake temperature,

$$\begin{aligned}60 = \left(1 - \frac{300}{\text T_1'}\right) \times 100 \\[1em] 0.6 = 1 - \frac{300}{\text T_1'}\end{aligned}$$

$\frac{300}{\text T_1'} = 0.4 \quad \Rightarrow \quad \text T_1' = \frac{300}{0.4} = 750\ \text K$

372 K and 310 K

Reason — Given,

• Efficiency in the first case, $\eta_1 = \frac{1}{6}$
• When T₂ is lowered by 62 K, the efficiency becomes $\eta_2 = \frac{1}{3}$

For the first case : The efficiency of a Carnot engine is

$\eta = 1 - \frac{\text T_2}{\text T_1}$

$\frac{1}{6} = 1 - \frac{\text T_2}{\text T_1} \quad \Rightarrow \quad \frac{\text T_2}{\text T_1} = 1 - \frac{1}{6} = \frac{5}{6} \qquad \text{...(i)}$

For the second case : The sink temperature becomes (T₂ − 62) K,

$\frac{1}{3} = 1 - \frac{\text T_2 - 62}{\text T_1} \quad \Rightarrow \quad \frac{\text T_2 - 62}{\text T_1} = 1 - \frac{1}{3} = \frac{2}{3} \qquad \text{...(ii)}$

Subtracting (ii) from (i),

$$\begin{aligned}\frac{\text T_2}{\text T_1} - \frac{\text T_2 - 62}{\text T_1} = \frac{5}{6} - \frac{2}{3} \\[1em] \frac{62}{\text T_1} = \frac{5 - 4}{6} = \frac{1}{6}\end{aligned}$$

$\text T_1 = 62 \times 6 = 372\ \text K$

Substituting this value in equation (i),

$\text T_2 = \frac{5}{6} \times 372 = 310\ \text K$

Hence T₁ = 372 K and T₂ = 310 K.

Given,

• Work delivered by the engine per minute, W = 5.4 × 10⁸ J
• Heat received from the boiler per minute, Q₁ = 3.6 × 10⁹ J

Efficiency of the engine : The efficiency is the ratio of the useful work delivered by the engine to the heat taken in from the hot reservoir,

$\eta = \frac{\text W}{\text Q_1} = \frac{5.4 \times 10^{8}\ \text{J min}^{-1}}{3.6 \times 10^{9}\ \text{J min}^{-1}}$

$= 0.15 = 15%$

Heat wasted per minute : The heat wasted is the heat rejected to the sink, which is the difference between the heat taken in and the work done,

$\text Q_2 = \text Q_1 - \text W = 3.6 \times 10^{9} - 5.4 \times 10^{8}$

$$\begin{aligned}= 3.6 \times 10^{9} - 0.54 \times 10^{9} \\[1em] = 3.06 \times 10^{9}\ \text{J min}^{-1} \approx 3.1 \times 10^{9}\ \text{J min}^{-1}\end{aligned}$$

Hence the efficiency of the engine is 15% and the heat wasted per minute is 3.1 × 10⁹ J.

Given,

• Temperature of the eatables kept inside (cold body), T₂ = 9°C = 9 + 273 = 282 K
• Room temperature (hot body), T₁ = 36°C = 36 + 273 = 309 K

The coefficient of performance of a refrigerator is defined as the ratio of the heat taken in from the cold body to the work needed to run the refrigerator, and for an ideal refrigerator it is given by

$\text K = \frac{\text T_2}{\text T_1 - \text T_2}$

where T₁ is the absolute temperature of the hot body to which heat is delivered and T₂ is the absolute temperature of the cold body from which heat is removed.

Substituting the values,

$\text K = \frac{282\ \text K}{309\ \text K - 282\ \text K} = \frac{282\ \text K}{27\ \text K}$

$= 10.4$

Hence, the coefficient of performance of the refrigerator is 10.4.

Given,

• Number of moles of hydrogen, μ = 3
• The gas is initially at STP
• Final volume, $\text V_2 = \frac{\text V_1}{2}$
• γ = 1.4

Let P₁, V₁ be the original pressure and volume of the gas and P₂, V₂ those after the gas is compressed.

The cylinder is insulated and the piston is also insulated, so the whole system is well insulated and no heat can enter or leave the gas. The compression is therefore adiabatic and obeys Poisson's law,

$\text P_1\text V_1^{\gamma} = \text P_2\text V_2^{\gamma}$

Rearranging,

$\frac{\text P_2}{\text P_1} = \left(\frac{\text V_1}{\text V_2}\right)^{\gamma}$

Here $\text V_2 = \frac{\text V_1}{2}$, so

$\frac{\text P_2}{\text P_1} = \left(\frac{\text V_1}{\text V_1/2}\right)^{1.4} = (2)^{1.4}$

$= 2.64$

Hence, the pressure of the gas increases by a factor of 2.64.

Conservation of energy

Reason — The first law of thermodynamics states that if an amount of heat Q is given to a system, a part of it is used in increasing the internal energy (ΔU) of the system and the rest in doing work (W) by the system,

$\text Q = \Delta \text U + \text W$

Energy is thus neither created nor destroyed but only transferred or transformed from one form to another. Hence the first law is a statement of the conservation of energy.

zero

Reason — An isochoric process takes place at constant volume, so ΔV = 0. The work done by a system is

$\text{dW} = \text P \times \text{dV} = \text P \times 0 = 0$

Hence the work done by the system in an isochoric process is zero, and the entire heat given to the system is used in increasing its internal energy.

temperature

Reason — For an ideal gas there are no intermolecular forces, so the potential energy of the molecules is zero and the internal energy is wholly kinetic. The kinetic energy of the molecules depends only upon the temperature. Hence the internal energy of an ideal gas is a function of temperature alone, U = f(T), and does not depend on pressure or volume.

zero

Reason — An adiabatic process is one during which no exchange of heat takes place between the system and the surroundings, that is, Q = 0. For such a process to occur, the system must be perfectly insulated from the surroundings.

remains constant

Reason — A process taking place at a constant pressure is called an isobaric process (ΔP = 0). The volume and the temperature of the system may change, but the pressure stays the same. Heating of water in the boiler of a steam engine, formation of steam and freezing of water are isobaric processes.

zero

Reason — In an isothermal process the temperature of the system remains constant (ΔT = 0). The internal energy of an ideal gas depends only upon the temperature, so if the temperature does not change, the internal energy does not change either, that is, ΔU = 0. By the first law, the heat absorbed is then entirely used in the work done by the gas, Q = W.

isochoric

Reason — In an isochoric process the volume is kept constant, so W = P × ΔV = 0, that is, no work is done. Heat can still be transferred, and by the first law ΔU = Q, so the entire heat given to the system is used in increasing its internal energy.

entropy

Reason — The first law tells us only that energy is conserved; it does not tell us in which direction a process will occur. The second law of thermodynamics supplies this direction and introduces the concept of entropy, which is a measure of the disorder or randomness of a system and governs the direction of natural processes.

temperature

Reason — The internal energy of a system is the sum of the kinetic energy and the potential energy of its molecules. For an ideal gas the intermolecular forces are absent, so the energy is wholly kinetic and depends only upon the temperature. Thus U = f(T).

Conservation of energy

Reason — The first law of thermodynamics, ΔU = Q − W, expresses the fact that the energy given to a system appears partly as a rise in its internal energy and partly as the work done by it. No energy is created or destroyed in the process. Hence the first law is essentially a law of conservation of energy.

internal energy only

Reason — At constant volume the gas cannot expand, so no external work is done (W = P × ΔV = 0). By the first law, Q = ΔU. Thus all the heat supplied at constant volume goes into increasing the internal energy of the gas, and

$\text C_v = \frac{1}{\mu}\left(\frac{\text{dQ}}{\text{dT}}\right)_v$

is related to the internal energy only.

pressure remains constant

Reason — An isobaric process is one in which the pressure P is kept constant (ΔP = 0), while the volume and the temperature may change. The work done in such a process is W = P (V₂ − V₁).

zero

Reason — 'Isochoric' means constant volume. Hence in an isochoric process the change in the volume of the gas is zero, ΔV = 0, and consequently the work done, W = P × ΔV, is also zero.

the temperature remains constant

Reason — In an isothermal process the temperature of the system remains constant (ΔT = 0). For such a process to take place, the system must be surrounded by a perfectly conducting material so that any heat produced immediately goes out to the surroundings, or any heat absorbed comes in from the surroundings. Heat can therefore still be exchanged, and work is done.

No heat is transferred.

Reason — In an adiabatic process neither does heat enter the system nor does it leave the system, that is, Q = 0. The temperature of the system does change in such a process, because the work done alters the internal energy of the system.

at a constant temperature

Reason — In the first process of the Carnot cycle, the cylinder is placed on the source and the working substance expands infinitely slowly, taking in heat Q₁ from the source by conduction through the base. The expansion is therefore isothermal, taking place at the constant temperature T₁ of the source.

remains constant

Reason — The internal energy of an ideal gas depends only upon its temperature. Since the temperature is constant in an isothermal process, the internal energy remains constant, ΔU = 0.

insulated from its surroundings

Reason — For an adiabatic process no heat should enter or leave the system. This is possible only when the system is perfectly insulated from the surroundings, so that Q = 0. In practice, since no material is perfectly insulating, a process carried out very rapidly is very nearly adiabatic.

Adiabatic

Reason — In an adiabatic process no heat is exchanged with the surroundings (Q = 0), but the temperature does change. By the first law ΔU = − W, so when the gas does work its internal energy and hence its temperature falls, and when work is done on the gas its temperature rises.

the difference between heat added and work done by the system

Reason — The first law of thermodynamics is

$\Delta \text U = \text Q - \text W$

that is, the change in the internal energy of a system is equal to the heat added to the system minus the work done by the system.

ΔQ = ΔU + ΔW

Reason — If an amount of heat ΔQ is given to a system, a part of it is used in increasing the internal energy (ΔU) of the system and the rest in doing external work (ΔW) by the system. Hence

$\Delta \text Q = \Delta \text U + \Delta \text W$

which is the standard form of the first law of thermodynamics.

work is done on the system

Reason — By the first law, ΔU = Q − W. When work is done on the system, W is negative, so

$\Delta \text U = \text Q - (-\text W) = \text Q + \text W$

which is positive. Hence the internal energy increases. The internal energy also increases when heat is added to the system.

Q = ΔU + W

Reason — For a closed system the first law of thermodynamics is written as

$\text Q = \Delta \text U + \text W$

that is, the heat added to the system equals the change in its internal energy plus the work done by the system. Note that it is the change ΔU, and not U itself, that enters the equation.

the temperatures of the reservoirs

Reason — The efficiency of a reversible (Carnot) heat engine is

$\eta = 1 - \frac{\text T_2}{\text T_1}$

which is independent of the nature of the working substance and depends only upon the absolute temperatures of the source and the sink. The greater the temperature difference between the two reservoirs, the greater is the efficiency.

both reversible and irreversible processes

Reason — The first law of thermodynamics is simply a statement of the conservation of energy, and energy is conserved whatever be the nature of the process. Hence the first law applies to both reversible and irreversible processes, and to open as well as closed systems.

the heat added to the system

Reason — In an isothermal process the temperature is constant, and since the internal energy of an ideal gas depends only upon the temperature,

$\Delta \text U = 0$

Therefore, by the first law of thermodynamics ΔU = Q − W,

$\text Q = \text W$

Hence the work done is equal to, and therefore proportional to, the heat added to the system.

The work done in an isothermal change is

$\text W = 2.3026\mu\text{RT}\log_{10}\frac{\text V_2}{\text V_1}$

which depends on the ratio of the two volumes and not on their difference, so the work done is not proportional to the change in volume or to the change in pressure. The change in temperature is zero throughout.

Note: The printed answer key of the textbook gives option (c). The work done varies as the logarithm of the volume ratio, so it is not proportional to the change in volume; it is exactly equal to the heat added, since ΔU = 0. Option 4 is therefore the correct choice.

T_h is high and T_c is low

Reason — The efficiency of a heat engine working between the two reservoirs is

$\eta = 1 - \frac{\text T_c}{\text T_h}$

The efficiency is greater the smaller the ratio $\frac{\text T_c}{\text T_h}$, that is, the greater the temperature difference between the reservoirs. Hence the efficiency is maximum when T_h is high and T_c is low.

$η = 1 - \frac{\text T_c}{\text T_h}$

Reason — For a Carnot engine the ratio of the heat exchanged to the absolute temperature is the same for the source and the sink, $\frac{\text Q_1}{\text Q_2} = \frac{\text T_h}{\text T_c}$. Therefore

$\eta = \frac{\text W}{\text Q_1} = \frac{\text Q_1 - \text Q_2}{\text Q_1} = 1 - \frac{\text Q_2}{\text Q_1} = 1 - \frac{\text T_c}{\text T_h}$

zero

Reason — A reversible process is one which can be reversed in such a way that all the changes taking place in the direct process are exactly repeated in the inverse order, leaving no change in the bodies taking part in the process or in the surroundings. Since the process can be perfectly reversed, the total entropy change of the system and the surroundings taken together is zero.

remains constant

Reason — In an isothermal process the temperature of the gas is constant. As the internal energy of an ideal gas depends only upon its temperature, the internal energy remains constant and ΔU = 0.

be perfectly insulated

Reason — An adiabatic process demands that no heat should enter or leave the system, Q = 0. This requires the system to be perfectly insulated from the surroundings. Since no material is perfectly insulating, a perfect adiabatic process is an ideal conception; in practice a very rapid process is nearly adiabatic.

equal to the work done by the gas

Reason — During an isothermal expansion of an ideal gas the temperature is constant, so the internal energy does not change, ΔU = 0. By the first law ΔU = Q − W, we get

$0 = \text Q - \text W \quad \Rightarrow \quad \text Q = \text W$

Hence the heat absorbed by the gas is entirely used in the work done by the gas.

decreases

Reason — In an adiabatic expansion no heat is supplied to the gas (Q = 0), so by the first law ΔU = − W. The gas does positive work in expanding at the expense of its own internal energy. Since the internal energy decreases, the temperature of the gas falls.

changes due to work done

Reason — In an adiabatic process Q = 0, so ΔU = − W. Thus the internal energy, and hence the temperature of the gas, changes solely on account of the work done by or on the gas, and not because of any heat transfer.

Heat cannot be transferred from a colder body to a hotter body without work

Reason — This is the Clausius statement of the second law of thermodynamics : it is impossible to transfer heat from a cold body to a hot body without expenditure of work by an external energy source. In a refrigerator the transfer takes place only because an external agent does work on the working substance.

two isothermal and two adiabatic processes

Reason — Carnot's cycle consists of four processes — an isothermal expansion AB at the temperature T₁ of the source, an adiabatic expansion BC, an isothermal compression CD at the temperature T₂ of the sink, and an adiabatic compression DA which restores the substance to its initial state. Hence it consists of two isothermal and two adiabatic processes.

the temperatures of the heat reservoirs

Reason — The efficiency of Carnot's reversible engine is

$\eta = 1 - \frac{\text T_2}{\text T_1}$

This expression contains only the absolute temperatures of the sink and the source. The efficiency of Carnot's reversible engine is therefore independent of the working substance and depends only upon the absolute temperatures of the sink and the source.

Carnot cycle

Reason — By Carnot's theorem, no engine working between two given temperatures can be more efficient than a reversible engine working between the same two temperatures. The Carnot cycle, being completely reversible, therefore sets the maximum possible efficiency of any heat engine operating between two given temperatures.

heat is added to the system

Reason — Entropy is a measure of the disorder or randomness of a system. When heat is added to the system, the molecular motion becomes more random and the disorder increases. Hence the entropy of the system increases.

can never be 100%

Reason — By the Kelvin-Planck statement of the second law, it is impossible to convert 'all' the heat extracted from a hot body into work. Some heat Q₂ must always be given up to the sink, so

$\eta = 1 - \frac{\text Q_2}{\text Q_1} \lt 1$

Hence the efficiency of a heat engine can never be 100%.

