Kinetic Theory

Given,

• Diameter of an oxygen molecule, d = 3 Å = 3 × 10^-10 m
• Radius of an oxygen molecule, r = $\frac{\text d}{2}$ = 1.5 × 10^-10 m

At normal temperature and pressure, the volume of 1 mole of a gas is 22.4 L and this quantity of the gas contains 6.02 × 10²³ molecules (Avogadro's number).

Therefore, the actual volume occupied by the gas is

$\text V = 22.4\ \text L = 22.4 \times 10^{-3}\ \text m^3$

Considering an oxygen molecule to be a sphere of radius r, the volume of one molecule is $\frac{4}{3}\pi \text r^3$. Hence the molecular volume, that is, the volume occupied by the 6.02 × 10²³ molecules themselves, is

$\text V_{molecular} = \frac{4}{3}\pi \text r^3 \times 6.02 \times 10^{23}$

Substituting the values,

$$\begin{aligned}\text V_{molecular} = \frac{4}{3} \times 3.14 \times (1.5 \times 10^{-10})^3 \times 6.02 \times 10^{23} \\[1em] = 8.5 \times 10^{-6}\ \text m^3\end{aligned}$$

The required fraction is

$$\begin{aligned}\frac{\text V_{molecular}}{\text V} = \frac{8.5 \times 10^{-6}\ \text m^3}{22.4 \times 10^{-3}\ \text m^3} \\[1em] = 3.8 \times 10^{-4}\end{aligned}$$

Hence, the fraction of the molecular volume to the actual volume occupied by oxygen gas at STP is 3.8 × 10^-4.

The molecular volume is thus about four ten-thousandths of the volume of the gas, which justifies the postulate of the kinetic theory that the actual volume of the molecules is negligible compared to the total volume of the container.

Given,

• Number of moles, μ = 1 mol
• Pressure, P = 1 atm = 1.01 × 10⁵ N m^-2
• Temperature, T = 0°C = 273 K
• Gas constant, R = 8.31 J mol^-1 K^-1

The ideal gas equation for μ moles of a gas is

$\text{PV} = \mu \text{RT} \quad \Rightarrow \quad \text V = \frac{\mu \text{RT}}{\text P}$

For the molar volume we put μ = 1 mol. Substituting the values,

$$\begin{aligned}\text V = \frac{1 \times 8.31 \times 273}{1.01 \times 10^5} \\[1em] = \frac{2268.63}{1.01 \times 10^5} \\[1em] = 22.4 \times 10^{-3}\ \text m^3\end{aligned}$$

Since 1 litre = 10^-3 m³,

$\text V = 22.4\ \text L$

Hence, the molar volume of an ideal gas at STP is 22.4 L.

Given,

• Mass of oxygen gas, m = 1.00 × 10^-3 kg
• Molecular mass of oxygen, M = 32.0 g mol^-1 = 32.0 × 10^-3 kg mol^-1
• Molecular mass of hydrogen = 2.02 g mol^-1 = 2.02 × 10^-3 kg mol^-1
• Gas constant, R = 8.31 J mol^-1 K^-1

(i) The dotted plot is a straight line parallel to the P-axis, which means that the value of $\frac{\text{PV}}{\text T}$ stays the same at all pressures. For μ moles of a gas, PV = μRT gives $\frac{\text{PV}}{\text T} = \mu \text R$, a constant. Hence the dotted line AB signifies ideal gas behaviour.

(ii) The curve drawn at the temperature T₁ lies closer to the dotted line AB than the curve drawn at T₂. A real gas approaches ideal gas behaviour at higher temperatures, so the curve nearer to the ideal line corresponds to the higher temperature. Hence T₁ > T₂.

(iii) At the point A the two curves meet the dotted line, that is, oxygen behaves like an ideal gas there. For μ moles,

$\text{PV} = \mu \text{RT} \quad \Rightarrow \quad \frac{\text{PV}}{\text T} = \mu \text R$

The number of moles in 1.00 × 10^-3 kg of oxygen is

$\mu = \frac{\text m}{\text M} = \frac{1.00 \times 10^{-3}\ \text{kg}}{32.0 \times 10^{-3}\ \text{kg mol}^{-1}} = \frac{1}{32.0}\ \text{mol}$

Therefore,

$$\begin{aligned}\frac{\text{PV}}{\text T} = \left(\frac{1}{32.0}\right) \times 8.31 \\[1em] = 0.26\ \text{J K}^{-1}\end{aligned}$$

(iv) The value of $\frac{\text{PV}}{\text T}$ at the point A would not be the same for 1.00 × 10^-3 kg of hydrogen, because the molecular mass of hydrogen differs from that of oxygen and hence the same mass contains a different number of moles.

Let m' be the mass of hydrogen which gives the same value 0.26 J K^-1. Then

$\frac{\text{PV}}{\text T} = \frac{\text m'}{\text M_{\text H_2}} \times \text R$

$$\begin{aligned}\text m' = \frac{\left(\frac{\text{PV}}{\text T}\right) \times \text M_{\text H_2}}{\text R} = \frac{0.26 \times (2.02 \times 10^{-3})}{8.31} \\[1em] = 6.3 \times 10^{-5}\ \text{kg}\end{aligned}$$

Hence, a mass of 6.3 × 10^-5 kg of hydrogen would yield the same value of PV/T in the low pressure, high temperature region of the plot.

Given,

• Volume of the cylinder, V = 30 L = 30 × 10^-3 m³
• Initial gauge pressure = 15 atm
• Initial temperature, T₁ = 27°C = 27 + 273 = 300 K
• Final gauge pressure = 11 atm
• Final temperature, T₂ = 17°C = 17 + 273 = 290 K
• Atmospheric pressure = 1 atm = 1.01 × 10⁵ N m^-2
• Gas constant, R = 8.3 J mol^-1 K^-1
• Molecular mass of O₂, M = 32

The ideal gas equation PV = μRT holds for the absolute pressure of the gas, whereas a pressure gauge reads the pressure above the atmospheric pressure. Hence the pressures must first be converted,

$\text P_{abs} = \text P_{gauge} + \text P_{atm}$

$\text P_1 = (15 + 1)\ \text{atm} = 16 \times 1.01 \times 10^5\ \text N\thinspace\text m^{-2}$

$\text P_2 = (11 + 1)\ \text{atm} = 12 \times 1.01 \times 10^5\ \text N\thinspace\text m^{-2}$

The volume V of the cylinder remains unchanged. Let μ₁ be the number of moles of oxygen initially present and μ₂ the number of moles left after the withdrawal. Writing the gas equation for the two states,

$\text P_1 \text V = \mu_1 \text{RT}_1 \quad \text{and} \quad \text P_2 \text V = \mu_2 \text{RT}_2$

The number of moles initially present is

$$\begin{aligned}\mu_1 = \frac{\text P_1 \text V}{\text{RT}_1} = \frac{(16 \times 1.01 \times 10^5) \times (30 \times 10^{-3})}{8.3 \times 300} \\[1em] = \frac{4.848 \times 10^4}{2490} = 19.47\end{aligned}$$

The number of moles left is

$$\begin{aligned}\mu_2 = \frac{\text P_2 \text V}{\text{RT}_2} = \frac{(12 \times 1.01 \times 10^5) \times (30 \times 10^{-3})}{8.3 \times 290} \\[1em] = \frac{3.636 \times 10^4}{2407} = 15.11\end{aligned}$$

Therefore the number of moles of oxygen taken out of the cylinder is

$\mu_1 - \mu_2 = 19.47 - 15.11 = 4.36$

The mass of oxygen taken out is

$$\begin{aligned}\text m = (\mu_1 - \mu_2)\text M = 4.36 \times 32 \\[1em] = 139.6\ \text g \approx 140\ \text g\end{aligned}$$

Hence, the mass of oxygen taken out of the cylinder is nearly 140 g.

Note: The textbook substitutes the gauge pressures 15 atm and 11 atm directly into PV = μRT and obtains 141 g. Since the gas equation is valid only for the absolute pressure, the atmospheric pressure of 1 atm must be added to each gauge reading, which gives 140 g.*

Given,

• Volume of the bubble at the bottom, V₁ = 1.0 cm³ = 1.0 × 10^-6 m³
• Depth of the lake, h = 40 m
• Temperature at the bottom, T₁ = 12°C = 12 + 273 = 285 K
• Temperature at the surface, T₂ = 35°C = 35 + 273 = 308 K
• Atmospheric pressure, P = 1.0 × 10⁵ Pa
• Density of water, ρ = 10³ kg m^-3
• g = 9.8 N kg^-1

At the bottom of the lake the bubble is under the atmospheric pressure together with the pressure of the water column above it. Hence the pressure at the bottom is

$$\begin{aligned}\text P_1 = \text P + \text h\rho \text g \\[1em] = (1.0 \times 10^5) + (40 \times 10^3 \times 9.8) \\[1em] = (1.0 \times 10^5) + (3.92 \times 10^5) \\[1em] = 4.92 \times 10^5\ \text{Pa}\end{aligned}$$

At the surface the bubble is under the atmospheric pressure alone, so

$\text P_2 = 1.0 \times 10^5\ \text{Pa}$

The air enclosed in the bubble obeys the gas law,

$\frac{\text P_1 \text V_1}{\text T_1} = \frac{\text P_2 \text V_2}{\text T_2} \quad \Rightarrow \quad \text V_2 = \frac{\text P_1 \text V_1 \text T_2}{\text T_1 \text P_2}$

Substituting the values,

$$\begin{aligned}\text V_2 = \frac{(4.92 \times 10^5) \times (1.0 \times 10^{-6}) \times 308}{285 \times (1.0 \times 10^5)} \\[1em] = 5.3 \times 10^{-6}\ \text m^3\end{aligned}$$

$\text V_2 = 5.3\ \text{cm}^3$

Hence, the bubble grows to a volume of 5.3 cm³ on reaching the surface.

Given,

• Volume of the room, V = 25.0 m³
• Temperature, T = 27°C = 27 + 273 = 300 K
• Pressure, P = 1 atm = 1.01 × 10⁵ N m^-2
• Boltzmann constant, k = 1.38 × 10^-23 J K^-1

For a given mass of a gas containing n molecules, the gas equation in terms of the Boltzmann constant is

$\text{PV} = \text{nkT} \quad \Rightarrow \quad \text n = \frac{\text{PV}}{\text{kT}}$

This relation involves no property of the particular gas, so it holds for the mixture of oxygen, nitrogen, water vapour and the other constituents of air taken together.

Substituting the values,

$$\begin{aligned}\text n = \frac{(1.01 \times 10^5) \times 25.0}{(1.38 \times 10^{-23}) \times 300} \\[1em] = \frac{2.525 \times 10^6}{4.14 \times 10^{-21}} \\[1em] = 6.10 \times 10^{26}\end{aligned}$$

Hence, the total number of air molecules in the room is 6.10 × 10²⁶.

Given,

• Boltzmann constant, k = 1.38 × 10^-23 J K^-1

Helium is a monoatomic gas, and according to the kinetic theory the average thermal (kinetic) energy of a molecule of a gas at absolute temperature T is

$\overline{\text E} = \frac{3}{2}\text{kT}$

(i) At room temperature, T = 27°C = 300 K :

$$\begin{aligned}\overline{\text E} = \frac{3}{2} \times (1.38 \times 10^{-23}) \times 300 \\[1em] = 6.21 \times 10^{-21}\ \text J\end{aligned}$$

(ii) At the temperature on the surface of the sun, T = 6000 K :

$$\begin{aligned}\overline{\text E} = \frac{3}{2} \times (1.38 \times 10^{-23}) \times 6000 \\[1em] = 1.24 \times 10^{-19}\ \text J\end{aligned}$$

(iii) At the core temperature of a star, T = 10⁷ K :

$$\begin{aligned}\overline{\text E} = \frac{3}{2} \times (1.38 \times 10^{-23}) \times 10^7 \\[1em] = 2.07 \times 10^{-16}\ \text J\end{aligned}$$

Hence, the average thermal energy of a helium atom is 6.21 × 10^-21 J, 1.24 × 10^-19 J and 2.07 × 10^-16 J at the three temperatures respectively.

Number of molecules : For n molecules of a gas, the ideal gas equation is

$\text{PV} = \text{nkT} \quad \Rightarrow \quad \text n = \frac{\text{PV}}{\text{kT}}$

The three vessels are of equal capacity and the gases are at the same temperature and pressure, so P, V and T are the same for all three, and k is the Boltzmann constant. Hence the three vessels contain equal numbers of their respective molecules. This is Avogadro's law.

Root-mean-square speed : The rms speed of the molecules of a gas of molecular mass M at absolute temperature T is

$\text v_{rms} = \sqrt{\frac{3\text{RT}}{\text M}} \quad \Rightarrow \quad \text v_{rms} \propto \sqrt{\frac{1}{\text M}}$

Since the three gases have different molecular masses, the root-mean-square speed is not the same in the three cases. The rms speed is the largest for the gas of the smallest molecular mass.