It involves only reversible processes.

Reason — Each of the four processes of Carnot's cycle — the isothermal expansion, the adiabatic expansion, the isothermal compression and the adiabatic compression — is carried out infinitely slowly and in the complete absence of dissipative forces, so that every step is reversible. Hence the Carnot cycle involves only reversible processes, and this is exactly why it has the maximum efficiency.

Isothermal compression

Reason — In the third process of Carnot's cycle the cylinder is placed on the sink and the substance is compressed isothermally in infinitely small steps along the curve CD at the constant temperature T₂. During this process work is done on the substance and the heat Q₂ developed is given out to the sink.

No engine operating between two heat reservoirs can be more efficient than a Carnot engine.

Reason — Carnot's theorem states that no engine working between two given temperatures can be more efficient than a reversible engine working between the same two temperatures, and that all reversible engines working between the same two temperatures have the same efficiency, whatever the working substance.

It is the most efficient engine cycle.

Reason — Carnot's engine is an ideal conception which cannot be realised in practice, since bodies of infinitely large heat capacity, perfectly conducting and perfectly insulating materials and the complete absence of dissipative forces are not available. Its importance lies in the fact that it is the most efficient engine cycle and therefore serves as a standard against which the efficiencies of all real engines are compared.

the difference between heat absorbed and heat rejected

Reason — In one complete cycle the working substance returns to its original state, so there is no net change in its internal energy. Applying the first law of thermodynamics to one complete cycle,

$\text W = \text Q_1 - \text Q_2$

Hence the work output is the difference between the heat absorbed from the source and the heat rejected to the sink.

maximum

Reason — For the same expansion between two given volumes, the isothermal curve lies above the adiabatic curve, so the area enclosed between the isothermal curve and the volume-axis is the largest. Since the work done by a gas is measured by this area, the work done in an isothermal expansion is maximum as compared with the adiabatic and other expansions.

decreases

Reason — In an adiabatic process Q = 0, so the first law gives ΔU = − W. If the system does work, W is positive and therefore ΔU is negative, that is, the internal energy of the system decreases and its temperature falls.

work done by or on the system

Reason — In an adiabatic process no heat is exchanged, Q = 0. Hence the first law ΔU = Q − W reduces to

$\Delta \text U = -\text W$

Thus the change in the internal energy is equal in magnitude to the work done by or on the system. If work is done on the system the internal energy increases, and if work is done by the system it decreases.

80%

Reason — Given, T_h = 500 K and T_c = 100 K.

The efficiency of a Carnot engine is

$$\begin{aligned}\eta = 1 - \frac{\text T_c}{\text T_h} = 1 - \frac{100}{500} \\[1em] = 1 - 0.2 = 0.8 = 80%\end{aligned}$$

change in internal energy

Reason — In an adiabatic process no heat is exchanged with the surroundings, Q = 0. Therefore the first law ΔU = Q − W becomes

$\text W = -\Delta \text U$

Hence the work done by the system during an adiabatic process is equal in magnitude to the change in its internal energy, the work being done at the expense of the internal energy of the system.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: By the Kelvin-Planck statement of the second law of thermodynamics, it is impossible to convert 'all' the heat extracted from a hot body into work. A part of the heat taken in from the source must always be given up to the sink, so the efficiency

$\eta = 1 - \frac{\text Q_2}{\text Q_1}$

is always less than 1, that is, less than 100%.

Reason (R) is also correct: Since the engine must reject the heat Q₂ to the sink, some energy is always transferred to the surroundings as waste heat, and this is demanded by the second law of thermodynamics.

The waste heat rejected to the sink is precisely the reason why the whole of Q₁ cannot appear as work. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: In an ideal gas there are no intermolecular forces, so the potential energy of the molecules is zero and the internal energy is wholly the kinetic energy of the molecules. This kinetic energy depends only upon the temperature, so U = f(T).

Reason (R) is also correct: Because of the large molecular separation in an ideal gas the mutual potential energy is zero, and hence the internal energy has no dependence upon the pressure or the volume of the gas.

Since the internal energy has no term depending on pressure or volume, the only variable left on which it can depend is the temperature. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: The efficiency of Carnot's reversible engine is

$\eta = 1 - \frac{\text T_2}{\text T_1}$

which contains only the absolute temperatures of the source and the sink. It is independent of the nature of the working substance.

Reason (R) is also correct: The Carnot cycle is an ideal reversible cycle, and the greater the temperature difference between the hot and the cold reservoirs, the greater is the efficiency.

Because every step of the cycle is reversible, the heats exchanged bear the ratio $\frac{\text Q_2}{\text Q_1} = \frac{\text T_2}{\text T_1}$, and this is exactly what makes the efficiency depend on the two temperatures alone. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If assertion is true but reason is false.

Explanation

Assertion (A) is correct: An isothermal process is by definition one in which the temperature of the system remains constant (ΔT = 0). For this the system is surrounded by a perfectly conducting material and the process is carried out very slowly.

Reason (R) is false: The internal energy of an ideal gas depends only upon its temperature. Since the temperature is constant in an isothermal expansion, the internal energy does not change, ΔU = 0. By the first law, the heat supplied is used entirely in doing external work by the gas, not in increasing the internal energy.

Therefore, assertion is true but reason is false.

If assertion is true but reason is false.

Explanation

Assertion (A) is correct: An adiabatic process is defined as one during which no exchange of heat takes place between the system and the surroundings, that is, Q = 0. For this the system must be perfectly insulated from the surroundings.

Reason (R) is false: In an adiabatic process the internal energy does not remain constant. Since Q = 0, the first law gives ΔU = − W. The work done by or on the system therefore changes the internal energy, and hence the temperature of the system changes as well.

Therefore, assertion is true but reason is false.

If assertion is false but reason is true.

Explanation

Assertion (A) is false: The Clausius statement of the second law of thermodynamics is exactly the opposite — it is impossible to transfer heat from a cold body to a hot body without expenditure of work by an external energy source. Heat can never flow spontaneously from a colder to a hotter body.

Reason (R) is correct: The second law states that the entropy of an isolated system tends to increase or at best remains constant; it can never decrease.

Therefore, assertion is false but reason is true.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: For an isolated system the total entropy can never decrease. It increases in every irreversible (natural) process and remains constant in a reversible process.

Reason (R) is also correct: This is precisely the statement of the second law of thermodynamics in terms of entropy — the entropy of an isolated system tends to increase or remain constant.

The Assertion is simply a restatement of the law contained in the Reason, so the Reason explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: The work done by a system is dW = P dV. When the pressure P is constant it can be taken outside, and integrating between the initial and the final volumes gives

$\text W = \text P(\text V_2 - \text V_1) = \text P\Delta \text V$

Reason (R) is also correct: An isobaric process is by definition one in which the pressure remains constant while the volume of the system changes.

It is only because P stays constant that it can be taken out of the integral to give the simple product P ΔV. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If assertion is true but reason is false.

Explanation

Assertion (A) is correct: In one complete cycle the working substance returns to its initial state, so there is no net change in its internal energy (ΔU = 0). By the first law the net work done is then equal to the net heat absorbed in the cycle,

$\text W = \text Q_1 - \text Q_2$

Reason (R) is false: The Carnot cycle does not convert all the absorbed heat into work. The heat Q₂ is always rejected to the sink, and by the second law this rejection cannot be avoided. Hence its efficiency, though the maximum possible, is still less than 100%.

Therefore, assertion is true but reason is false.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: The first law of thermodynamics asserts that the energy given to a system appears partly as an increase in its internal energy and partly as the work done by it, so that no energy is created or destroyed. It is therefore a statement of the conservation of energy.

Reason (R) is also correct: The mathematical form of the law is

$\Delta \text U = \text Q - \text W$

that is, the change in the internal energy equals the heat added to the system minus the work done by the system.

The equation in the Reason is exactly the book-keeping of energy that makes the law a conservation law. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If assertion is false but reason is true.

Explanation

Assertion (A) is false: An isochoric process is one which takes place at constant volume (ΔV = 0), and not at constant pressure. A process taking place at constant pressure is called an isobaric process.

Reason (R) is correct: Since the volume does not change in an isochoric process, the work done is

$\text W = \text P \times \Delta \text V = \text P \times 0 = 0$

Therefore, assertion is false but reason is true.

Note: The textbook answer key gives option (c), but the correct option is (d). In an isochoric process, volume remains constant, not pressure.

If assertion is false but reason is true.

Explanation

Assertion (A) is false: No heat at all is added during an adiabatic process, since Q = 0 by definition. It is therefore meaningless to speak of heat added in such a process. The internal energy changes only because of the work done, ΔU = − W.

Reason (R) is correct: An adiabatic process is precisely one in which no heat is exchanged with the surroundings, the system being perfectly insulated.

Therefore, assertion is false but reason is true.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: For an ideal gas undergoing an isothermal process the change in the internal energy is zero, ΔU = 0, and by the first law the heat absorbed is entirely used in the work done by the gas.

Reason (R) is also correct: An isothermal process is by definition one in which the temperature of the system remains constant.

The internal energy of an ideal gas is a function of temperature alone, so a constant temperature necessarily means a constant internal energy. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: The first law places no restriction on the direction of a process. The second law supplies this direction, and in doing so it introduces the state variable entropy.

Reason (R) is also correct: Entropy is the measure of the disorder or randomness in a system, often related to the number of ways a system's energy can be arranged.

It is because entropy measures disorder, and because disorder tends to increase in natural processes, that the second law can use entropy to fix the direction of a process. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: When a gas is compressed adiabatically its temperature rises. This is why the barrel of a bicycle pump becomes hot when a tube is pumped.

Reason (R) is also correct: In an adiabatic compression Q = 0 and work is done on the gas, so W is negative. By the first law

$\Delta \text U = -(-\text W) = \text W$

which is positive, that is, the internal energy of the gas increases.

Since the internal energy of a gas is determined by its temperature, an increase in internal energy means a rise in temperature. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: Since a part of the heat taken in from the source is always rejected to the sink, Q₂ is never zero, and hence

$\eta = 1 - \frac{\text Q_2}{\text Q_1} \lt 1$

Reason (R) is also correct: By the Kelvin-Planck statement of the second law, it is impossible to convert all the heat extracted from a hot body into work, so no heat engine can be 100% efficient.

The impossibility stated in the Reason is exactly what forces Q₂ to be non-zero and the efficiency to be less than 1. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If assertion is false but reason is true.

Explanation

Assertion (A) is false: In an isothermal process the temperature of the gas remains constant. Since the internal energy of an ideal gas depends only upon the temperature, the internal energy does not change, that is, ΔU = 0.

Reason (R) is correct: For an ideal gas there are no intermolecular forces, so the internal energy is wholly kinetic and depends only upon the temperature.

Therefore, assertion is false but reason is true.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: An adiabatic process is by definition one in which no heat transfer takes place between the system and the surroundings, Q = 0, though the temperature of the system does change.

Reason (R) is also correct: For such a process the system has to be perfectly insulated from the surroundings, so that heat can neither enter nor leave it.

It is precisely the perfect insulation described in the Reason that makes the absence of heat transfer possible. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: By Carnot's theorem, no engine working between two given temperatures can be more efficient than a reversible engine working between the same two temperatures. A reversible process is therefore more efficient than an irreversible one.

Reason (R) is also correct: In a reversible process all the changes of the direct process are exactly repeated in the inverse order, so no entropy is generated and the entropy remains constant.

Irreversible processes generate entropy through dissipative effects such as friction, and this generated entropy is what lowers the efficiency. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: Carnot's engine requires a source and a sink of infinitely large heat capacities, perfectly insulating walls with a perfectly conducting base, a frictionless piston and infinitely slow processes. These conditions can never be actually realised, so Carnot's engine is an ideal conception.

Reason (R) is also correct: All real processes involve dissipative forces such as friction, viscosity and electrical resistance, and are therefore irreversible.

The unavoidable irreversibilities and frictional losses of the Reason are exactly what prevent the ideal conditions of the Carnot engine from being met. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: In a Carnot cycle the working substance takes in heat Q₁ from the source and gives out heat Q₂ to the sink, the difference appearing as the useful work W = Q₁ − Q₂. Since W is positive, Q₂ is always less than Q₁.

Reason (R) is also correct: The Carnot cycle operates between two temperatures T₁ and T₂, and by the second law some heat must always be rejected to the cold reservoir.

Because $\frac{\text Q_2}{\text Q_1} = \frac{\text T_2}{\text T_1}$ and T₂ < T₁, the heat rejected is necessarily less than the heat absorbed. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: In a cyclic process the system, after passing through different states, returns to its original state. Hence there is no net change in its internal energy, ΔU = 0, and by the first law the heat given to the system equals the net work done by it, Q = W.

Reason (R) is also correct: The internal energy is a state function; its value depends only upon the state of the system and not upon the path by which that state was reached.

Since the initial and the final states of a cycle are the same state, a quantity that depends only on the state must return to its original value, making ΔU zero. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: This is Carnot's theorem — no engine working between two given temperatures can be more efficient than a reversible engine working between the same two temperatures.

Reason (R) is also correct: Each of the four processes of Carnot's cycle is carried out infinitely slowly and in the complete absence of dissipative forces, so the cycle consists of reversible processes only.

It is the complete reversibility of every step that removes all dissipative losses and gives the cycle the maximum possible efficiency. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: In an isochoric process the work done by the system is

$\text W = \text P \times \Delta \text V = \text P \times 0 = 0$

so the entire heat given to the system is used in increasing its internal energy, Q = ΔU.

Reason (R) is also correct: An isochoric process is by definition one in which the volume of the system is kept constant, so ΔV = 0.

Since the work done is the product of the pressure and the change in volume, a zero change in volume makes the work zero. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: For an isolated system the entropy either increases, in an irreversible process, or remains constant, in a reversible process. It can never decrease.

Reason (R) is also correct: This is the statement of the second law of thermodynamics in terms of entropy — the total entropy of an isolated system can only increase or remain constant.

The Assertion is a direct consequence of the law quoted in the Reason, so the Reason explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: In an adiabatic process the first law reduces to ΔU = − W, so the change in the internal energy is determined entirely by the work done by or on the system.

Reason (R) is also correct: An adiabatic process is one in which no heat is exchanged with the surroundings, Q = 0.

Putting Q = 0 in ΔU = Q − W is precisely what leaves the work as the only quantity governing the change in internal energy. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: For an isothermal process in an ideal gas the first law gives 0 = Q − W, so

$\text Q = \text W$

that is, the heat supplied is entirely used in the work done by the gas.

Reason (R) is also correct: In an isothermal process the temperature is constant, and since the internal energy of an ideal gas depends only upon the temperature, the internal energy remains constant, ΔU = 0.

It is exactly because ΔU vanishes that the whole of Q appears as W. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: A heat pump, like a refrigerator, is essentially a heat engine running backwards. It removes heat from a cold body and delivers a larger amount of heat to a hot body.

Reason (R) is also correct: By the Clausius statement of the second law, heat cannot flow from a cold body to a hot body without the expenditure of work by an external energy source. The heat pump therefore needs an external agent to do work on the working substance.

The transfer is against the natural direction of heat flow, and it is the external work described in the Reason that makes such a transfer possible. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: In an adiabatic expansion no heat is supplied to the gas, so it does external work at the expense of its own internal energy. The internal energy therefore decreases and the temperature falls.

Reason (R) is also correct: The internal energy of an ideal gas is wholly kinetic and depends only upon the temperature.

Since the internal energy is fixed by the temperature alone, a fall in internal energy must show itself as a fall in temperature. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: The efficiency of a Carnot engine is

$\eta = 1 - \frac{\text T_2}{\text T_1}$

As the temperature T₂ of the cold reservoir is lowered, the ratio $\frac{\text T_2}{\text T_1}$ decreases and the efficiency increases.