Of the three gases, neon has the least molecular mass, chlorine has a larger one and uranium hexafluoride the largest. Hence v_rms is the largest for neon.

Given,

• Atomic mass of argon, M_A = 39.9
• Atomic mass of helium, M_He = 4.0
• Temperature of helium, T_He = −20°C = −20 + 273 = 253 K
• Temperature of argon, T_A = ?

According to the kinetic theory, the rms speed of the atoms of a gas of molecular mass M at absolute temperature T is

$\text v_{rms} = \sqrt{\frac{3\text{RT}}{\text M}}$

For the two gases,

$(\text v_{rms})_{\text A} = \sqrt{\frac{3\text{RT}_{\text A}}{\text M_{\text A}}} \quad \text{and} \quad (\text v_{rms})_{\text{He}} = \sqrt{\frac{3\text{RT}_{\text{He}}}{\text M_{\text{He}}}}$

The two rms speeds are to be equal, so

$\sqrt{\frac{3\text{RT}_{\text A}}{\text M_{\text A}}} = \sqrt{\frac{3\text{RT}_{\text{He}}}{\text M_{\text{He}}}}$

Squaring both sides and cancelling 3R,

$\frac{\text T_{\text A}}{\text M_{\text A}} = \frac{\text T_{\text{He}}}{\text M_{\text{He}}} \quad \Rightarrow \quad \text T_{\text A} = \frac{\text T_{\text{He}} \times \text M_{\text A}}{\text M_{\text{He}}}$

Substituting the values,

$$\begin{aligned}\text T_{\text A} = \frac{253 \times 39.9}{4.0} \\[1em] = 2523.7\ \text K\end{aligned}$$

Hence, at 2523.7 K the rms speed of an argon atom equals the rms speed of a helium atom at −20°C.

Given,

• Pressure, P = 2.0 atm = 2 × 1.013 × 10⁵ = 2.026 × 10⁵ N m^-2
• Temperature, T = 17°C = 17 + 273 = 290 K
• Radius of a nitrogen molecule, r = 1.0 Å = 1.0 × 10^-10 m, so d = 2r = 2.0 × 10^-10 m
• Molecular mass of N₂, M = 28.0 u = 28 × 10^-3 kg mol^-1

Number of molecules per unit volume : For 1 mole of an ideal gas,

$$\begin{aligned}\text V = \frac{\text{RT}}{\text P} = \frac{8.31 \times 290}{2.026 \times 10^5} \\[1em] = 1.189 \times 10^{-2}\ \text m^3\end{aligned}$$

$$\begin{aligned}\text n = \frac{\text N}{\text V} = \frac{6.02 \times 10^{23}}{1.189 \times 10^{-2}} \\[1em] = 5.0 \times 10^{25}\ \text m^{-3}\end{aligned}$$

Mean free path : The mean free path of a molecule is

$\lambda = \frac{1}{\sqrt2\pi \text d^2 \text n} = \frac{1}{\sqrt2\pi (2\text r)^2 \text n}$

$$\begin{aligned}\lambda = \frac{1}{1.414 \times 3.14 \times (2.0 \times 10^{-10})^2 \times 5.0 \times 10^{25}} \\[1em] = 1.0 \times 10^{-7}\ \text m\end{aligned}$$

Root-mean-square speed :

$$\begin{aligned}\text v_{rms} = \sqrt{\frac{3\text{RT}}{\text M}} = \sqrt{\frac{3 \times 8.31 \times 290}{28 \times 10^{-3}}} \\[1em] = 5.1 \times 10^2\ \text{m s}^{-1}\end{aligned}$$

Collision frequency : The collision frequency is the number of collisions made by a molecule per second,

$$\begin{aligned}\nu = \frac{\text v_{rms}}{\lambda} = \frac{5.1 \times 10^2}{1.0 \times 10^{-7}} \\[1em] = 5.1 \times 10^9\ \text s^{-1}\end{aligned}$$

Collision time : The collision time is the time for which a molecule is actually in contact during a collision, that is, the time taken to cover a distance equal to its own diameter,

$$\begin{aligned}\text t_{collision} = \frac{\text d}{\text v_{rms}} = \frac{2.0 \times 10^{-10}}{5.1 \times 10^2} \\[1em] = 4 \times 10^{-13}\ \text s\end{aligned}$$

Time between two successive collisions :

$$\begin{aligned}\text t_{free} = \frac{\lambda}{\text v_{rms}} = \frac{1.0 \times 10^{-7}}{5.1 \times 10^2} \\[1em] = 2 \times 10^{-10}\ \text s\end{aligned}$$

Comparison :

$\frac{\text t_{free}}{\text t_{collision}} = \frac{2 \times 10^{-10}}{4 \times 10^{-13}} = 500$

Hence, the mean free path is 1.0 × 10^-7 m, the collision frequency is 5.1 × 10⁹ s^-1, and the time a molecule moves freely between two successive collisions is about 500 times the collision time.

Given,

• Height of the cylindrical vessel = 500 mm
• Steady height of the water column after opening the orifice = 200 mm
• Atmospheric pressure, P_atm = 1.0 × 10⁵ N/m²
• Density of water, ρ = 1000 kg/m³
• g = 10 m/s²

Let H be the initial height of the water column when the orifice is closed. The air enclosed above the water is then at the atmospheric pressure, since the vessel was open before it was sealed with the cap.

Initial state of the enclosed air :

$\text P_i = 1.0 \times 10^5\ \text N/\text m^2, \qquad \text V_i = \text A(500 - \text H)$

where A is the area of cross-section of the vessel.

Final state of the enclosed air : When the orifice is opened, some water flows out and the level becomes steady at 200 mm. The water column then stands still, so the pressure at the level of the orifice due to the enclosed air together with the water column must equal the atmospheric pressure. Hence the final pressure of the enclosed air is

$$\begin{aligned}\text P_f = \text P_{atm} - \rho \text{gh} \\[1em] = (1.0 \times 10^5) - (1000 \times 10 \times 0.2) \\[1em] = (1.0 \times 10^5) - (0.02 \times 10^5) \\[1em] = 0.98 \times 10^5\ \text N/\text m^2\end{aligned}$$

and its volume is

$\text V_f = \text A(500 - 200) = 300\text A$

Applying Boyle's law : The temperature stays constant, so

$\text P_i \text V_i = \text P_f \text V_f$

$(1.0 \times 10^5) \times \text A(500 - \text H) = (0.98 \times 10^5) \times (300\text A)$

$500 - \text H = 0.98 \times 300 = 294\ \text{mm}$

$\text H = 500 - 294 = 206\ \text{mm}$

Fall in the height of the water level :

$$\begin{aligned}= \text H - 200 = 206 - 200 \\[1em] = 6\ \text{mm}\end{aligned}$$

Hence, the fall in the height of the water level due to opening of the orifice is 6 mm.