Reason (R) is also correct: The efficiency depends upon the temperatures of the hot and the cold reservoirs, and the greater the temperature difference between them, the greater is the efficiency.

Lowering T₂ widens the temperature difference, which is exactly why the efficiency rises. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

The internal energy of a system is present in the following forms :

(i) Translational kinetic energy of the molecules.

(ii) Internal rotational and vibrational energy of the molecules.

(iii) Potential energy of the molecules due to the inter-atomic forces.

For an adiabatic change in an ideal gas, the relation between the volume V and the absolute temperature T is

$\text{TV}^{\gamma - 1} = \text{constant}$

where $\gamma = \frac{\text C_p}{\text C_v}$ is the ratio of the two specific heats of the gas.

The work done by μ moles of an ideal gas in changing its volume from V₁ to V₂ at the constant temperature T is

$\text W = \mu\text{RT}\log_e\frac{\text V_2}{\text V_1} = 2.3026\mu\text{RT}\log_{10}\frac{\text V_2}{\text V_1}$

where R is the universal gas constant.

Two types of internal energy are present in a molecule of carbon monoxide, namely rotational and vibrational energy.

Carbon monoxide (CO) is a diatomic molecule, so besides the translational motion its molecules can also rotate and vibrate.

The internal energy of an ideal gas can be changed by the following methods :

(i) By heating or cooling the gas in a closed vessel. The volume being constant, no work is done and the whole of the heat given goes into the internal energy of the gas.

(ii) By adiabatic compression or expansion of the gas. In adiabatic compression work is done on the gas and its internal energy increases, while in adiabatic expansion the gas does work at the expense of its internal energy, which therefore decreases.

The internal energy of an ideal gas increases with a rise in temperature.

In an ideal gas the intermolecular forces are absent, so the internal energy is wholly the kinetic energy of the molecules, and this kinetic energy is directly proportional to the absolute temperature of the gas.

The increase in the internal energy will again be C_v ΔT, whatever be the process.

The internal energy of an ideal gas depends only upon its temperature. Hence for the same temperature-rise ΔT the increase in internal energy is the same, whether the gas is heated at constant volume or at constant pressure. At constant pressure a greater amount of heat has to be supplied, but the extra heat is used in doing external work by the expanding gas and not in raising the internal energy.

First law of thermodynamics : If an amount of heat Q is given to a thermodynamic system, a part of it is used in increasing the internal energy (ΔU) of the system and the rest in doing external work (W) by the system. Thus

$\Delta \text U = \text Q - \text W$

This law is based on the conservation of energy.

The first law of thermodynamics is written as

$\Delta \text U = \text Q - \text W \qquad \text{or} \qquad \text Q = \Delta \text U + \text W$

where

• Q is the heat given to the system. It is taken positive when heat is taken in by the system and negative when heat is given out by the system.
• ΔU is the change in the internal energy of the system.
• W is the work done. It is taken positive when work is done by the system and negative when work is done on the system by an external agency.

All the three quantities must be expressed in the same unit.

When a gas is heated at constant pressure, the heat taken by it is used not only in raising its temperature but also in doing external work by the expanding gas against the external pressure.

On the other hand, when the gas is heated at constant volume, the gas cannot expand and so no work is done; the entire heat is used in raising the temperature.

Hence, for the same temperature-rise, the heat required at constant pressure is greater, that is, C_p > C_v.

(i) No, the air of the room will not be cooled; in fact it will be heated. The fan only sets the air of the room into motion, so the speed of the air molecules increases. We feel cold because the moving air increases the rate of evaporation of sweat from our body, and the heat required for this evaporation is taken from the body.

(ii) The internal energy will remain the same. The internal energy of an ideal gas depends only upon its temperature, and here the temperature is constant.

(iii) Yes, the gas will do external work at the expense of its internal energy. Since no energy is supplied from outside, the energy for the work comes from the internal energy of the gas itself, and consequently the temperature of the gas falls.

(iv) No. For a given mass of a gas the pressure, volume and temperature are connected by the gas equation PV = μRT. If both P and V are kept constant, then T is also fixed automatically and cannot be changed.

(v) Yes, provided no physical or chemical change is taking place in the system. An isolated system can neither do work nor take or give heat, so ΔU = 0 and the temperature remains constant.

(vi) Yes, the room becomes hotter. The refrigerator does not create cooling of itself. Its motor does mechanical work on the working substance, which takes heat from the inside of the fridge and gives out a larger amount of heat to the room through the radiator at the back, the extra amount being equal to the work supplied by the motor. Hence, if the door of the fridge is opened in a room, the heat given to the room is greater than the heat removed from the inside, and therefore the room becomes hotter.

Given,

• Heat absorbed in going from the first state to the second state = Q₁
• Heat liberated in coming back from the second state to the first state = Q₂

The substance finally returns to its initial state, so the process is a cyclic process.

The internal energy is a state function, that is, it depends only upon the state of the system. Since the initial and the final states are the same,

$\Delta \text U = 0$

By the first law of thermodynamics, Q = ΔU + W. The net heat absorbed in the complete cycle is (Q₁ − Q₂), therefore

$\text Q_1 - \text Q_2 = 0 + \text W \quad \Rightarrow \quad \text W = \text Q_1 - \text Q_2$

Hence the work done by the substance is (Q₁ − Q₂) and the change in its internal energy is zero.

The translational motion of the molecules is responsible for the internal energy of a monoatomic gas.

A monoatomic gas has single-atom molecules which possess only translational motion; they have no rotational or vibrational energy of consequence.

There will be no change in the internal energy of the gas.

The molecules of an 'ideal' gas possess kinetic energy only, since the intermolecular forces and hence the potential energy are absent. This kinetic energy depends only upon the temperature. As the gas is compressed at constant temperature, its internal energy remains unchanged.

In a compressed (real) gas the molecules come closer, so the mutual attraction between the molecules increases. Thus potential energy is added to the internal energy of the gas. But this potential energy is negative, so the total internal energy of the compressed gas decreases.

The kinetic energy of the molecules is the same in both the cases, since both the gases are at the same temperature.

Hence the internal energy of a compressed gas is less than that of a rarefied gas at the same temperature.

The specific heat of the liquid at this time will be infinite.

During vaporisation at the boiling point the temperature of the liquid remains constant, so ΔT = 0. Hence the specific heat is

$\text c = \frac{\text Q}{\text m \times \Delta \text T} = \frac{\text Q}{\text m \times 0} = \infty$

No, it is not possible.

By the Kelvin-Planck statement of the second law of thermodynamics, no engine can convert the whole of the heat extracted from the source into mechanical work. It has to reject a part of the heat to the colder surroundings. A sink is therefore necessary for obtaining continuous work.

The two essential features of Carnot's ideal heat engine are :

(i) The source and the sink are bodies of infinitely large heat capacities, so that the working substance takes in all the heat at a constant temperature and gives up all the heat at another constant temperature.

(ii) Each process of the engine's cycle is fully reversible, being carried out infinitely slowly and in the complete absence of dissipative forces.

For an adiabatic change in an ideal gas, the relation between the absolute temperature T and the pressure P is

$\frac{\text T^{\gamma}}{\text P^{\gamma - 1}} = \text{constant}$

where $\gamma = \frac{\text C_p}{\text C_v}$ is the ratio of the two specific heats of the gas.

The superheating of steam is an isobaric process.

The process takes place at the constant pressure of the boiler, while during the heating the temperature of the steam does not remain constant but goes on rising. Hence it is isobaric and not isothermal.

There are two isothermal changes in this process :

(i) The conversion of ice at 0°C into water at 0°C, which takes place at the constant temperature of 0°C.

(ii) The conversion of water at 100°C into steam at 100°C, which takes place at the constant temperature of 100°C.

During each of these changes of state the heat supplied is used as latent heat, so the temperature does not change.

In an isothermal process the internal energy does not change. The internal energy of an ideal gas depends only upon its temperature, and the temperature is constant in an isothermal process, so ΔU = 0.

In an adiabatic process the internal energy does change. Here Q = 0, so by the first law ΔU = − W. The work done by or on the gas alters its internal energy, and hence the temperature of the gas changes.

Yes, the internal energy of the gas decreases.

In an adiabatic expansion no heat is given to or taken from the gas, so Q = 0. Therefore, according to the first law of thermodynamics (ΔU = Q − W),

$\Delta \text U = -\text W$

Since the gas expands, the work W done by the gas is positive, and so ΔU is negative. Thus the internal energy is decreased by an amount equivalent to the work done by the gas in expansion, and the temperature of the gas falls.

Yes, it is possible.

The temperature of a gas can be raised without giving any heat to it by compressing the gas adiabatically. In an adiabatic compression Q = 0 and work is done on the gas, so by the first law ΔU = − W, which is positive. The internal energy of the gas therefore increases and its temperature rises.

When the gas is expanded suddenly, the change is adiabatic, since the heat finds no time to flow in or out.

In this expansion the gas does work against the external pressure. As no heat is supplied from outside, a part of the internal energy of the gas is used in doing this work. The internal energy therefore decreases, and hence the temperature of the gas falls.

As the air of the atmosphere rises up, the atmospheric pressure decreases, so the air expands. This expansion is so rapid that the air finds no time to take heat from the surroundings, that is, it is an adiabatic expansion.

In expanding, the air does work against the surrounding pressure at the expense of its own internal energy. Hence its internal energy decreases and the air cools.

When the valve of a bicycle-tube is removed, the compressed air inside rushes out suddenly, so it expands adiabatically.

The escaping air does work against the atmospheric pressure at the expense of its own internal energy. The internal energy therefore decreases and the temperature of the air falls. Hence the escaping air appears cold.

This phenomenon represents an adiabatic change.

When the balloon bursts suddenly, the air inside expands so rapidly that no heat can be exchanged with the surroundings. Work is done against the external pressure at the expense of the internal energy of the gas, which is therefore cooled.

No, two isothermal curves cannot intersect each other.

If they were to intersect, then at the point of intersection the volume and the pressure of the gas would be the same for two different temperatures, which is impossible.

Yes, the value of γ depends upon the atomicity of the gas.

For a monoatomic gas, γ = 1.67.

For a diatomic gas, γ = 1.41.

There will be the same rise in temperature of both the balls.

The whole of the kinetic energy of a ball is used in heating it, so

$\frac{1}{2}\text{mv}^2 = \text{mc}\Delta \text T \quad \Rightarrow \quad \Delta \text T = \frac{\text v^2}{2\text c}$

The mass m cancels out. Since both the balls are of the same metal (same specific heat c) and strike the target with the same velocity v, the rise in temperature ΔT is the same for both, irrespective of their masses.

The molecules of water at 0°C have greater potential energy.

Heat has to be given to ice at 0°C to melt it into water at 0°C. Since the temperature does not change, the kinetic energy of the molecules remains the same. Hence this latent heat is used up in increasing the potential energy of the molecules of the water formed at 0°C.

The gas has the greatest internal energy and the solid has the least.

In a gas the molecules are far apart, so the (negative) potential energy of its molecules is very small. In a solid the molecules are very close together, so the (negative) potential energy of its molecules is very large. Since the kinetic energy is the same for all three at the same temperature, the total internal energy is greatest for the gas and least for the solid.

(i) No, heat has not been given to the water. The thermos-bottle is closed and thermally insulated, so no heat can enter it from outside. Hence Q = 0.

(ii) Yes, work has been done on the water. In shaking the bottle, work is done on the water against the viscous forces present in it. Thus W is negative.

(iii) Yes, the internal energy of the water has increased. According to the first law of thermodynamics, ΔU = Q − W. Here Q = 0 and W is negative, so ΔU is positive. That is why the temperature of the water rises.

The efficiency of the engine will decrease.

The efficiency of a Carnot engine is

$\eta = 1 - \frac{\text T_2}{\text T_1}$

where T₂ is the absolute temperature of the sink. On raising T₂, the ratio $\frac{\text T_2}{\text T_1}$ increases and hence the efficiency η decreases.

The specific heat of a gas is

$\text c = \frac{\text Q}{\text m \times \Delta \text T}$

Specific heat is zero in an adiabatic process, because in it Q = 0, so that c = 0.

Specific heat is infinite in an isothermal process, because in it ΔT = 0, so that c = ∞.

Given,

• Work done on the gas = 400 J, so W = − 400 J
• The change is adiabatic, so Q = 0

Heat absorbed by the gas : In an adiabatic process no heat is exchanged with the surroundings, so the heat absorbed by the gas is zero.

Change in internal energy : By the first law of thermodynamics,

$\Delta \text U = \text Q - \text W = 0 - (-400)$

$= +400\ \text J$

Hence the internal energy of the gas increases by 400 J and the heat absorbed by the gas is zero.

The final pressure will be more than 5 times the initial pressure.

For an adiabatic compression Poisson's law gives

$\text P_1\text V_1^{\gamma} = \text P_2\text V_2^{\gamma} \quad \Rightarrow \quad \frac{\text P_2}{\text P_1} = \left(\frac{\text V_1}{\text V_2}\right)^{\gamma}$

Here $\text V_2 = \frac{\text V_1}{5}$, so

$\frac{\text P_2}{\text P_1} = (5)^{\gamma}$

Since γ is always greater than 1, the quantity 5^γ is greater than 5. Physically, in adiabatic compression work is done on the gas, so its temperature also rises, and this rise in temperature raises the pressure further than the compression alone would.

When a system is taken from an initial state to other different states and is finally brought back to the initial state, then the process is called a cyclic process.

Since the internal energy is a function only of the state of the system, and the initial and the final states are the same, there is no change in the internal energy in a cyclic process, that is, ΔU = 0. Hence by the first law of thermodynamics,

$0 = \text Q - \text W \quad \Rightarrow \quad \text Q = \text W$

Thus, in a cyclic process the heat given to a system equals the net work done by the system. On a P-V diagram the cyclic process is represented by a closed curve, and the net work done is equal to the area enclosed by that curve.

Molar specific heat of a gas at constant volume (C_v) : The amount of heat required to raise the temperature of one mole of a gas by 1 K at constant volume is called the molar specific heat capacity at constant volume.

$\text C_v = \frac{1}{\mu}\left(\frac{\text{dQ}}{\text{dT}}\right)_v$

where μ is the number of moles of the gas. Since the volume is fixed, the gas cannot expand and hence all the heat supplied increases the internal energy of the gas.

Isothermal process : If during a thermodynamic process taking place in a system the temperature remains constant, then the process is 'isothermal'.

In such a process all the three quantities Q, W and ΔU may change. However, in the case of an ideal gas the internal energy depends only upon the temperature of the gas. Therefore, if an ideal gas undergoes an isothermal process, there will be no change in its internal energy,

$\Delta \text U = 0$

Then, according to the first law of thermodynamics ΔU = Q − W, we have

$0 = \text Q - \text W \qquad \text{or} \qquad \text Q = \text W$

Hence, for an isothermal process in an ideal gas, the heat absorbed by the gas is entirely used in the work done by the gas. This is true for an ideal gas only and not for real systems.

Adiabatic process : If during a process taking place in a system, heat neither enters the system nor leaves it (Q = 0), then the process is 'adiabatic'.

From the first law of thermodynamics ΔU = Q − W, we have

$\Delta \text U = 0 - \text W \qquad \text{or} \qquad \Delta \text U = -\text W$

When work is done on the system (that is, W is negative), then

$\Delta \text U = -(-\text W) = \text W$

that is, the internal energy of the system increases and its temperature rises.

When work is done by the system (that is, W is positive), then

$\Delta \text U = -(+\text W) = -\text W$

that is, the internal energy of the system decreases and its temperature falls.

Internal energy (U) of a gas : Every bulk system consists of a large number of molecules. The internal energy of the system is the sum of the kinetic energy and the potential energy of these molecules.

Why U is a unique function : Suppose a system is taken from an initial state A to a state B by a process 1. Let Q₁ be the heat taken by the system and W₁ the work done by the system. If the system is taken from A to B by other different processes 2, 3, 4, ..., then the heat transferred and the work done are found to be different in each case, but the difference is the same for all of them,

$\text Q_1 - \text W_1 = \text Q_2 - \text W_2 = \text Q_3 - \text W_3 = \ldots = \text{constant}$

This difference (Q − W) is defined as the change in the internal energy of the system.

Thus the internal energy U of a thermodynamic system is a characteristic property of the state of the system; it does not matter how that state has been obtained. U is therefore said to be a unique function, because it depends only upon the state of the system and not upon the path adopted to reach that state.

First law of thermodynamics : If an amount of heat Q is given to a system, a part of it will be used in increasing the internal energy (ΔU) of the system and the rest in doing work (W) by the system. Thus

$\text Q = \Delta \text U + \text W \qquad \text{or} \qquad \Delta \text U = \text Q - \text W$

Concept of internal energy on the basis of this law : Let a system be taken from an initial state A to a state B by a process 1, in which Q₁ is the heat taken by the system and W₁ the work done by the system. Now let the same change of state be brought about by other processes 2, 3, 4, ..., for which the corresponding quantities are Q₂, W₂; Q₃, W₃ and so on. It is found that

$\text Q_1 - \text W_1 = \text Q_2 - \text W_2 = \text Q_3 - \text W_3 = \ldots = \text{constant}$

Thus, although the heat transferred Q and the work done W are different for different processes, their difference (Q − W) is the same for all of them. This means that the difference depends only upon the two states of the system and not upon the path adopted between them.

This quantity (Q − W) is defined as the change in the internal energy of the system, denoted by ΔU. Hence the internal energy is a characteristic property of the state of the system.

For the melting of ice : When ice melts, its volume decreases, so work is done on the system by the atmosphere. During the change of state the heat absorbed from an external source is called the total latent heat. Since both the total latent heat and the work done on the system remain in the system, the internal latent heat of the system increases. Hence the internal latent heat of melting of ice is greater than its total latent heat.

For the formation of steam : When water is converted into steam, its volume increases considerably, so now the work is done by the system against the atmospheric pressure. In this case a part of the total latent heat is used in doing this external work, and only the rest remains in the steam. Hence the internal latent heat of steam is less than its total latent heat.

Only the energy of the translatory motion of the molecules is represented by temperature. The other forms of energy, such as the intermolecular potential energy and the energy of the binding of the atoms within a molecule, are not represented by temperature.

Therefore, at absolute zero, the translational motion of the molecules ceases and the temperature becomes zero, but the substance still possesses these other forms of energy, which do not become zero. (A small residual energy, called the zero-point energy, also remains even at absolute zero.) Hence the internal energy of the substance is still not zero.

Hence absolute zero temperature is not the temperature of zero energy.

Water fall — irreversible. The potential energy of the falling water is not fully converted into kinetic energy; a part is converted into heat (and sound also) on striking the ground. The energy dissipated as heat cannot be recovered.

Rusting of iron — irreversible. In rusting, iron gets oxidised by air, which is a chemical change that cannot be reversed of itself.

Electrolysis — approximately reversible. On reversing the direction of the current, the motion of the ions is reversed.

Heat conduction from a hot to a cold body — irreversible. Heat cannot be transferred back from the cold body to the hot body without any external aid, since that would violate the second law of thermodynamics.

The basic requirements of Carnot's engine cannot be fulfilled in practice. For example :

(i) Bodies of infinitely large heat capacities are not available for the source and the sink.

(ii) Perfectly conducting and perfectly insulating materials for the cylinder and the piston do not exist.

(iii) The complete absence of dissipative forces such as friction is not possible.

(iv) An ideal gas as the working substance is only an imaginary conception.

Hence Carnot's engine cannot be realised in practice.

The efficiency of an 'ideal' heat engine is given by

$\eta = 1 - \frac{\text T_2}{\text T_1}$

The efficiency will be 1, that is 100%, only when the temperature T₂ of the sink is 0 K. Since absolute zero temperature cannot be achieved, a heat engine of 100% efficiency is a theoretical impossibility.

Carnot's engine is an imaginary conception based on ideal situations, such as the complete absence of dissipative forces, perfectly conducting and perfectly insulating materials, and infinitely slow processes. Hence, for the given working temperatures, it has the theoretically possible maximum efficiency.

Every real engine suffers from friction and other dissipative losses and its processes are irreversible, so its efficiency must be smaller. Carnot's theorem is the mathematical argument for this fact.

A refrigerator is a heat engine working in the reverse direction.

Heat engine	Refrigerator
The working substance takes in heat from a body at a higher temperature (source).	The working substance takes in heat from a body at a lower temperature (inner space of the refrigerator).
It converts a part of this heat into mechanical work.	It has a net amount of work done on it by an external agent.
It gives out the rest of the heat to a body at a lower temperature (sink).	It gives out a larger amount of heat to a hot body (the atmosphere).
Heat flows from the hot body to the cold body, that is, in the natural direction.	Heat is transferred from the cold body to the hot body, that is, against the natural direction.

No, a room cannot be cooled by leaving the door of an electric refrigerator open. On the contrary, the room will warm up.

The refrigerator removes heat from its interior and expels it through the radiator provided at the back into the surrounding air, thus warming the air. For doing this, additional energy is supplied to the refrigerator by an electric motor. The heat expelled into the air is therefore the sum of the energy from the motor and the heat removed from the interior of the refrigerator.

In other words, the refrigerator adds more heat into the room than it removes from its interior. On opening its door, the refrigerator will run continuously and hence add even more heat to the room than when its door is closed.

Yes, a refrigerator can be used as a heat pump to warm a room.

A refrigerator removes heat from its interior and expels it into the outer surroundings through its radiator at the back, this being done by the expenditure of the electrical energy supplied to it. Therefore, if the refrigerator is installed with its back (radiator) inside the room, the remaining part being outside the room, the room will be warmed up.

The work done in a thermodynamic process is equal to the area enclosed between the P-V curve and the volume-axis. It is taken positive when the volume increases (work done by the gas) and negative when the volume decreases (work done on the gas).

(a) The graph is a straight line parallel to the pressure-axis at V = 1.5 m³, so the volume does not change, ΔV = 0. Therefore

$\text W = \text P \times \Delta \text V = 0$

Hence the work done is zero.

(b) The closed figure is a triangle with vertices (2, 1), (2, 3) and (4, 1), traced clockwise. The work done is the area of this triangle,

$\text W = \frac{1}{2} \times (4 - 2)\ \text m^3 \times (3 - 1)\ \text{N m}^{-2} = \frac{1}{2} \times 2 \times 2$

$= 2\ \text J$

Hence 2 J of work is done by the gas.

(c) The graph is a straight line parallel to the volume-axis at the constant pressure P = 3 N m^-2, the volume increasing from 1 m³ to 4 m³. Therefore

$\text W = \text P(\text V_2 - \text V_1) = 3 \times (4 - 1)$

$= 9\ \text J$

Hence 9 J of work is done by the gas.

(d) The closed figure is a rectangle between V = 1 m³ and V = 4 m³ and between P = 1 N m^-2 and P = 3 N m^-2, traced anticlockwise. The magnitude of the work is the area of the rectangle,

$\text W = (4 - 1)\ \text m^3 \times (3 - 1)\ \text{N m}^{-2} = 3 \times 2 = 6\ \text J$

Since the cycle is traced anticlockwise, the net work is done on the gas.

Hence 6 J of work is done on the gas.

More work will have to be done in the second process, that is, in first increasing the pressure only and then increasing the volume.

The work done is measured by the area enclosed between the P-V curve and the volume-axis.

First process (i a f) : The pressure is first kept at the low value P_i while the volume increases from V_i to V_f, and then the pressure is raised at the constant volume V_f, during which no work is done. Hence

$\text W_{iaf} = \text{area } iadc$

Second process (i b f) : The pressure is first raised at the constant volume V_i, during which no work is done, and then the volume increases from V_i to V_f at the higher constant pressure P_f. Hence

$\text W_{ibf} = \text{area } bfdc$

The area bfdc is larger than the area iadc, because in the second process the expansion takes place against the greater pressure P_f. Hence more work is done in the second process.

Change in internal energy : The internal energy is a state function, so its change depends only upon the initial state i and the final state f, which are the same in both the cases. Hence the change in the internal energy of the gas will be the same in the two processes.

By the first law of thermodynamics,

$\text Q = \Delta \text U + \text W \qquad \text{and} \qquad \Delta \text U = \text U_f - \text U_i$

Process 1 : Given Q = 35 J, W = − 15 J and U_f = − 10 J.

$\Delta \text U = \text Q - \text W = 35 - (-15) = 50\ \text J$

$\text U_i = \text U_f - \Delta \text U = -10 - 50 = -60\ \text J$

Process 2 : Given Q = − 15 J, U_f = 45 J and ΔU = − 20 J.

$\text W = \text Q - \Delta \text U = -15 - (-20) = 5\ \text J$

$\text U_i = \text U_f - \Delta \text U = 45 - (-20) = 65\ \text J$

Process 3 : Given W = − 20 J, U_i = 50 J and ΔU = 20 J.

$\text Q = \Delta \text U + \text W = 20 + (-20) = 0$

$\text U_f = \text U_i + \Delta \text U = 50 + 20 = 70\ \text J$

Hence the completed table is :

Given, R = 8.31 J mol^-1 K^-1.

By Mayer's relation, the difference between the two gram-molecular heats of an ideal gas is

$\text C_p - \text C_v = \text R$

This difference is equal to that amount of heat which is used in doing work against the external pressure in raising the temperature of a gram-molecule of an ideal gas through 1°C at constant pressure.

Expressing R in calorie, and using J = 4.18 J cal^-1,

$\text C_p - \text C_v = \frac{\text R}{\text J} = \frac{8.31\ \text{J mol}^{-1}\ \text K^{-1}}{4.18\ \text{J cal}^{-1}}$

$= 1.99\ \text{cal mol}^{-1}\ \text K^{-1}$

Hence the difference in the gram-molecular heats of an ideal gas is nearly 2 calorie/(mole-K).

The difference between C_p and C_v is equal to that amount of heat which is used in doing work against the external pressure in raising the temperature of one gram-molecule of an ideal gas through 1°C at constant pressure. Hence the heat required for the external work is

$\text C_p - \text C_v = \text R = 8.31\ \text J\ \text{(mol-}°\text C)^{-1}$

Converting into calorie, since 4.18 J = 1 cal,

$\text{Heat required} = \frac{8.31\ \text J\ \text{(mol-}°\text C)^{-1}}{4.18\ \text{J cal}^{-1}}$

$= 1.99\ \text{cal (mol-}°\text C)^{-1}$

Hence approximately 2 calorie of heat is required for the external work.

Oxygen will undergo a greater temperature-rise.

Equal masses of the two gases are taken, so the rise in temperature is

$\Delta \text T = \frac{\text Q}{\text mc_v}$

where c_v is the specific heat per unit mass. For equal masses and equal heat, the gas having the smaller specific heat per unit mass will show the greater rise in temperature.

Helium is monoatomic, so its molar specific heat is $\text C_v = \frac{3}{2}\text R$, and its molar mass is only 4 g mol^-1. Oxygen is diatomic, so $\text C_v = \frac{5}{2}\text R$, but its molar mass is 32 g mol^-1. Hence

$c_v = \frac{\text C_v}{\text M}$

Although the molar specific heat of oxygen is the greater of the two, its molar mass is eight times that of helium. Therefore the specific heat per unit mass of helium is much the larger, and one kilogram of helium contains far more molecules than one kilogram of oxygen.

Hence, for equal masses given equal quantities of heat, the temperature-rise of oxygen is greater than that of helium.

Note: The answer given in the textbook is helium. That answer holds when equal numbers of moles are compared, for which the molar specific heats decide the result. The question as set compares equal masses, and then the specific heat per unit mass decides it, which reverses the conclusion in favour of oxygen.

(i) Yes, heat has been transferred from the system to the water. The mixture burns in the vessel and gives out heat, which is taken by the surrounding water; that is why the temperature of the water rises.

(ii) No, the external work has not been done, because the vessel is closed and so the volume of the system remained constant, giving W = P × ΔV = 0.

(iii) Yes, the internal energy of the water has increased. Applying the first law of thermodynamics ΔU = Q − W to the water, here Q is positive (heat has been taken by the water) and W = 0. Thus ΔU is positive, that is, the internal energy of the water has increased.

For an ideal gas : In this expansion neither any exchange of heat takes place (Q = 0), because the box is thermally insulated, nor any external work is done (W = P × ΔV = 0), because the expansion takes place into vacuum where P = 0. Hence by the first law of thermodynamics,

$\Delta \text U = \text Q - \text W = 0 - 0 = 0$

Thus there is no change in the internal energy of the gas. Since the internal energy of an 'ideal' gas depends only upon the temperature, the temperature also remains unchanged.

For a real gas : Among the molecules of a real gas there is mutual attraction, and so internal work will be done against it in the expansion of the gas. Hence the internal energy will decrease and the temperature will also fall.

The method of decreasing the temperature T₂ of the sink is preferable.

The efficiency of Carnot's engine is

$\eta = 1 - \frac{\text T_2}{\text T_1}$

The efficiency η can be increased either by increasing T₁ or by decreasing T₂. Since T₂ is smaller than T₁, a given decrease in T₂ changes the ratio $\frac{\text T_2}{\text T_1}$ by a larger amount than an equal increase in T₁.

Hence a decrease in T₂ will be more effective than an equal increase in T₁.

It is an attractive idea to drive a ship on the energy drawn from the internal energy of the water of the ocean. At the start of its cycle the engine of the ship will draw some heat Q₁ from the water and convert a part of it into work, but where would it reject the rest?

By the second law of thermodynamics, the engine must reject some heat to a colder reservoir, but none is available at hand. Theoretically it is possible if we can arrange some conveyance to the cold upper atmosphere, but the practical difficulties make it almost impossible.

From Carnot's cycle also we see that the efficiency is

$\eta = 1 - \frac{\text T_2}{\text T_1}$

Thus η = 0 if T₂ = T₁, that is, without a temperature-difference the conversion of thermal energy into mechanical work is impossible.

No, it is not possible.

The coefficient of performance of an 'ideal' refrigerator is given by

$\text K = \frac{\text T_2}{\text T_1 - \text T_2}$

Obviously, K becomes vanishingly small as the temperature T₂ of the freezer, from which heat is taken, approaches absolute zero. This means that as the freezer becomes more and more cold, the refrigerator finds it more and more difficult to run.

Therefore it is impossible to take the freezer down to T₂ = 0 K.

No, they do not violate the laws of thermodynamics.

A heat pump transfers heat from some cold place to a hotter place at the expense of energy supplied externally by a motor. The Clausius statement of the second law forbids such a transfer only when it takes place without the expenditure of work by an external energy source.

Since the necessary work is here supplied by the motor, the heat pump does not violate any law of thermodynamics.

Gas C is ideal.

For a given mass of an ideal gas at a constant temperature, Boyle's law gives

$\text{PV} = \text{constant}$

that is, the product PV should not change however the pressure P is varied. In the graph, the line for the gas C is parallel to the pressure-axis, showing that PV remains constant for it at all pressures.

For the gases A and B the product PV changes with the pressure, so they do not obey Boyle's law at all pressures and are therefore not ideal.

A truly reversible process is an ideal conception, not realisable in practice. However, the following two processes are approximately reversible :

(i) An extremely slow isothermal expansion or compression of a so-called perfect gas contained in a cylinder at room temperature and fitted with a smoothly-moving piston.

(ii) An extremely slowly carried extension or contraction of an elastic spring.