Given,

• Lower compartment : 2 moles of an ideal monoatomic gas at 700 K, C_v = $\frac{3}{2}$ R, C_p = $\frac{5}{2}$ R
• Upper compartment : 2 moles of an ideal diatomic gas at 400 K, C_v = $\frac{5}{2}$ R, C_p = $\frac{7}{2}$ R

The container and the piston are perfectly insulating, so no heat is exchanged with the surroundings. The partition is thermally conducting, so heat flows from the hotter lower gas to the colder upper gas until both reach a common temperature T.

(i) 490 K

The partition is rigidly fixed, so the volume of the lower gas cannot change and the heat is given out by it at constant volume,

$\text Q_{lower} = \mu \text C_v \Delta \text T = 2 \times \frac{3}{2}\text R(700 - \text T) \qquad \ldots(\text i)$

The piston on the top is free to move, so the upper gas receives heat at constant pressure,

$\text Q_{upper} = \mu \text C_p \Delta \text T = 2 \times \frac{7}{2}\text R(\text T - 400) \qquad \ldots(\text{ii})$

Since the container is insulated, the heat given out by the lower gas is entirely absorbed by the upper gas. Equating (i) and (ii),

$3\text R(700 - \text T) = 7\text R(\text T - 400)$

$2100 - 3\text T = 7\text T - 2800$

$10\text T = 4900 \quad \Rightarrow \quad \text T = 490\ \text K$

Hence, the final temperature of the gases is 490 K.

(ii) -100 R

Now the partition is also free to move, so the pressure of the gas in both the compartments is the same and both the gases undergo their changes at constant pressure.

$\text Q_{lower} = 2 \times \frac{5}{2}\text R(700 - \text T) \qquad \ldots(\text i)$

$\text Q_{upper} = 2 \times \frac{7}{2}\text R(\text T - 400) \qquad \ldots(\text{ii})$

Equating (i) and (ii),

$5(700 - \text T) = 7(\text T - 400)$

$3500 - 5\text T = 7\text T - 2800$

$12\text T = 6300 \quad \Rightarrow \quad \text T = 525\ \text K$

The work done by a gas at constant pressure is W = μRΔT. For the upper gas,

$\text W_{upper} = 2\text R(525 - 400) = 250\text R$

For the lower gas,

$\text W_{lower} = 2\text R(525 - 700) = -350\text R$

The total work done by the gases is

$$\begin{aligned}\text W_{net} = \text W_{upper} + \text W_{lower} = 250\text R + (-350\text R) \\[1em] = -100\text R\end{aligned}$$

Hence, the total work done by the gases till they achieve equilibrium is −100 R.

8 : 9

Given,

• Ratio of the atomic masses, M₁ : M₂ = 2 : 3
• Ratio of the partial pressures, P₁ : P₂ = 4 : 3
• Both the gases are in the same vessel at a constant temperature

For μ moles of an ideal gas of molecular mass M and mass m,

$\text{PV} = \mu \text{RT} = \frac{\text m}{\text M}\text{RT}$

$\text {PM} = \frac{\text m}{\text V}\text{RT} = \rho \text{RT} \quad \Rightarrow \quad \rho = \frac{\text{PM}}{\text{RT}}$

Since the temperature T is the same for both the gases,

$\frac{\rho_1}{\rho_2} = \left(\frac{\text P_1}{\text P_2}\right)\left(\frac{\text M_1}{\text M_2}\right)$

Substituting the values,

$$\begin{aligned}\frac{\rho_1}{\rho_2} = \frac{4}{3} \times \frac{2}{3} \\[1em] = \frac{8}{9}\end{aligned}$$

Hence, the ratio of the densities of the two gases is 8 : 9.

3.16

Given,

• Atomic mass of helium, M_He = 4 amu
• Atomic mass of argon, M_Ar = 40 amu
• Temperature, T = 300 K (the same for both the gases)

The rms speed of the molecules of a gas of molecular mass M at absolute temperature T is

$\text v_{rms} = \sqrt{\frac{3\text{RT}}{\text M}}$

Both the gases are in the same container at the same temperature. Hence

$\frac{(\text v_{rms})_{\text{He}}}{(\text v_{rms})_{\text{Ar}}} = \frac{\sqrt{\frac{3\text{RT}}{\text M_{\text{He}}}}}{\sqrt{\frac{3\text{RT}}{\text M_{\text{Ar}}}}} = \sqrt{\frac{\text M_{\text{Ar}}}{\text M_{\text{He}}}}$

Substituting the values,

$$\begin{aligned}\frac{(\text v_{rms})_{\text{He}}}{(\text v_{rms})_{\text{Ar}}} = \sqrt{\frac{40}{4}} = \sqrt{10} \\[1em] = 3.16\end{aligned}$$

Hence, the required ratio of the rms speeds is 3.16.

The number of moles of the two gases does not enter the result, because the rms speed depends only on the temperature and the molecular mass of the gas.

$\frac{3}{\text T}$

Given,

• The gas expands such that PT² = constant = K

For 1 mole of an ideal gas,

$\text{PV} = \text{RT} \quad \Rightarrow \quad \text P = \frac{\text{RT}}{\text V}$

Substituting this value of P in PT² = K,

$\left(\frac{\text{RT}}{\text V}\right)\text T^2 = \text K \quad \Rightarrow \quad \text T^3 = \frac{\text{KV}}{\text R}$

Differentiating both sides with respect to T,

$3\text T^2\text{dT} = \frac{\text K}{\text R}\text{dV}$

$\frac{\text{dV}}{\text{dT}} = \frac{3\text T^2 \text R}{\text K}$

Putting back K = PT²,

$\frac{\text{dV}}{\text{dT}} = \frac{3\text T^2 \text R}{\text{PT}^2} = \frac{3\text R}{\text P}$

The coefficient of volume expansion of the gas is

$\gamma = \frac{1}{\text V}\frac{\text{dV}}{\text{dT}} = \frac{1}{\text V} \times \frac{3\text R}{\text P} = \frac{3\text R}{\text{PV}}$

Since PV = RT for 1 mole,

$\gamma = \frac{3\text R}{\text{RT}} = \frac{3}{\text T}$

Hence, the coefficient of volume expansion of the gas is $\frac{3}{\text T}$.

pressure is low and temperature is high

A real gas behaves like an ideal gas when the two chief properties assumed for an ideal gas are nearly satisfied, namely that the volume of the molecules is negligible and that there is no force of attraction between them.

At low pressure the volume of the gas is large, so the actual volume of the molecules is negligible in comparison with the volume of the gas.

At high temperature the molecules have large kinetic energy, so the effect of the intermolecular force on their motion is negligible.

Hence, at low pressure and high temperature the behaviour of a real gas is approximately ideal.

$\frac{\text n_1 \text T_1 + \text n_2 \text T_2 + \text n_3 \text T_3}{\text n_1 + \text n_2 + \text n_3}$

Given,

• Absolute temperatures of the three gases : T₁, T₂, T₃
• Number of molecules : n₁, n₂, n₃
• There is no loss of energy on mixing

Assumption: the three gases have the same number of degrees of freedom, or only the translational kinetic energy of the molecules is being considered.

According to the kinetic theory, the average translational kinetic energy of one molecule of a gas at absolute temperature T is $\frac{3}{2}\text{kT}$, where k is the Boltzmann constant. Hence, for a gas containing n molecules, the total translational kinetic energy is

$\text E = \frac{3}{2}\text{nkT}$

Before mixing, the total energy of the three gases is

$\frac{3}{2}\text n_1 \text{kT}_1 + \frac{3}{2}\text n_2 \text{kT}_2 + \frac{3}{2}\text n_3 \text{kT}_3$

After mixing, the mixture contains (n₁ + n₂ + n₃) molecules at the common temperature T, so its energy is

$\frac{3}{2}(\text n_1 + \text n_2 + \text n_3)\text{kT}$

Since there is no loss of energy on mixing, the two are equal,

$\frac{3}{2}\text n_1 \text{kT}_1 + \frac{3}{2}\text n_2 \text{kT}_2 + \frac{3}{2}\text n_3 \text{kT}_3 = \frac{3}{2}(\text n_1 + \text n_2 + \text n_3)\text{kT}$

Cancelling the common factor $\frac{3}{2}\text k$ throughout,

$\text n_1 \text T_1 + \text n_2 \text T_2 + \text n_3 \text T_3 = (\text n_1 + \text n_2 + \text n_3)\text T$

$\text T = \frac{\text n_1 \text T_1 + \text n_2 \text T_2 + \text n_3 \text T_3}{\text n_1 + \text n_2 + \text n_3}$

Hence, the final temperature of the mixture is $\frac{\text n_1 \text T_1 + \text n_2 \text T_2 + \text n_3 \text T_3}{\text n_1 + \text n_2 + \text n_3}$.

The masses m₁, m₂ and m₃ of the molecules do not appear in the result, because the average kinetic energy per molecule depends only on the temperature and not on the mass of the molecule.

$\frac{γ + 1}{2}$

The average time of collision, that is, the average time between two successive collisions of a molecule, is

$\tau = \frac{\lambda}{\text v_{rms}}$

where the mean free path is $\lambda = \frac{1}{\sqrt2\pi \text n \text d^2}$ and $\text v_{rms} = \sqrt{\frac{3\text{RT}}{\text M}}$.

Since the number of molecules per unit volume is $\text n = \frac{\text N}{\text V}$, the mean free path is proportional to V. Therefore

$\tau = \frac{1}{\sqrt2\pi\left(\frac{\text N}{\text V}\right)\text d^2\sqrt{\frac{3\text{RT}}{\text M}}} \quad \Rightarrow \quad \tau \propto \frac{\text V}{\sqrt{\text T}}$

For an adiabatic expansion of an ideal gas,

$\text{TV}^{\gamma - 1} = \text{constant} \quad \Rightarrow \quad \text T \propto \text V^{1-\gamma}$

$\sqrt{\text T} \propto \text V^{\frac{1-\gamma}{2}} \quad \Rightarrow \quad \frac{1}{\sqrt{\text T}} \propto \text V^{\frac{\gamma - 1}{2}}$

Substituting this in the expression for τ,

$\tau \propto \text V \times \text V^{\frac{\gamma - 1}{2}} = \text V^{1 + \frac{\gamma - 1}{2}} = \text V^{\frac{\gamma + 1}{2}}$

Comparing with τ ∝ V^q,

$\text q = \frac{\gamma + 1}{2}$

Hence, the value of q is $\frac{γ + 1}{2}$.

Given,

• Length of the tube = 100 cm, closed at one end
• Length of the mercury thread = 76 cm
• Length of the trapped air column = 15 cm
• Atmospheric pressure = 76 cm of mercury

When the tube is horizontal : The trapped air column lies next to the closed end, the mercury thread is 76 cm long, and the length of the open portion of the tube is

$100 - 76 - 15 = 9\ \text{cm}$

The mercury thread is horizontal, so it exerts no pressure of its own on the enclosed air. Hence

$\text P_1 = 76\ \text{cm of mercury}, \qquad \text V_1 = 15\text A\ \text{cm}^3$

where A is the area of cross-section of the tube.

When the tube is held vertically with the open end at the bottom : The mercury tends to run down, so the enclosed air expands into the 9 cm of empty tube and its length would become 15 + 9 = 24 cm. In this position the pressure of the enclosed air column together with the 76 cm mercury column exceeds the atmospheric pressure, and therefore some mercury flows out of the open end.

Let h cm of mercury flow out. Then the length of the mercury thread left in the tube is (76 − h) cm and the length of the enclosed air column becomes (24 + h) cm.

For the mercury to be in equilibrium, the pressure of the enclosed air together with the mercury column must balance the atmospheric pressure,

$\text P_2 + (76 - \text h) = 76 \quad \Rightarrow \quad \text P_2 = \text h\ \text{cm of mercury}$

$\text V_2 = (24 + \text h)\text A\ \text{cm}^3$

Applying Boyle's law : The temperature remains constant, so

$\text P_1 \text V_1 = \text P_2 \text V_2$

$76 \times 15\text A = \text h(24 + \text h)\text A$

$\text h^2 + 24\text h - 1140 = 0$

Solving this quadratic equation,

$$\begin{aligned}\text h = \frac{-24 \pm \sqrt{(24)^2 - 4 \times 1 \times (-1140)}}{2} \\[1em] = \frac{-24 \pm \sqrt{576 + 4560}}{2} = \frac{-24 \pm \sqrt{5136}}{2}\end{aligned}$$

$\text h = \frac{-24 \pm 71.7}{2} = 23.85\ \text{cm} \quad \text{or} \quad -47.85\ \text{cm}$

A negative length has no meaning, so h = 23.85 cm.

Hence, when the tube is held vertically with the open end at the bottom, nearly 24 cm of mercury flows out of the tube, leaving a mercury column of about 52 cm and an enclosed air column of about 48 cm, which together keep the atmospheric pressure in balance.