Suppose there are two bodies 1 and 2 at temperatures T₁ and T₂, where T₁ > T₂. When they are brought into contact, heat flows by conduction from body 1 to body 2 till they reach a common temperature.

Heat cannot flow in the reverse direction, from body 2 to body 1, because heat-flow by itself from a cold body to a hot body is not allowed by the second law of thermodynamics.

Thus heat-conduction through a finite temperature difference is an irreversible phenomenon.

AB is the isobaric process, since along it the pressure remains constant, the curve being parallel to the volume-axis.

AC is the isothermal process, since its slope is smaller than that of the adiabatic curve.

AD is the adiabatic process, since the slope of an adiabatic curve is greater than the slope of an isothermal curve at the same point, the ratio of the slopes being γ.

The product P ΔV denotes the work done on the gas or by the gas.

The work done is measured by the area enclosed between the curve and the volume-axis. Since for the same change of volume from V₁ to V₂ the curve AB encloses the largest area, the work done is maximum for the isobaric process AB.

For a gas the difference between C_p and C_v should be nearly 2 cal/(mol-K), by Mayer's relation, and the ratio $\frac{\text C_p}{\text C_v}$ should be either 1.67, for a monoatomic gas, or 1.41, for a diatomic gas.

(i) C_p = 5.0, C_v = 3.0 :

$\text C_p - \text C_v = 5.0 - 3.0 = 2.0 \quad \text{and} \quad \frac{\text C_p}{\text C_v} = \frac{5.0}{3.0} = 1.67$

Both the conditions are satisfied.

(ii) C_p = 6.0, C_v = 4.0 :

$\text C_p - \text C_v = 6.0 - 4.0 = 2.0 \quad \text{and} \quad \frac{\text C_p}{\text C_v} = \frac{6.0}{4.0} = 1.5$

The difference is correct, but the ratio is neither 1.67 nor 1.41.

(iii) C_p = 4.2, C_v = 3.0 :

$\text C_p - \text C_v = 4.2 - 3.0 = 1.2 \quad \text{and} \quad \frac{\text C_p}{\text C_v} = \frac{4.2}{3.0} = 1.4$

The ratio is correct, but the difference is not nearly 2.

Hence the first set (i) is the most reliable.

Isothermal process : When a system undergoes a physical change under the condition that the temperature of the system remains constant, then such a process is called an 'isothermal process'. For such a process to take place, it is necessary that the system be surrounded by a perfectly conducting material and that the process be carried out very slowly, so that heat finds sufficient time to flow.

Examples : The melting of ice at 0°C and the boiling of water at 100°C are isothermal changes, since the temperature remains constant throughout.

Relation between pressure and volume : For an ideal gas an isothermal process obeys Boyle's law, that is, for a given mass of gas at constant temperature,

$\text{PV} = \text{constant}$

Adiabatic process : When a system undergoes a change under the condition that no exchange of heat takes place between the system and the surroundings, then such a process is called an 'adiabatic process'. For such a process to take place, it is necessary that the system be perfectly insulated from the surroundings, or that the process be carried out very rapidly, so that the heat does not find time to flow in or out.

Examples : The sudden bursting of a bicycle tube and the sudden expansion of the air coming out of a burst balloon are adiabatic changes.

Relation between pressure and volume : For an ideal gas an adiabatic process obeys Poisson's law, that is,

$\text{PV}^{\gamma} = \text{constant}$

where $\gamma = \frac{\text C_p}{\text C_v}$ is the ratio of the two specific heats of the gas.

The curve PR represents the adiabatic expansion.

Of the two curves, the slope of the adiabatic curve is greater than the slope of the isothermal curve at the same point, the ratio of the two slopes being

$\frac{\text{slope of adiabatic curve}}{\text{slope of isothermal curve}} = \gamma$

and γ is always greater than 1.

The reason is that in both the isothermal and the adiabatic expansions the pressure of the gas falls, but for the same fall in pressure the increase in the volume of the gas during the adiabatic expansion is less than that during the isothermal expansion, because during the adiabatic expansion the temperature of the gas also falls.

Since the curve PR is steeper than the curve PQ, PR is the adiabatic expansion.

(i) The transfer and transformation of energy

Thermodynamics is concerned with how energy is transferred and transformed within systems, particularly those involving temperature, pressure and volume. The motion of objects is the subject of mechanics, not of thermodynamics.

(ii) Heat transfer

Thermal physics focuses on the behaviour of heat, temperature and energy exchanges, and explains how heat flows from a hotter object to a cooler one. Work, kinetic energy and potential energy are the key concepts of mechanics.

(iii) Work

Thermodynamics combines thermal physics and mechanics by investigating how thermal energy can be transformed into mechanical work. For instance, the expansion of a gas in a piston converts heat energy into mechanical work, which is exactly where the two fields meet.

(iv) Second Law of Thermodynamics

The second law states that heat flows naturally from a hotter object to a cooler one until thermal equilibrium is reached, and that the reverse flow cannot take place of itself without external work.

(v) Heat transfer stops

When a system reaches thermal equilibrium, the temperatures of the systems in contact become equal and there is no further net transfer of heat energy between them.

(i) Heat flows naturally from hot to cold.

The second law of thermodynamics asserts that heat flows of itself from a hotter body to a cooler body, and that it is impossible to transfer heat from a cooler body to a hotter body unless external work is performed.

(ii) Isothermal expansion

In the first process of the Carnot cycle the gas absorbs heat from the hot reservoir and does work on its surroundings while its temperature remains constant. This is the isothermal expansion.

(iii) Irreversibility of real processes

Real processes are not reversible and involve friction, heat loss and other inefficiencies. These irreversibilities reduce the efficiency of a real engine below that of the ideal Carnot engine.

(iv) Decreases

In an adiabatic expansion no heat is exchanged, so the gas does work on its surroundings at the expense of its own internal energy. The internal energy therefore falls and the temperature decreases.

(v) It defines the maximum theoretical efficiency any heat engine can achieve.

The Carnot cycle sets the upper limit for the efficiency of any heat engine working between the same two temperatures. Real engines can approach this limit but can never exceed it, because of practical limitations.

Given,

• Temperature of the hot reservoir, T₁ = 500 K
• Temperature of the cold reservoir, T₂ = 300 K
• Heat absorbed from the hot reservoir, Q₁ = 600 J
• Heat released to the cold reservoir, Q₂ = 400 J

(i) 200 J

In one complete cycle the working substance returns to its original state, so there is no net change in its internal energy. By the first law of thermodynamics, the work done by the engine is the difference between the heat absorbed and the heat released,

$$\begin{aligned}\text W = \text Q_1 - \text Q_2 = 600\ \text J - 400\ \text J \\[1em] = 200\ \text J\end{aligned}$$

(ii) 33.3%

The efficiency is the ratio of the work done to the heat absorbed during the isothermal expansion,

$\eta = \frac{\text W}{\text Q_1} = \frac{200\ \text J}{600\ \text J} = \frac{1}{3}$

$= 0.333 = 33.3%$

(iii) It states that some energy is always lost as heat.

The second law ensures that not all the absorbed heat can be converted into work; some heat is always rejected to the cold reservoir. Hence the efficiency can never be 100%.

(iv) Adiabatic expansion

In an adiabatic process no heat is exchanged with the surroundings, though the temperature of the gas changes. The isothermal expansion and the isothermal compression both involve heat exchange.

(v) 40%

The Carnot efficiency for an engine operating between the two temperatures is

$\eta = 1 - \frac{\text T_2}{\text T_1} = 1 - \frac{300\ \text K}{500\ \text K}$

$= 1 - 0.6 = 0.4 = 40%$

(i) Given, work done on the system = 50 J, so W = − 50 J. The process is adiabatic, so the exchange of heat Q = 0.

From the first law of thermodynamics,

$\Delta \text U = \text Q - \text W = 0 - (-50)$

$= 50\ \text J$

Hence the internal energy of the system increases by 50 J.

(ii) Given, heat given to the system, Q = 200 J, and work done by the system, W = 100 J.

From the first law of thermodynamics,

$\Delta \text U = \text Q - \text W = 200 - 100$

$= 100\ \text J$

Hence the internal energy of the system increases by 100 J.

(iii) In a monoatomic gas the molecules consist of single atoms, so only translational motion takes place. Hence the translational motion of the molecules is responsible for the internal energy of a monoatomic gas.

Given,

• The Carnot engine works between 0°C and 100°C
• Temperature of the source, T₁ = 100°C = 100 + 273 = 373 K
• Temperature of the sink, T₂ = 0°C = 0 + 273 = 273 K
• Heat taken from the source per cycle, Q₁ = 746 J

(i) Efficiency of the engine : The efficiency of a Carnot engine is

$\eta = 1 - \frac{\text T_2}{\text T_1} = 1 - \frac{273}{373}$

$$\begin{aligned}= \frac{373 - 273}{373} = \frac{100}{373} \\[1em] = 0.268 = 26.8%\end{aligned}$$

(ii) Work done by the engine per cycle : The efficiency is the ratio of the work done to the heat taken in from the source,

$\eta = \frac{\text W}{\text Q_1} \quad \Rightarrow \quad \text W = \eta \times \text Q_1$

$$\begin{aligned}\text W = \frac{26.8}{100} \times 746 \\[1em] \approx 200\ \text J\end{aligned}$$

(iii) Heat rejected to the sink per cycle : The heat rejected is the difference between the heat taken in and the work done,

$\text Q_2 = \text Q_1 - \text W = 746\ \text J - 200\ \text J$

$= 546\ \text J$

Hence the efficiency is 26.8%, the work done per cycle is 200 J and the heat rejected to the sink per cycle is 546 J.

Value of the mechanical equivalent of heat J :

$\text J = 4.18\ \text{joule calorie}^{-1}\ (\text{J cal}^{-1})$

$\text J = 4.18 \times 10^{3}\ \text{joule kilocalorie}^{-1}\ (\text{J kcal}^{-1})$

J is not a physical quantity; it is simply a 'conversion factor'.

Reason : Joule performed a number of experiments to establish a relation between heat and work. He found that 4.18 × 10³ joule of work produces the same temperature-rise as 1 kilocalorie of heat, and he obtained the same result in every case, whatever the manner of doing the work and whatever the substance used. Thus he concluded that

$4.18 \times 10^{3}\ \text{joule of work is equivalent to 1 kilocalorie of heat.}$

Hence, if Q calorie of heat is equivalent to W joule of work, then

$\text W = 4.18\ \text Q \qquad \text{or} \qquad \text W = \text{JQ}$

Here Q is in calorie. Therefore, in order to express Q calorie of heat in joule, it is multiplied by 4.18. Since heat and work are both forms of energy, and are therefore the same physical quantity, the number 4.18 merely converts a measurement expressed in one unit into the same measurement expressed in another unit.

Hence J is not a physical quantity; it is called the 'mechanical equivalent of heat' and is simply a conversion factor.

Explanation of the equation dQ = dU + P dV :

Suppose an infinitesimally small amount of heat dQ is given to a system. A part of it is used in increasing the internal energy of the system by dU, and the rest is used by the system in doing external work. If the system expands by a volume dV against a constant external pressure P, then the external work done is P dV. Hence

$\text{dQ} = \text{dU} + \text P\text{dV}$

This equation expresses the first law of thermodynamics. It is simply the law of conservation of energy applied to a thermodynamic system.

Concept of internal energy on the basis of this equation :

Writing the equation as dU = dQ − P dV, let a system be taken from an initial state A to a final state B by a process 1. Let Q₁ be the heat taken by the system and W₁ the work done by the system, so that the difference is (Q₁ − W₁).

Now let the same change of state A → B be brought about by other different processes 2, 3, 4, ..., for which the corresponding differences are (Q₂ − W₂), (Q₃ − W₃) and so on. It is found experimentally that

$\text Q_1 - \text W_1 = \text Q_2 - \text W_2 = \text Q_3 - \text W_3 = \ldots = \text{constant}$

Thus, although the heat transferred Q and the work done W are individually different for different processes, their difference is the same for all of them. This means that the difference (Q − W) for two given states of a system depends only upon those states and does not depend upon the path adopted by the system between the two states.

This quantity is defined as the change in the internal energy of the system, ΔU. Hence the internal energy U of a thermodynamic system is a characteristic property of the state of the system; it does not matter how that state has been obtained. U is a unique function, because it depends only upon the state of the system.

Let a mass m of water freeze into ice at the freezing point.

Heat given up : In freezing, the water gives up heat. If L be the latent heat of freezing, the heat given up by the water is

$\text Q = -\text{mL}$

the negative sign appearing because the heat is given up by the water and not taken by it.

Work done : When water freezes, its volume increases, since the volume of ice is greater than the volume of water of the same mass. Therefore the water has to do work against the external atmospheric pressure. If P be the atmospheric pressure and ΔV the increase in volume, then the work done by the system is

$\text W = \text P \times \Delta \text V$

which is positive, since the volume increases.

Applying the first law of thermodynamics : The change in the internal energy of the water is

$\Delta \text U = \text Q - \text W = -\text{mL} - \text P\Delta \text V$

Both the terms on the right-hand side are negative, so ΔU is negative.

Hence, when water freezes its internal energy decreases, and this decrease is greater than the heat given up by the water in freezing, the extra decrease being on account of the work done by the water against the atmospheric pressure in expanding.

The thermodynamic equation dU = dQ − P dV can be written as

$\text{dQ} = \text{dU} + \text P\text{dV}$

This means that when an amount of heat dQ is given to an ideal gas, a part of it (dU) is used in increasing the internal energy of the gas and the rest (P dV) is used in doing external work by the expanding gas.

Now, Mayer's formula for an ideal gas is

$\text C_p - \text C_v = \text R$

which may be written as

$\text C_p\text{dT} = \text C_v\text{dT} + \text R\text{dT}$

Comparing the two equations term by term :

• C_p dT is the heat given to one mole of an ideal gas at constant pressure, and so corresponds to dQ.
• C_v dT is the part used in increasing the temperature, that is, in increasing the internal energy of the gas, and so corresponds to dU.
• R dT is the part used in doing work against the external pressure, and so corresponds to P dV.

Hence Mayer's formula is simply the first law of thermodynamics applied to one mole of an ideal gas heated through 1 K at constant pressure, and the universal gas constant R represents the external work done in that process.

Reversible process : A reversible process is one which can be reversed in such a way that all changes taking place in the direct process are exactly repeated in the inverse order and opposite sense, and no changes are left in any of the bodies taking part in the process or in the surroundings.

For example, if an amount of heat is supplied to a system and an amount of work is obtained from it in the direct process, the same amount of heat should be obtainable by doing the same amount of work on the system in the reverse process.

Examples : An extremely slow isothermal expansion or compression of a perfect gas contained in a cylinder fitted with a smoothly-moving piston, and an extremely slowly carried extension or contraction of an elastic spring, are approximately reversible.

Irreversible process : Any process which is not reversible exactly is an irreversible process.

Examples : Free expansion, electrical heating of a wire, diffusion of liquids or gases, heat conduction, radiation and radioactive decay are all irreversible. Thus irreversibility is a rule.

Conditions for reversibility : A process can be reversible only when it satisfies the following two conditions :

(i) Dissipative forces such as friction, viscosity, inelasticity, electrical resistance and magnetic hysteresis must be completely absent. Suppose a gas is contained in a cylinder fitted with a piston and placed in contact with a furnace. If the load on the piston is decreased, the gas expands, doing external work in pushing up the piston and also in overcoming the friction between the piston and the walls of the cylinder. If now the load on the piston is increased, the gas is compressed. The work used in pushing up the piston during the expansion is recovered, but not the work used against the friction; on the contrary, more work has to be done against the friction. Thus, due to friction, neither the work done in expansion is fully recovered in compression, nor is the heat taken from the furnace in expansion fully returned to the furnace. The expansion is therefore irreversible.