Given,

• Diffusion rate of hydrogen, R₁ = 28.7 cm³ s^-1
• Diffusion rate of the unknown gas, R₂ = 7.2 cm³ s^-1
• Molecular mass of hydrogen, M₁ = 2

The rate of diffusion of a gas is directly proportional to the rms speed of its molecules,

$\text R \propto \text v_{rms}$

Since $\text v_{rms} = \sqrt{\frac{3\text{RT}}{\text M}}$, at the same temperature the rms speed is inversely proportional to the square-root of the molecular mass. Therefore, for hydrogen and the unknown gas,

$\frac{\text R_1}{\text R_2} = \frac{\text v_{1rms}}{\text v_{2rms}} = \sqrt{\frac{\text M_2}{\text M_1}}$

Squaring both sides,

$$\begin{aligned}\frac{\text M_2}{\text M_1} = \left(\frac{\text R_1}{\text R_2}\right)^2 = \frac{(28.7)^2}{(7.2)^2} \\[1em] = \frac{823.69}{51.84} = 16\end{aligned}$$

$\text M_2 = 16\text M_1 = 16 \times 2 = 32$

Hence, the molecular mass of the unknown gas is 32, and the gas is oxygen.

The law of atmospheres for a gas column under gravity is

$\text n_2 = \text n_1 \exp^{[-\text m \text g(\text h_2 - \text h_1)/\text k_\text B \text T]} \qquad \ldots(\text i)$

Here the quantity mg in the exponent is the weight of one particle of the column.

Apparent weight of a suspended particle : When a particle of mass m and density ρ is suspended in a liquid of density ρ', it is buoyed up by the weight of the liquid displaced. The volume of the liquid displaced is equal to the volume of the particle,

$\text V = \frac{\text m}{\rho}$

Hence the apparent weight of the suspended particle is

$$\begin{aligned}\text W_{app} = \text{mg} - \text V\rho'\text g = \text{mg} - \left(\frac{\text m}{\rho}\right)\rho'\text g \\[1em] = \text{mg}\left(1 - \frac{\rho'}{\rho}\right)\end{aligned}$$

Sedimentation equilibrium : For a suspension in a liquid column the particle settles under its apparent weight, so mg in equation (i) is to be replaced by W_app,

$\text n_2 = \text n_1 \exp^{\left[-\text{mg}\left(1 - \frac{\rho'}{\rho}\right)(\text h_2 - \text h_1)/\text k_\text B \text T\right]}$

$\text n_2 = \text n_1 \exp^{[-\text{mg}(\rho - \rho')(\text h_2 - \text h_1)/\rho \text k_\text B \text T]}$

Replacing the Boltzmann constant : Since $\text k_\text B = \frac{\text R}{\text N_\text A}$, where R is the universal gas constant and N_A is Avogadro's number,

$\frac{1}{\text k_\text B} = \frac{\text N_\text A}{\text R}$

Substituting this value,

$\text n_2 = \text n_1 \exp^{[-\text m \text g \text N_\text A (\rho - \rho')(\text h_2 - \text h_1)/(\rho \text R \text T)]}$

Hence, the equation for sedimentation equilibrium of a suspension in a liquid column is established.

Let M be the atomic mass of the substance in gram, so that the mass of 1 mole is M × 10^-3 kg. If N_A is Avogadro's number, the mass of one atom is

$\text m = \frac{\text M \times 10^{-3}}{\text N_\text A}\ \text{kg}$

Assuming an atom to be a sphere of radius r, its volume is $\frac{4}{3}\pi \text r^3$ and its mass is

$\text m = \text V \times \rho = \frac{4}{3}\pi \text r^3 \rho$

Equating the two expressions for m,

$\frac{4}{3}\pi \text r^3 \rho = \frac{\text M \times 10^{-3}}{\text N_\text A}$

$\text r^3 = \frac{3\text M \times 10^{-3}}{4\pi \rho \text N_\text A}$

$\text r = \left(\frac{3\text M \times 10^{-3}}{4\pi \rho \text N_\text A}\right)^{\frac{1}{3}}$

Putting N_A = 6 × 10²³ and simplifying,

$\text r = \left(\frac{3\text M \times 10^{-3}}{4\pi \rho \times 6 \times 10^{23}}\right)^{\frac{1}{3}} = \left(\frac{\text M \times 10^{-26}}{8\pi \rho}\right)^{\frac{1}{3}}$

For carbon (diamond) : M = 12.01 and ρ = 2.22 × 10³ kg m^-3,

$\text r = \left(\frac{12.01 \times 10^{-26}}{8 \times 3.14 \times 2.22 \times 10^3}\right)^{\frac{1}{3}}$

$= 1.29 \times 10^{-10}\ \text m = 1.29\ \text{Å}$

Working in the same way for the other substances,

Substance	Atomic mass (u)	Density (103 kg m-3)	Size of atom
Carbon (diamond)	12.01	2.22	1.29 Å
Gold	197.00	19.32	1.59 Å
Nitrogen (liquid)	14.01	1.00	1.77 Å
Lithium	6.94	0.53	1.73 Å
Fluorine (liquid)	19.00	1.14	1.88 Å

Hence, the sizes of the atoms of these different elements are all of the order of 1 Å.

are in constant random motion

Reason — According to the first postulate of the kinetic theory of gases, gas molecules move in continuous, random motion in all directions. They travel in straight lines until they collide with other gas molecules or with the walls of the container. The molecules are therefore never stationary, they do not exert forces on one another except during collisions, and the collisions being perfectly elastic, no kinetic energy is lost in them.

collisions of gas molecules with the walls of the container

Reason — The molecules of a gas keep on colliding with the walls of the vessel and thus transfer momentum to them. By Newton's law of motion, this rate of change of momentum appears as a steady force acting perpendicular to the walls of the vessel. The magnitude of this force per unit area of the walls is called the pressure of the gas. The attraction or repulsion between the molecules is negligible except during collisions, so it plays no part in producing the pressure.

temperature

Reason — According to the kinetic theory, the average kinetic energy of a gas molecule at absolute temperature T is

$\overline{\text E} = \frac{3}{2}\text{kT}$

where k is the Boltzmann constant. Since $\frac{3}{2}$ k is a constant, the average kinetic energy of the molecules is directly proportional to the absolute temperature of the gas, and it does not depend on the volume, the pressure or the number of molecules.

the square root of the temperature

Reason — From the kinetic theory, the root-mean-square speed of the molecules of a gas of molecular weight M at absolute temperature T is

$\text v_{rms} = \sqrt{\frac{3\text{RT}}{\text M}} \quad \Rightarrow \quad \text v_{rms} \propto \sqrt{\text T}$

Since 3R/M is a constant for a given gas, the root-mean-square speed of the molecules is directly proportional to the square-root of the absolute temperature of the gas. This is the kinetic interpretation of temperature.

doubles

Reason — The average kinetic energy of a gas molecule is $\overline{\text E} = \frac{3}{2}\text{kT}$, so $\overline{\text E} \propto \text T$. If the absolute temperature is doubled,

$\frac{\overline{\text E}_2}{\overline{\text E}_1} = \frac{\text T_2}{\text T_1} = \frac{2\text T}{\text T} = 2$

Hence the average kinetic energy of the molecules also becomes twice its original value.

average kinetic energy of gas molecules

Reason — According to the kinetic theory, the average kinetic energy of a molecule of an ideal gas is $\overline{\text E} = \frac{3}{2}\text{kT}$, which gives

$\text T = \frac{2\overline{\text E}}{3\text k}$

Thus the absolute temperature of a gas is directly proportional to the average kinetic energy of its molecules. This is called the kinetic interpretation of temperature. In an ideal gas there are no intermolecular forces, so there is no internal potential energy.

have zero kinetic energy

Reason — Since $\overline{\text E} = \frac{3}{2}\text{kT}$, the average kinetic energy of the molecules becomes zero when T = 0. Also $\text v_{rms} = \sqrt{\frac{3\text{RT}}{\text M}}$ becomes zero at T = 0. Hence absolute zero is that temperature at which the motion of all the molecules of a gas stops. As the kinetic energy can never be negative, no temperature is possible below the absolute zero.

inversely proportional to pressure

Reason — Boyle's law states that for a fixed amount of gas at constant temperature, the volume V of a gas is inversely proportional to its pressure P,

$\text P \propto \frac{1}{\text V} \quad \text{or} \quad \text P_1 \text V_1 = \text P_2 \text V_2$

Thus, at a constant temperature, if the pressure of a gas increases its volume decreases, and vice-versa.

directly proportional to temperature

Reason — Charles' law states that the volume of a given mass of gas at constant pressure is directly proportional to its absolute temperature,

$\text V \propto \text T \quad \text{or} \quad \frac{\text V}{\text T} = \text{a constant}$

Thus, at constant pressure, a gas expands on heating and contracts on cooling.

directly proportional to temperature

Reason — The pressure law (Gay-Lussac's law) states that for a fixed amount of gas at constant volume, the pressure P of the gas is directly proportional to its absolute temperature T,

$\text P \propto \text T \quad \text{or} \quad \frac{\text P}{\text T} = \text{a constant}$

Thus, when the volume is kept constant, heating a gas raises its pressure in the same ratio as the absolute temperature.

3

Reason — The molecule of a monoatomic gas, such as helium or argon, consists of a single atom. Its translational motion can take place in any direction in space and can therefore be resolved along the three coordinate axes, giving three independent motions. Hence a monoatomic gas has three degrees of freedom, all translational. A monoatomic molecule can rotate also, but its moment of inertia is so small that the kinetic energy of rotation is insignificant.

5 degrees of freedom

Reason — The molecule of a diatomic gas, such as H₂ or O₂, is made up of two atoms joined rigidly through a bond. It can not only move bodily, giving three translational degrees of freedom, but can also rotate about any of the three coordinate axes. However, its moment of inertia about the axis joining the two atoms is negligible, so it can have only two rotational motions. Hence a diatomic molecule has five degrees of freedom, three translational and two rotational. The vibrational motion does not usually occur at ordinary temperatures.

greater than 5

Reason — A polyatomic molecule, such as H₂O, H₂S or CO₂, can rotate about any of the three coordinate axes. Hence a polyatomic gas has six degrees of freedom, three translational and three rotational, which is greater than 5. If the atoms of the molecule vibrate with respect to each other, it would possess vibrational degrees of freedom also.

$\frac{1}{2}$ kT

Reason — According to the law of equipartition of energy, in a classical system of particles which is in equilibrium at absolute temperature T, the average internal (kinetic) energy per particle associated with each degree of freedom is $\frac{1}{2}$ kT, where k is Boltzmann's constant. If the particle has f degrees of freedom, its average kinetic energy would be $\frac{1}{2}$ f kT.

$\frac{5}{2}$ kT

Reason — By the law of equipartition of energy, the average energy per molecule is $\frac{1}{2}$ f kT, where f is the number of degrees of freedom. A diatomic gas molecule has f = 5, three translational and two rotational degrees of freedom. Hence the average energy per molecule of a diatomic gas is

$\overline{\text E} = \frac{1}{2}\text{fkT} = \frac{5}{2}\text{kT}$

Note: The value $\frac{5}{2}$ kT obtained above is not offered among the four options. The printed answer key marks option 2, that is, $\frac{1}{2}$ kT, which is the energy associated with one degree of freedom of a molecule and not the average energy of a diatomic molecule. The textbook's own explanation for this question also states the value as $\frac{5}{2}$ kT, so the options appear to be misprinted.*

Collisions between gas molecules are elastic

Reason — One of the postulates of the kinetic theory is that the collisions between the gas molecules, and between the molecules and the walls of the container, are perfectly elastic, so that no kinetic energy is lost in the collisions. The other statements contradict the postulates, since gas molecules do not exert attractive or repulsive forces on each other except during collisions, they have negligible potential energy, and their speeds vary over a wide range.

the average distance a molecule travels before colliding with another molecule

Reason — The molecules of a gas move in straight lines with constant speeds between two successive collisions, so a particular molecule is found to follow short zig-zag paths of different lengths. These are called the 'free paths' of the molecule. The mean free path is the average distance travelled by a molecule between two successive collisions with other molecules.

the number density of the gas

Reason — The mean free path of a molecule is

$\lambda = \frac{1}{\sqrt2\pi \text n \text d^2}$

where n is the number of molecules per unit volume, that is, the number density, and d is the diameter of a molecule. Hence the mean free path is inversely proportional to the number density of the gas, which is why the mean free path varies inversely as the density of the gas.

molecular collisions with the walls of the container

Reason — The pressure exerted by a gas on the walls of its container is due to the collisions of the gas molecules with the walls. Each collision imparts a small force on the wall, and the cumulative effect of a very large number of such collisions results in the measurable pressure. The collisions of the molecules with one another do not contribute to the pressure on the walls.