(ii) The process should take place infinitely slowly. In the above example, when the gas expands, an amount of work is done by the gas to give kinetic energy to the piston. This work cannot be recovered during the reverse process; on the contrary, more work is to be done, again to give kinetic energy to the piston. Hence, in order to make the expansion of the gas reversible, the pressure of the gas on the piston should be only infinitesimally different from the pressure exerted by the piston on the gas. This will be so when the expansion or compression takes place infinitely slowly, so that no kinetic energy is produced.

These conditions are never realised in practice. Hence a reversible process is only an ideal conception.

Thermodynamic system : A thermodynamic system refers to that part of the universe on which we are focusing our attention, such as a gas inside a piston or water inside a kettle. Everything outside the system under consideration is called the surroundings. The state of a system is described by the state variables pressure P, volume V, temperature T and internal energy U.

Cyclic process : When a system is taken from an initial state to other different states and is finally brought back to the initial state, then this is called a 'cyclic process'. Since the internal energy is a function only of the state of the system, there is no change in the internal energy in a cyclic process, ΔU = 0. Hence by the first law of thermodynamics,

$0 = \text Q - \text W \qquad \text{or} \qquad \text Q = \text W$

Work done in a cyclic process :

Suppose a gas is in an initial pressure-volume state A and expands to a state B along the curve ACB. Since the volume increases, the work is done by the gas and is equal to the area enclosed between the curve ACB and the volume-axis,

$\text W_{\text{AB}} = \text{area } \text{ACBB}'\text A' \qquad \text{...(i)}$

Now suppose the gas is compressed and brought back to the state A along the different curve BDA. Since the volume decreases, the work is now done on the gas by an external agent and is equal to the area enclosed between the curve BDA and the volume-axis,

$\text W_{\text{BA}} = \text{area } \text{BDAA}'\text B' \qquad \text{...(ii)}$

The area BDAA′B′ is larger than the area ACBB′A′. Hence the work done on the gas is greater than the work done by the gas, and the net work done on the gas is

$$\begin{aligned}\text W = \text W_{\text{BA}} - \text W_{\text{AB}} \\[1em] = \text{area } \text{BDAA}'\text B' - \text{area } \text{ACBB}'\text A' \\[1em] = \text{area } \text{ACBDA}\end{aligned}$$

Hence the total work done by a thermodynamic system in a cyclic process is equal to the area enclosed by the closed curve.

If the closed P-V curve is traced clockwise, the net work is done by the system; if it is traced anticlockwise, the net work is done on the system.

First law of thermodynamics : If an amount of heat Q is given to a system, a part of it will be used in increasing the internal energy (ΔU) of the system and the rest in doing work (W) by the system. Thus

$\text Q = \Delta \text U + \text W \qquad \text{or} \qquad \Delta \text U = \text Q - \text W$

If in a process an infinitesimally small heat dQ is given to a system and an infinitesimally small work dW is done by the system, then the change in the internal energy dU will also be infinitesimally small, and

$\text{dU} = \text{dQ} - \text{dW}$

(i) Isothermal process : If during a process taking place in a system the temperature remains constant, then the process is 'isothermal'. In the case of an ideal gas the internal energy depends only upon the temperature of the gas. Therefore, if an ideal gas undergoes an isothermal process, there will be no change in its internal energy, ΔU = 0. Then, from ΔU = Q − W,

$0 = \text Q - \text W \qquad \text{or} \qquad \text Q = \text W$

Hence, for an isothermal process in an ideal gas, the heat absorbed by the gas is entirely used in the work done by the gas.

(ii) Isochoric process : If a process takes place in a system at a constant volume, then the process is 'isochoric'. Since there is no change in volume (ΔV = 0), no work is done by the system,

$\text W = \text P \times \Delta \text V = 0$

Therefore, from ΔU = Q − W,

$\Delta \text U = \text Q$

Hence, in an isochoric process, the entire heat given to the system is used in increasing the internal energy of the system. Explosions in gases are isochoric.

(iii) Adiabatic process : If during a process taking place in a system, heat neither enters the system nor leaves it (Q = 0), then the process is 'adiabatic'. From ΔU = Q − W,

$\Delta \text U = 0 - \text W \qquad \text{or} \qquad \Delta \text U = -\text W$

Hence, if work is done on the system (that is, W is negative), then

$\Delta \text U = -(-\text W) = \text W$

that is, the internal energy of the system increases. On the other hand, if work is done by the system (that is, W is positive), then

$\Delta \text U = -(+\text W) = -\text W$

that is, the internal energy of the system decreases.

Why there are two specific heats of a gas : In solids and liquids the specific heat capacity is the amount of heat required to raise the temperature of unit mass of the substance by 1 K. When heat is supplied, the temperature rises uniformly and there is no difference whether the heating takes place at constant volume or at constant pressure, since solids and liquids are nearly incompressible.

For gases, however, the situation is different. Because gases are compressible and can expand significantly when heated, the heat supplied may partly raise the temperature of the gas and partly do external work of expansion in pushing back the surroundings. Hence the amount of heat required depends on the mode of heating — whether the volume is kept constant or the pressure is kept constant. Therefore gases have two specific heat capacities, C_v and C_p.

Deduction of C_p − C_v = R :

Let us consider 1 mole of an ideal gas in equilibrium at a pressure P, volume V and absolute temperature T. Let C_v and C_p be the molar specific heats of the gas at constant volume and at constant pressure respectively.

Heating at constant volume : Let the gas be heated at constant volume so that its temperature is raised by an infinitesimal amount dT. The heat supplied will be C_v dT. As the volume remains constant, the external work done is zero. From the differential form of the first law of thermodynamics,

$\text{dQ} = \text{dU} + \text{dW}$

Here dQ = C_v dT and dW = 0, so that

$\text C_v\text{dT} = \text{dU} \qquad \text{...(i)}$

Heating at constant pressure : Let the same gas now be heated at constant pressure P, until the temperature is raised by the same amount dT. The heat supplied will be C_p dT. Now the gas would expand and do external work against the pressure P. If dV be the change in volume of the gas, the external work would be P dV. Thus, for this process dQ = C_p dT and dW = P dV. Hence, from the first law of thermodynamics,

$\text C_p\text{dT} = \text{dU} + \text P\text{dV} \qquad \text{...(ii)}$

The temperature-change is the same in both the processes. Since the internal energy U of an ideal gas depends only on the temperature, the change in internal energy dU is the same in both the processes. Then, eliminating dU from equations (i) and (ii), we get

$(\text C_p - \text C_v)\text{dT} = \text P\text{dV} \qquad \text{...(iii)}$

Now, the equation of state for 1 mole of an ideal gas is

$\text{PV} = \text{RT}$

where R is the universal gas constant. Differentiating it, keeping P constant, we get

$\text P\text{dV} = \text R\text{dT}$

Substituting this value of P dV in equation (iii),

$(\text C_p - \text C_v)\text{dT} = \text R\text{dT}$

$\text C_p - \text C_v = \text R$

Hence the difference between the two molar specific heats of an ideal gas is equal to the universal gas constant. This relation is known as Mayer's relation.

Heat engine : Any 'cyclic' device by which heat is converted into mechanical work is called a heat engine.

There are three main parts in an engine : a hot body called the source, a working substance and a cold body called the sink. The working substance takes in heat from the source, converts a part of it into useful work and gives out the rest to the sink. This series of processes is called a 'cycle', since the working substance returns to its original state. By repeating the same cycle over and over again, work can be continuously obtained.

Efficiency : Suppose the working substance takes in an amount of heat Q₁ from the source and gives out an amount Q₂ to the sink, and suppose W is the amount of work obtained. The net amount of heat absorbed by the substance is (Q₁ − Q₂), which has been actually converted into work. Applying the first law of thermodynamics to one complete cycle, in which there is no net change in internal energy, we get

$\text Q_1 - \text Q_2 = \text W$

The efficiency η of an engine is defined as the ratio of the net work done by the engine during one cycle to the heat taken in from the source in one cycle. Thus

$\eta = \frac{\text W}{\text Q_1} = \frac{\text Q_1 - \text Q_2}{\text Q_1} = 1 - \frac{\text Q_2}{\text Q_1}$

The engine cannot convert 'all' the heat into work :

From the above expression, the efficiency of the heat engine would be unity, that is 100%, only when

$\frac{\text Q_2}{\text Q_1} = 0 \qquad \text{that is,} \qquad \text Q_2 = 0$

This means that no heat at all should be given out to the sink. But by the Kelvin-Planck statement of the second law of thermodynamics, it is impossible to convert 'all' the heat extracted from a hot body into work. No engine has ever been designed which may convert all the heat taken from the source into work, without giving any heat to the sink. In other words, for obtaining continuous work, a sink is necessary.

Since Q₂ is always less than Q₁ but never zero, the value of η is always less than 1.

Hence the engine cannot convert all the heat taken from a hot source into work.

Carnot's reversible heat engine : Carnot developed the plan of an idealised heat engine, free from all the imperfectness of an actual engine.

The Carnot's engine consists of four components :

(i) A cylinder with perfectly heat-insulating walls but a perfectly conducting base and closed with a tight-fitting perfectly insulating and frictionless piston. A fixed mass of a gas (working substance) is filled in the cylinder, and some weights are placed on the piston of the cylinder.

(ii) A hot body of infinitely large heat capacity at a constant temperature T₁, serving as the 'source'.

(iii) A cold body of infinitely large heat capacity at a constant temperature T₂, serving as the 'sink'.

(iv) A perfectly heat-insulating stand.

The working substance is imagined to go through a cycle of four processes, known as the "Carnot's cycle". Suppose it is in an initial state represented by the point A on the P-V diagram.

Process 1 : The cylinder is placed on the source and the weights on the piston are removed in infinitely small steps. The working substance thus expands infinitely slowly, doing work in raising the piston. During this process the substance takes in heat Q₁ from the source by conduction through the base. Thus the expansion is isothermal. It is continued until the state represented by the point B is reached. The curve AB represents the isothermal expansion of the substance at the constant temperature T₁.

Process 2 : The cylinder is removed from the source and placed on the heat-insulating stand. The weights are further removed so that the substance is further expanded. This expansion is adiabatic, because now no heat can leave or enter the substance through the insulating cylinder. The substance does work in raising the piston and its temperature falls, until it becomes T₂. This adiabatic expansion is represented by the curve BC.

Process 3 : The cylinder is now removed from the stand and placed on the sink. Weights are now placed on the piston in infinitely small steps. The substance is thus compressed isothermally until the state represented by the point D is reached. During this process work is done on the substance and the heat Q₂ developed is given out to the sink. The curve CD represents the isothermal compression at the constant temperature T₂.

Process 4 : The cylinder is once more put on the heat-insulating stand and, by further placing weights on the piston, the substance is compressed adiabatically. A further amount of work is done on the substance and its temperature rises, until it rises once more to T₁ and the state A is recovered. The curve DA represents the adiabatic compression.

The area ABCD represents the net external work W done by the substance in one complete cycle. Since the initial and final states of the substance are the same, there is no net change in its internal energy. Hence, by the first law of thermodynamics,

$\text W = \text Q_1 - \text Q_2$

Expression for the efficiency :

Let (P_a, V_a), (P_b, V_b), (P_c, V_c) and (P_d, V_d) be the co-ordinates of the points A, B, C and D respectively, and let the working substance be 1 mole of an ideal gas.

Since the internal energy of an ideal gas remains unchanged during the isothermal expansion AB, the heat Q₁ is equivalent only to the external work done by the gas in expanding from A to B at the temperature T₁,

$\text Q_1 = \int_{\text V_a}^{\text V_b}\text P\text{dV} = \text{RT}_1\int_{\text V_a}^{\text V_b}\frac{\text{dV}}{\text V} = \text{RT}_1\log_e\frac{\text V_b}{\text V_a} \qquad \text{...(i)}$

Similarly, Q₂ is equivalent to the external work done on the gas during the compression from C to D at the temperature T₂,

$\text Q_2 = -\int_{\text V_c}^{\text V_d}\text P\text{dV} = \text{RT}_2\log_e\frac{\text V_c}{\text V_d} \qquad \text{...(ii)}$

Dividing (i) by (ii),

$\frac{\text Q_1}{\text Q_2} = \frac{\text T_1}{\text T_2} \times \frac{\log_e\frac{\text V_b}{\text V_a}}{\log_e\frac{\text V_c}{\text V_d}} \qquad \text{...(iii)}$

The points B and C, and similarly the points D and A, lie on the same adiabatic. Therefore, from Poisson's law,

$\text T_1\text V_b^{\gamma - 1} = \text T_2\text V_c^{\gamma - 1} \qquad \text{...(iv)}$

$\text T_1\text V_a^{\gamma - 1} = \text T_2\text V_d^{\gamma - 1} \qquad \text{...(v)}$

Dividing equation (iv) by equation (v), we get

$\frac{\text V_b}{\text V_a} = \frac{\text V_c}{\text V_d}$

Using this result in equation (iii), the logarithmic terms cancel and we get

$\frac{\text Q_1}{\text Q_2} = \frac{\text T_1}{\text T_2}$

Hence the efficiency of Carnot's engine is

$\eta = 1 - \frac{\text Q_2}{\text Q_1} = 1 - \frac{\text T_2}{\text T_1}$

Thus the efficiency of Carnot's reversible engine is independent of the working substance and depends only upon the absolute temperatures of the sink and the source.

Can it be realised in practice? No. Carnot's engine has two ideal features :

(i) The source and the sink are bodies of infinitely large heat capacities, so that the working substance takes in all the heat at a constant temperature and gives up all the heat at another constant temperature.

(ii) Each process in Carnot's cycle is completely reversible, which requires that the working substance be contained in a cylinder with perfectly insulating walls and a perfectly insulating frictionless piston so that dissipative forces are absent, that the base of the cylinder be perfectly conducting, and that the gas be expanded and compressed infinitely slowly.

These features cannot be actually realised. Hence it is not possible to obtain a Carnot's engine in practice.

Carnot's ideal refrigerator : Any device for removing heat from a cold place and adding it to a hotter place is called a "refrigerator" or a heat pump. It is essentially a heat engine running backwards.

In a heat engine the working substance takes in heat from a body at a higher temperature, converts a part of it into mechanical work and gives out the rest to a body at a lower temperature. In a refrigerator, a working substance takes in heat from a body at a lower temperature, has a net amount of work done on it by an external agent and gives out a larger amount of heat to a hot body. Thus it continually transfers heat from a cold to a hot body at the expense of mechanical energy supplied to it by an external agent. The working substance is called a "refrigerant".

The engine employing the Carnot cycle may be adopted as a refrigerator, since each step in the cycle is reversible and therefore the entire cycle can be reversed.

Expression for the coefficient of performance :

Let Q₂ be the heat removed from the cold body at temperature T₂, W the net work done on the refrigerant, and Q₁ the heat delivered to the hot body at temperature T₁. Then

$\text Q_1 = \text Q_2 + \text W$

$\text W = \text Q_1 - \text Q_2 = \text Q_2\left(\frac{\text Q_1}{\text Q_2} - 1\right)$

As in Carnot's engine, if we use an ideal gas as the working substance, we can show that

$\frac{\text Q_1}{\text Q_2} = \frac{\text T_1}{\text T_2}$

Therefore

$\text W = \text Q_2\left(\frac{\text T_1}{\text T_2} - 1\right) = \text Q_2\frac{\text T_1 - \text T_2}{\text T_2}$

This is the expression for the work that must be supplied to run the refrigerator.

Coefficient of performance : The purpose of the refrigerator is to remove as much heat (Q₂) as possible from the cold body with the expenditure of as little work (W) as possible. Therefore, a measure of the performance of the refrigerator is expressed by the 'coefficient of performance' K, which is defined as the ratio of the heat taken in from the cold body to the work needed to run the refrigerator. That is

$\text K = \frac{\text Q_2}{\text W}$

Substituting the value of W obtained above,

$\text K = \frac{\text T_2}{\text T_1 - \text T_2}$

This is the expression for the coefficient of performance. A good refrigerator should have a high coefficient of performance, typically 5 or 6. Thicker and high-quality insulation tends to increase the coefficient of performance.

Adiabatic process : When a system undergoes a change under the condition that no exchange of heat takes place between the system and the surroundings, then such a process is called an 'adiabatic process'. For such a process to take place, it is necessary that the system be perfectly insulated from the surroundings.

Relation between T and V for an adiabatic change :

For an ideal gas, an adiabatic process obeys Poisson's law, that is, for a given mass of an ideal gas,

$\text{PV}^{\gamma} = \text{constant}$

where γ is the ratio of the two specific heats C_p and C_v of the gas.

The equation of state for 1 mole of an ideal gas is

$\text{PV} = \text{RT} \qquad \Rightarrow \qquad \text P = \frac{\text{RT}}{\text V}$

Substituting this value of P in Poisson's law,

$\frac{\text{RT}}{\text V} \times \text V^{\gamma} = \text{constant}$

$\text{RT}\text V^{\gamma - 1} = \text{constant}$

Since R is also a constant, it can be absorbed into the constant on the right-hand side. Hence

$\text{TV}^{\gamma - 1} = \text{constant}$

This is the required relation connecting the absolute temperature T and the volume V for an adiabatic change in a perfect gas.

It may also be written for the initial and the final states as

$\text T_1\text V_1^{\gamma - 1} = \text T_2\text V_2^{\gamma - 1}$

Adiabatic process : When a system undergoes a change under the condition that no exchange of heat takes place between the system and the surroundings, then such a process is called an 'adiabatic process'. For such a process to take place, it is necessary that the system be perfectly insulated from the surroundings.

Relation between T and P for an adiabatic change :

For an ideal gas, an adiabatic process obeys Poisson's law,

$\text{PV}^{\gamma} = \text{constant}$

where γ is the ratio of the two specific heats C_p and C_v of the gas.

The equation of state for 1 mole of an ideal gas is

$\text{PV} = \text{RT} \qquad \Rightarrow \qquad \text V = \frac{\text{RT}}{\text P}$

Substituting this value of V in Poisson's law,

$\text P\left(\frac{\text{RT}}{\text P}\right)^{\gamma} = \text{constant}$

$\text P \times \frac{\text R^{\gamma}\text T^{\gamma}}{\text P^{\gamma}} = \text{constant}$

$\text R^{\gamma}\frac{\text T^{\gamma}}{\text P^{\gamma - 1}} = \text{constant}$

Since R is also a constant, it can be absorbed into the constant on the right-hand side. Hence

$\frac{\text T^{\gamma}}{\text P^{\gamma - 1}} = \text{constant}$

This is the required relation between the absolute temperature T and the pressure P of an ideal gas for an adiabatic change.

It may also be written for the initial and the final states as

$\frac{\text T_1^{\gamma}}{\text P_1^{\gamma - 1}} = \frac{\text T_2^{\gamma}}{\text P_2^{\gamma - 1}}$

Isothermal process : When a system undergoes a physical change under the condition that the temperature of the system remains constant, then such a process is called an 'isothermal process'. For such a process to take place, it is necessary that the system be surrounded by a perfectly conducting material, so that any heat produced in the process immediately goes out from the system to the surroundings, or any heat absorbed comes from the surroundings into the system, and the temperature of the system remains constant.

Proof of the expression for the work done :

Let μ moles of an ideal gas at the constant absolute temperature T expand from an initial volume V_i to a final volume V_f. Suppose the gas expands by an infinitesimal volume dV against the instantaneous pressure P. The infinitesimal work done by the gas is

$\text{dW} = \text P\text{dV}$

The total work done in expanding from V_i to V_f is obtained by integrating between these limits,

$\text W = \int_{\text V_i}^{\text V_f}\text P\text{dV} \qquad \text{...(i)}$

From the ideal gas equation for μ moles,

$\text{PV} = \mu\text{RT} \qquad \Rightarrow \qquad \text P = \frac{\mu\text{RT}}{\text V}$

Substituting this value of P in equation (i),

$\text W = \int_{\text V_i}^{\text V_f}\frac{\mu\text{RT}}{\text V}\text{dV}$

Since the process is isothermal, the temperature T is constant, and μ and R are also constants. Hence they can all be taken outside the integral,

$\text W = \mu\text{RT}\int_{\text V_i}^{\text V_f}\frac{\text{dV}}{\text V}$

Carrying out the integration,

$\text W = \mu\text{RT}\left[\log_e \text V\right]_{\text V_i}^{\text V_f}$

Evaluating between the limits,

$\text W = \mu\text{RT}\left(\log_e \text V_f - \log_e \text V_i\right) = \mu\text{RT}\log_e\frac{\text V_f}{\text V_i}$

Converting the natural logarithm into the common logarithm, using log_e x = 2.3026 log₁₀ x,

$\text W = 2.3026\mu\text{RT}\log_{10}\frac{\text V_f}{\text V_i}$

Hence proved.

Internal energy as a characteristic property of the state :

Every bulk system consists of a large number of molecules. The internal energy of the system is simply the sum of the kinetic energy and the potential energy of these molecules, and is denoted by U.

Suppose a system is in an initial state A and is taken to a state B by some process 1. Let Q₁ joule be the heat taken by the system and W₁ joule the work done by the system in this process. We can compute the difference

$\text Q_1 - \text W_1$

Now, suppose we again take the system from the state A to the state B by another process 2. Let Q₂ be the heat taken by the system and W₂ the work done by the system in this process. We can compute the difference Q₂ − W₂. Similarly, by taking the system from A to B by other different processes 3, 4, ..., we can compute Q₃ − W₃, Q₄ − W₄, and so on. In each case we find that

$\text Q_1 - \text W_1 = \text Q_2 - \text W_2 = \text Q_3 - \text W_3 = \text Q_4 - \text W_4 = \text{constant}$

From this we conclude that in taking a system from one state to another by different processes, the heat transferred Q and the work done W are different, but their difference Q − W is the same for all the processes. This means that the difference Q − W for two given states of a system depends only upon those states; it does not depend upon the path adopted by the system between the two states.

The quantity Q − W is defined as the 'change in the internal energy' of the system and is denoted by ΔU.

Thus, the internal energy U of a thermodynamic system is a characteristic property of the state of the system; it does not matter how that state has been obtained. U is a unique function, because it depends only upon the state of the system. If a system starting from one state passes through different states and finally returns to its initial state, then the change in its internal energy will be zero.

First law of thermodynamics : If an amount of heat Q is given to a system, a part of it will be used in increasing the internal energy (ΔU) of the system and the rest in doing work (W) by the system. Thus

$\text Q = \Delta \text U + \text W \qquad \text{or} \qquad \Delta \text U = \text Q - \text W$

Types of motion in the molecules of a substance : There are three types of motion in the molecules of a substance :

(i) Translational motion : The molecules move from one place to another in straight lines between collisions. All molecules — of monoatomic, diatomic and polyatomic gases — possess translational motion.

(ii) Rotational motion : The molecules rotate about an axis passing through their centre of mass. Only the molecules having two or more atoms, that is, diatomic and polyatomic molecules, possess rotational motion.

(iii) Vibrational motion : The atoms within a molecule vibrate about their mean positions, as if they were joined by a spring. This motion also occurs only in molecules having two or more atoms, and becomes significant at higher temperatures.

Which motion represents the temperature : Only the energy of the translational motion of the molecules is represented by temperature. The other forms of energy, such as the rotational and vibrational energies and the intermolecular potential energy, are not represented by temperature. This is why, at absolute zero, only the translational motion of the molecules ceases, while the other forms of molecular energy do not become zero.

Why the internal energy of a gas depends only upon the temperature : In an 'ideal' gas the molecules are so far apart that the intermolecular forces of attraction between them are negligible. Hence the mutual potential energy of the molecules is zero, and the internal energy of the gas is wholly the kinetic energy of its molecules. This kinetic energy is directly proportional to the absolute temperature of the gas.

Hence the internal energy of an ideal gas depends only upon its temperature, U = f(T), and not upon its pressure or volume.

For real gases, however, the intermolecular forces are appreciable. When the volume changes at a constant temperature, the distance between the molecules changes and so does their potential energy. Hence for a real gas the internal energy depends upon both the temperature and the volume, U = f(T, V).

The first law of thermodynamics states that if an amount of heat Q is given to a system, a part of it is used in increasing the internal energy (ΔU) of the system and the rest in doing work (W) by the system,

$\Delta \text U = \text Q - \text W$

(i) For an isovolumetric (isochoric) process :

An isovolumetric process is one which takes place at a constant volume. Since there is no change in the volume of the system,

$\Delta \text V = 0$

Therefore the work done by the system is

$\text W = \text P \times \Delta \text V = \text P \times 0 = 0$

Substituting W = 0 in the first law of thermodynamics,

$\Delta \text U = \text Q - 0$

$\Delta \text U = \text Q$

Hence, in an isovolumetric process the change in the internal energy of the system is equal to the heat given to or the heat taken from the system. If heat is given to the system, Q is positive and the internal energy increases; if heat is taken from the system, Q is negative and the internal energy decreases.

(ii) For an adiabatic process :

An adiabatic process is one in which heat neither enters the system nor leaves it, that is,

$\text Q = 0$

Substituting Q = 0 in the first law of thermodynamics,

$\Delta \text U = 0 - \text W$

$\Delta \text U = -\text W$

Hence, in an adiabatic process the change in the internal energy of the system is equal in magnitude to the work done by the system or the work done on the system.

If work is done on the system, W is negative and

$\Delta \text U = -(-\text W) = \text W$

so the internal energy of the system increases. If work is done by the system, W is positive and

$\Delta \text U = -(+\text W) = -\text W$

so the internal energy of the system decreases.

Equation for the first law of thermodynamics : If an amount of heat Q is given to a system, a part of it will be used in increasing the internal energy (ΔU) of the system and the rest in doing work (W) by the system. Thus

$\text Q = \Delta \text U + \text W \qquad \text{or} \qquad \Delta \text U = \text Q - \text W$

In the differential form, for an infinitesimally small heat dQ given to the system,

$\text{dQ} = \text{dU} + \text{dW} = \text{dU} + \text P\text{dV}$

Proof that C_p > C_v :

Let us consider 1 mole of an ideal gas in equilibrium at a pressure P, volume V and absolute temperature T. Let C_v and C_p be the molar specific heats of the gas at constant volume and at constant pressure respectively.

Heating at constant volume : Let the gas be heated at constant volume so that its temperature is raised by an infinitesimal amount dT. The heat supplied will be C_v dT. As the volume remains constant, the gas cannot expand and the external work done is zero, dW = 0. From the first law of thermodynamics,

$\text C_v\text{dT} = \text{dU} \qquad \text{...(i)}$

Heating at constant pressure : Let the same gas now be heated at the constant pressure P, until the temperature is raised by the same amount dT. The heat supplied will be C_p dT. Now the gas would expand and do external work against the pressure P. If dV be the change in volume, the external work would be P dV. Hence, from the first law of thermodynamics,

$\text C_p\text{dT} = \text{dU} + \text P\text{dV} \qquad \text{...(ii)}$

Comparing equations (i) and (ii), we see that in heating at constant pressure the gas requires an extra amount of heat P dV, over and above the heat dU needed to raise its temperature, this extra heat being used in doing external work against the pressure in expansion. In heating at constant volume no such extra heat is needed.

Hence, for the same temperature-rise, more heat is required at constant pressure than at constant volume, that is, C_p > C_v.

Deduction of Mayer's formula :

The temperature-change is the same in both the processes. Since the internal energy U of an ideal gas depends only on the temperature, the change in internal energy dU is the same in both the processes. Then, eliminating dU from equations (i) and (ii), we get

$(\text C_p - \text C_v)\text{dT} = \text P\text{dV} \qquad \text{...(iii)}$

Now, the equation of state for 1 mole of an ideal gas is

$\text{PV} = \text{RT}$

where R is the universal gas constant. Differentiating it, keeping P constant, we get

$\text P\text{dV} = \text R\text{dT}$

Substituting this value of P dV in equation (iii),

$(\text C_p - \text C_v)\text{dT} = \text R\text{dT}$

$\text C_p - \text C_v = \text R$

This relation is known as Mayer's formula.

Given,

• Heat available, Q = 100 calories
• J = 4.2 J cal^-1

The maximum work is obtained when the whole of the heat is converted into work. If Q calorie of heat is equivalent to W joule of work, then

$\text W = \text{JQ}$

Substituting the values,

$$\begin{aligned}\text W = 4.2\ \text{J cal}^{-1} \times 100\ \text{cal} \\[1em] = 420\ \text J\end{aligned}$$

Hence, the maximum work that can be done by 100 calories of heat is 420 J.

Given,

• Latent heat of ice, L = 80 cal g^-1
• J = 4.18 J cal^-1

Converting the calorie into joule and the gram into kilogram,

$\text L = 80\ \frac{\text{cal}}{\text g} = 80 \times \frac{4.18\ \text J}{10^{-3}\ \text{kg}}$

$$\begin{aligned}= 80 \times 4.18 \times 10^{3}\ \text{J kg}^{-1} \\[1em] = 3.34 \times 10^{5}\ \text{J kg}^{-1}\end{aligned}$$

Hence, the latent heat of ice is 3.34 × 10⁵ J/kg.

Given,

• Specific heat of water, c = 1 cal (g-°C)^-1
• J = 4.2 J cal^-1

Converting the calorie into joule and the gram into kilogram,

$\text c = 1\ \frac{\text{cal}}{\text g\text{-}°\text C} = \frac{4.2\ \text J}{10^{-3}\ \text{kg-}°\text C}$

$= 4.2 \times 10^{3}\ \text J\ (\text{kg-}°\text C)^{-1}$

Hence, the specific heat of water is 4.2 × 10³ J/(kg-°C).

Given,

• Mass of the car, m = 800 kg
• Heat produced, Q = 41.8 kcal
• J = 4.18 × 10³ J kcal^-1

When the brakes are applied, the whole of the kinetic energy of the car is converted into heat. Hence

$\frac{1}{2}\text{mv}^2 = \text{JQ}$

Substituting the values,

$\frac{1}{2} \times 800 \times \text v^2 = (4.18 \times 10^{3}\ \text{J kcal}^{-1}) \times (41.8\ \text{kcal})$

$400\text v^2 = 1.747 \times 10^{5}$

$\text v^2 = \frac{1.747 \times 10^{5}}{400} = 436.8$

$\text v = \sqrt{436.8} = 20.9\ \text{m/s}$

Hence, the speed of the car before applying the brakes was 20.9 m/s.

Given,

• Increase in volume, ΔV = 0.25 m³
• Constant pressure, P = 10³ N/m²

Since the pressure remains constant, the work done by the gas is

$\text W = \text P \times \Delta \text V$

Substituting the values,

$$\begin{aligned}\text W = (10^{3}\ \text{N/m}^2) \times (0.25\ \text m^3) \\[1em] = 250\ \text{N m} = 250\ \text J\end{aligned}$$

Hence, the work done is 250 J.

Given,

• Initial volume, V₁ = 75 L = 75 × 10^-3 m³
• Final volume, V₂ = 125 L = 125 × 10^-3 m³
• Constant pressure, P = 4 atmosphere = 4 × (1.0 × 10⁵) = 4.0 × 10⁵ N/m²

Since the pressure remains constant, the work done by the gas is

$\text W = \text P(\text V_2 - \text V_1)$

The increase in volume is

$\Delta \text V = (125 - 75) \times 10^{-3} = 50 \times 10^{-3}\ \text m^3$

Substituting the values,

$\text W = (4.0 \times 10^{5}\ \text{N/m}^2) \times (50 \times 10^{-3}\ \text m^3)$

$$\begin{aligned}= 4.0 \times 10^{5} \times 5.0 \times 10^{-2} \\[1em] = 2.0 \times 10^{4}\ \text J\end{aligned}$$

Hence, the work done by the gas during this expansion is 2.0 × 10⁴ J.

Given, from the graph,

• Constant pressure along AB, P = 3 N/m²
• Initial volume at A, V₁ = 1 m³
• Final volume at B, V₂ = 4 m³

Along the path AB the pressure remains constant, so the process is isobaric and the work done is

$\text W = \text P \times \Delta \text V = \text P(\text V_2 - \text V_1)$

Substituting the values,

$\text W = (3\ \text{N/m}^2) \times (4 - 1)\ \text m^3$

$= 3 \times 3 = 9\ \text{N m} = 9\ \text J$

Hence, 9 J of work will have to be done in taking the gas from the state A to the state B.

Given, from the graph,

• At A : P = 4 × 10⁴ N/m², V = 10 m³
• At B : P = 8 × 10⁴ N/m², V = 10 m³
• At C : P = 8 × 10⁴ N/m², V = 20 m³
• At D : P = 4 × 10⁴ N/m², V = 20 m³

(i) Work done by the gas in the process ABC :

Along A → B the volume remains constant at 10 m³, so the process is isochoric and

$\text W_{\text{AB}} = \text P \times \Delta \text V = 0$

Along B → C the pressure remains constant at 8 × 10⁴ N/m² while the volume increases from 10 m³ to 20 m³, so

$\text W_{\text{BC}} = \text P(\text V_{\text C} - \text V_{\text B}) = (8 \times 10^{4}) \times (20 - 10)$

$= 8 \times 10^{4} \times 10 = 8 \times 10^{5}\ \text J$

Therefore

$\text W_{\text{ABC}} = 0 + 8 \times 10^{5} = 8 \times 10^{5}\ \text J$

(ii) Work done by the gas in the process ADC :

Along A → D the pressure remains constant at 4 × 10⁴ N/m² while the volume increases from 10 m³ to 20 m³, so

$\text W_{\text{AD}} = (4 \times 10^{4}) \times (20 - 10) = 4 \times 10^{5}\ \text J$

Along D → C the volume remains constant at 20 m³, so

$\text W_{\text{DC}} = 0$

Therefore

$\text W_{\text{ADC}} = 4 \times 10^{5} + 0 = 4 \times 10^{5}\ \text J$

Hence, the work done by the gas in the process ABC is 8 × 10⁵ J and in the process ADC is 4 × 10⁵ J.

The work done is different along the two paths, although the initial and the final states are the same. This shows that the work done by a system depends upon the path adopted.

Given, from the graph,

• At A : P = 14 N/m², V = 10 L = 10 × 10^-3 m³
• At B : P = 14 N/m², V = 30 L = 30 × 10^-3 m³
• At C : P = 8 N/m², V = 30 L = 30 × 10^-3 m³
• At D : P = 8 N/m², V = 10 L = 10 × 10^-3 m³

Process A → B : The pressure is constant at 14 N/m² and the volume increases,

$\text W_{\text{AB}} = \text P\Delta \text V = 14 \times (30 - 10) \times 10^{-3}$

$= 14 \times 20 \times 10^{-3} = 0.28\ \text J$

Process B → C : The volume remains constant at 30 L, so

$\text W_{\text{BC}} = 0$

Process C → D : The pressure is constant at 8 N/m² and the volume decreases,

$\text W_{\text{CD}} = \text P\Delta \text V = 8 \times (10 - 30) \times 10^{-3}$

$= 8 \times (-20) \times 10^{-3} = -0.16\ \text J$

Process D → A : The volume remains constant at 10 L, so

$\text W_{\text{DA}} = 0$

Work done in the complete cycle :

$\text W = \text W_{\text{AB}} + \text W_{\text{BC}} + \text W_{\text{CD}} + \text W_{\text{DA}}$

$= 0.28 + 0 - 0.16 + 0 = 0.12\ \text J$

Hence, the work done in A → B is 0.28 J, in B → C is zero, in C → D is − 0.16 J, in D → A is zero, and in the complete cycle it is 0.12 J. Since the cycle is traced clockwise, this work is done by the gas.

Given, from the graph,

• At A : P = 10 N/m², V = 50 m³
• At B : P = 10 N/m², V = 10 m³
• At C : P = 40 N/m², V = 50 m³

(i) Work done in carrying the gas from A to B :

Along A → B the pressure remains constant at 10 N/m² while the volume decreases from 50 m³ to 10 m³. Hence

$\text W_{\text{AB}} = \text P(\text V_{\text B} - \text V_{\text A}) = 10 \times (10 - 50)$

$= 10 \times (-40) = -400\ \text J$

The negative sign shows that the volume decreases, so the work is done on the gas.

(ii) Work done in carrying the gas from B to C :

The path BC is a straight line from B (10 m³, 10 N/m²) to C (50 m³, 40 N/m²). The work done is the area of the trapezium under this line,

$\text W_{\text{BC}} = \frac{1}{2}(\text P_{\text B} + \text P_{\text C})(\text V_{\text C} - \text V_{\text B})$

$= \frac{1}{2}(10 + 40)(50 - 10) = \frac{1}{2} \times 50 \times 40$

$= 1000\ \text J$

Since the volume increases, the work is done by the gas.

Hence, 400 J of work is done on the gas from A to B, and 1000 J of work is done by the gas from B to C.

Given, from the graph,

• At A : P = 40 N m^-2, V = 5 m³
• At B : P = 10 N m^-2, V = 5 m³
• At C : P = 10 N m^-2, V = 20 m³

Process A → B : The volume remains constant at 5 m³, so the process is isochoric and

$\text W_{\text{AB}} = \text P \times \Delta \text V = 0$

Process B → C : The pressure remains constant at 10 N m^-2 while the volume increases from 5 m³ to 20 m³,

$\text W_{\text{BC}} = \text P(\text V_{\text C} - \text V_{\text B}) = 10 \times (20 - 5)$

$= 10 \times 15 = 150\ \text J$

Since the volume increases, this work is done by the gas.

Process C → A : The path CA is a straight line from C (20 m³, 10 N m^-2) to A (5 m³, 40 N m^-2). The magnitude of the work is the area of the trapezium under this line,

$|\text W_{\text{CA}}| = \frac{1}{2}(\text P_{\text C} + \text P_{\text A})(\text V_{\text C} - \text V_{\text A})$

$= \frac{1}{2}(10 + 40)(20 - 5) = \frac{1}{2} \times 50 \times 15 = 375\ \text J$

Since the volume decreases, the work is done on the gas, so W_CA = − 375 J.

Work done in the complete cycle ABCA :

$\text W = \text W_{\text{AB}} + \text W_{\text{BC}} + \text W_{\text{CA}} = 0 + 150 - 375$

$= -225\ \text J$

Hence, the work done in A → B is zero, in B → C it is 150 J by the gas, in C → A it is 375 J on the gas, and in the complete cycle ABCA it is 225 J on the gas. The negative sign, and the fact that the cycle is traced anticlockwise, both show that the net work is done on the gas.

Given,

• Initial volume, V₁ = 2.0 L = 2.0 × 10^-3 m³
• Final volume, V₂ = 2.5 L = 2.5 × 10^-3 m³
• Heat given to the gas, Q = 300 J
• Atmospheric pressure, P = 1.0 × 10⁵ N/m²

The volume increases at the same (atmospheric) pressure, so the process is isobaric and the work done by the gas is

$\text W = \text P(\text V_2 - \text V_1)$

$= (1.0 \times 10^{5}) \times (2.5 - 2.0) \times 10^{-3}$

$= (1.0 \times 10^{5}) \times (0.5 \times 10^{-3}) = 50\ \text J$

By the first law of thermodynamics, the change in the internal energy of the gas is

$\Delta \text U = \text Q - \text W = 300 - 50$

$= 250\ \text J$

Hence, the internal energy of the gas increases by 250 J.

Given,

• Mass of the bullet, m = 20 g = 20 × 10^-3 kg
• Velocity of the bullet, v = 100 m/s
• Rise in temperature, ΔT = 25°C
• 50% of the kinetic energy raises the temperature of the bullet

The kinetic energy of the bullet is

$\text K = \frac{1}{2}\text{mv}^2 = \frac{1}{2} \times (20 \times 10^{-3}) \times (100)^2$

$= \frac{1}{2} \times 0.02 \times 10^{4} = 100\ \text J$

(i) Increase in the internal energy of the bullet :

Only 50% of this kinetic energy is used in raising the temperature of the bullet, and this appears as the increase in its internal energy,

$$\begin{aligned}\Delta \text U = \frac{50}{100} \times 100 \\[1em] = 50\ \text J\end{aligned}$$

(ii) Specific heat of the material of the bullet :

The heat used in raising the temperature is Q = m c ΔT, so

$\text c = \frac{\Delta \text U}{\text m \times \Delta \text T} = \frac{50\ \text J}{(20 \times 10^{-3}\ \text{kg}) \times (25\ °\text C)}$

$= \frac{50}{0.5} = 100\ \text J\ (\text{kg-}°\text C)^{-1}$

Expressing it in calorie, using J = 4.2 J cal^-1,

$\text c = \frac{100}{4.2 \times 10^{3}} = 0.0238\ \text{cal}\ (\text g\text{-}°\text C)^{-1}$

Hence, the increase in the internal energy of the bullet is 50 J and the specific heat of its material is 100 J/(kg-°C) or 0.0238 cal/(g-°C).

Given,

• Speed of the bullet, v = 150 m/s
• 80% of the kinetic energy remains in the bullet
• Specific heat of lead, c = 30 cal (kg-°C)^-1
• J = 4.2 J cal^-1

Let m be the mass of the bullet. Its kinetic energy is

$\text K = \frac{1}{2}\text{mv}^2 = \frac{1}{2}\text m(150)^2 = 11250\text m\ \text J$

Only 80% of this energy remains in the bullet as heat, so the heat retained is

$\text Q = \frac{80}{100} \times 11250\text m = 9000\text m\ \text J$

Expressing this heat in calorie,

$\text Q = \frac{9000\text m}{4.2}\ \text{cal}$

This heat raises the temperature of the bullet, so

$\text Q = \text m\text c\Delta \text T$

$\frac{9000\text m}{4.2} = \text m \times 30 \times \Delta \text T$

The mass m cancels from both the sides,

$\Delta \text T = \frac{9000}{4.2 \times 30} = \frac{9000}{126}$

$= 71.43\ °\text C$

Hence, the rise in the temperature of the bullet is 71.43°C.

Given,

• Initial temperature of the bullet = 27°C
• Melting point of lead = 327°C
• Specific heat of lead, c = 0.03 cal (g-°C)^-1
• Latent heat of melting of lead, L = 6 cal g^-1
• 25% of the heat produced is absorbed by the obstacle, so 75% is retained by the bullet
• J = 4.2 J cal^-1

Let m gram be the mass of the bullet and v m/s its velocity.

Heat needed by the bullet : The bullet must first be heated from 27°C to its melting point 327°C, and then just melted. Hence

$\text Q = \text{mc}\Delta \text T + \text{mL}$

$= \text m \times 0.03 \times (327 - 27) + \text m \times 6$

$= \text m \times 0.03 \times 300 + 6\text m = 9\text m + 6\text m = 15\text m\ \text{cal}$

Heat produced by the bullet : The kinetic energy of the bullet is converted into heat, of which only 75% is retained by the bullet,

$\text Q = \frac{75}{100} \times \frac{1}{2}\frac{\text{mv}^2}{\text J}$

Here m must be in kilogram for the energy to be in joule, so m gram = m × 10^-3 kg. Equating the two expressions for Q,

$15\text m = \frac{75}{100} \times \frac{1}{2} \times \frac{(\text m \times 10^{-3})\text v^2}{4.2}$

The mass m cancels from both the sides,

$15 = 0.75 \times \frac{10^{-3}\text v^2}{2 \times 4.2}$

$15 = \frac{0.75 \times 10^{-3}\text v^2}{8.4}$

$\text v^2 = \frac{15 \times 8.4}{0.75 \times 10^{-3}} = \frac{126}{0.75 \times 10^{-3}}$

$= 1.68 \times 10^{5}$

$\text v = \sqrt{1.68 \times 10^{5}} = 410\ \text{m/s}$

Hence, the velocity of the bullet is 410 m/s.

Given,

• Mass of the sphere, m = 2 kg
• Height of fall, h = 3 m
• g = 9.8 m/s²
• 1 cal = 4.2 J

The whole of the potential energy of the sphere is converted into heat. The loss in potential energy is

$\text U = \text{mgh} = 2 \times 9.8 \times 3$

$= 58.8\ \text J$

Converting this energy into calorie,

$\text Q = \frac{58.8\ \text J}{4.2\ \text{J cal}^{-1}}$

$= 14\ \text{cal}$

Hence, 14 cal of heat will be produced.

Given,

• Mass of the hammer, M = 5 kg
• Velocity of the hammer, v = 1 m/s
• Mass of the lead, m = 300 g = 0.3 kg
• Initial temperature of the lead = 27°C
• Melting point of lead = 327°C
• Specific heat of lead, c = 0.03 kcal (kg-°C)^-1
• Latent heat of lead, L = 6 kcal kg^-1
• J = 4.18 × 10³ J kcal^-1

Heat required to melt the lead : The lead must first be heated from 27°C to 327°C and then melted,

$\text Q = \text{mc}\Delta \text T + \text{mL}$

$= 0.3 \times 0.03 \times (327 - 27) + 0.3 \times 6$

$= 0.3 \times 0.03 \times 300 + 1.8 = 2.7 + 1.8 = 4.5\ \text{kcal}$

Expressing this heat in joule,

$\text Q = 4.5 \times 4.18 \times 10^{3} = 1.881 \times 10^{4}\ \text J$

Energy delivered in one stroke : The kinetic energy of the falling hammer is

$\text K = \frac{1}{2}\text{Mv}^2 = \frac{1}{2} \times 5 \times (1)^2 = 2.5\ \text J$

Number of strokes required :

$\text n = \frac{\text Q}{\text K} = \frac{1.881 \times 10^{4}\ \text J}{2.5\ \text J}$

$= 7524$

Hence, 7524 strokes will be required for melting the lead.

Given,

• Mass of the lead ball, m = 2.0 kg
• Height of fall, h = 30.0 m
• Specific heat of lead, c = 125 J (kg-°C)^-1
• g = 9.8 m/s²

The whole of the kinetic energy of the ball is converted into heat and remains in it. On striking the ground the kinetic energy of the ball equals the potential energy lost,

$\text Q = \text{mgh} = 2.0 \times 9.8 \times 30.0$

$= 588\ \text J$

Rise in temperature : This heat raises the temperature of the ball, so

$\text Q = \text{mc}\Delta \text T$

$\Delta \text T = \frac{\text Q}{\text{mc}} = \frac{588}{2.0 \times 125} = \frac{588}{250}$

$= 2.35\ °\text C$

Change in internal energy : No heat is given to the ball from outside and the whole of the energy remains in the ball. Hence the internal energy of the ball increases by the whole of this amount,

$\Delta \text U = 588\ \text J$

Hence, the rise in the temperature of the ball is 2.35°C and its internal energy increases by 588 J.

Given,

• Height of fall, h = 20 m
• Rate of fall of water, m = 100 kg per second
• g = 9.8 m/s²
• J = 4.18 × 10³ J kcal^-1

The whole of the energy is converted into heat. The potential energy lost by the water per second is

$\text U = \text{mgh} = 100 \times 9.8 \times 20$

$= 19600\ \text J\ \text{per second}$

Converting this energy into kilocalorie,

$\text Q = \frac{19600\ \text J}{4.18 \times 10^{3}\ \text{J kcal}^{-1}}$

$= 4.7\ \text{kcal per second}$

Hence, 4.7 kcal (that is, 4.7 × 10³ calorie) of heat will be produced per second.