$\frac{3}{2}$ RT

Reason — For 1 mole of a gas of molecular weight M, the kinetic theory gives $\text{PV} = \frac{1}{3}\text M\overline{\text v^2}$, and the ideal gas equation gives PV = RT. Hence the kinetic energy of 1 mole of the gas is

$\frac{1}{2}\text M\overline{\text v^2} = \frac{3}{2}\text{RT}$

Thus the kinetic energy of one mole of an ideal gas at absolute temperature T is $\frac{3}{2}$ RT.

become half

Reason — By Boyle's law, at constant temperature P₁V₁ = P₂V₂. Putting P₂ = 2P₁,

$\text V_2 = \frac{\text P_1 \text V_1}{\text P_2} = \frac{\text P_1 \text V_1}{2\text P_1} = \frac{\text V_1}{2}$

Hence, on doubling the pressure at constant temperature, the volume of the gas becomes half.

double

Reason — By Charles' law, at constant pressure $\frac{\text V_1}{\text T_1} = \frac{\text V_2}{\text T_2}$. Putting T₂ = 2T₁,

$\text V_2 = \text V_1 \times \frac{\text T_2}{\text T_1} = \text V_1 \times \frac{2\text T_1}{\text T_1} = 2\text V_1$

Hence, on doubling the absolute temperature at constant pressure, the volume of the gas also doubles.

double

Reason — By the pressure law, at constant volume $\frac{\text P_1}{\text T_1} = \frac{\text P_2}{\text T_2}$. Putting T₂ = 2T₁,

$\text P_2 = \text P_1 \times \frac{\text T_2}{\text T_1} = 2\text P_1$

Hence, on doubling the absolute temperature at constant volume, the pressure of the gas doubles.

the temperature only

Reason — In an ideal gas there are no intermolecular forces, so there is no internal potential energy and the total internal energy of the gas is entirely the kinetic energy of its molecules. For μ moles of a gas having f degrees of freedom,

$\text U = \frac{1}{2}\mu \text{fRT}$

For a monoatomic gas f = 3, so $\text U = \frac{3}{2}\mu \text{RT}$. Thus, for a given amount of the gas, the internal energy depends only on the temperature and not on the pressure or the volume.

remains constant

Reason — An isothermal process is one which is carried out at a constant temperature, and Boyle's law applies to such a process. For an ideal gas the internal energy depends only on the temperature, so in an isothermal process the internal energy also remains unchanged, and changes in pressure do not affect the energy of the gas directly.

$\sqrt{\text T}$

Reason — The average speed of the molecules of a gas, like the root-mean-square speed, is proportional to the square-root of the absolute temperature of the gas,

$\text v \propto \sqrt{\text T}$

This follows from the kinetic theory, since $\text v_{rms} = \sqrt{\frac{3\text{RT}}{\text M}}$ and the average speed differs from the rms speed only by a numerical factor.

kinetic energy of molecules

Reason — From the kinetic theory, the pressure of a gas is

$\text P = \frac{1}{3}\rho\overline{\text v^2} = \frac{2}{3}\left(\frac{1}{2}\rho\overline{\text v^2}\right) = \frac{2}{3}\text E$

where $\text E \left(= \frac{1}{2}\rho\overline{\text v^2}\right)$ is the translational kinetic energy of the gas per unit volume. Thus the pressure of a gas is equal to two-thirds of its translational kinetic energy per unit volume, that is, the pressure is proportional to the kinetic energy of the molecules.

low pressure and high temperature

Reason — At low pressure the volume of the gas is large, so the actual volume of the molecules is negligible in comparison with it. At high temperature the molecules have large kinetic energy, so the effect of the intermolecular force on their motion is negligible. Since these are exactly the two properties assumed for a perfect gas, the behaviour of a real gas is most nearly ideal at low pressure and high temperature.

decrease

Reason — The mean free path of the molecules of a gas is

$\lambda = \frac{\text{kT}}{\sqrt2\pi \text d^2 \text P}$

so at a constant temperature $\lambda \propto \frac{1}{\text P}$. On increasing the pressure the molecules are packed more closely together, so a molecule collides with another after covering a shorter distance and the mean free path decreases.

P = $\frac{1}{3}$ ρv_rms²

Reason — From the kinetic theory, the pressure exerted by an ideal gas on the walls of the container is

$\text P = \frac{1}{3}\frac{\text{mn}}{\text V}\overline{\text v^2}$

Since mn is the mass of the whole gas and V its volume, mn/V is the density ρ of the gas. Hence

$\text P = \frac{1}{3}\rho \text v_{rms}^2$

The relation PV = nRT is the equation of state of the gas, and P = F/A is only the general definition of pressure; neither is derived from the kinetic theory.

remain the same

Reason — For n molecules of a gas, PV = nkT, so at constant volume

$\text P = \frac{\text{nkT}}{\text V} \quad \Rightarrow \quad \text P \propto \text{nT}$

Here the temperature is doubled and the number of molecules is halved, so

$\frac{\text P_2}{\text P_1} = \frac{\left(\frac{\text n}{2}\right)(2\text T)}{\text{nT}} = 1$

The two changes cancel each other, and hence the pressure of the gas remains the same.

quadruple

Reason — From the ideal gas equation PV = μRT, for a given mass of gas

$\frac{\text P_1 \text V_1}{\text T_1} = \frac{\text P_2 \text V_2}{\text T_2}$

Here $\text V_2 = \frac{\text V_1}{2}$ and T₂ = 2T₁. Therefore

$\text P_2 = \text P_1 \times \frac{\text V_1}{\text V_2} \times \frac{\text T_2}{\text T_1} = \text P_1 \times 2 \times 2 = 4\text P_1$

Hence the pressure of the gas becomes four times, that is, it quadruples.

2/3

Reason — By Boyle's law, at constant temperature P₁V₁ = P₂V₂, that is,

$\frac{\text P_2}{\text P_1} = \frac{\text V_1}{\text V_2}$

Given that $\frac{\text V_2}{\text V_1} = \frac{3}{1}$,

$\frac{\text P_2}{\text P_1} = \frac{1}{3} \quad \Rightarrow \quad \text P_2 = \frac{\text P_1}{3}$

The fall in the pressure is therefore

$\text P_1 - \text P_2 = \text P_1 - \frac{\text P_1}{3} = \frac{2}{3}\text P_1$

Hence, the pressure has been decreased by $\frac{2}{3}$ of its initial value, the final pressure being one-third of the initial pressure.

Note: The printed answer key marks option 3, "3 times". Since the pressure falls to one-third of its initial value, the decrease is $\frac{2}{3}$ of the initial pressure.

2/5

Reason — By the law of equipartition of energy, each degree of freedom contributes an equal amount $\frac{1}{2}$ kT to the energy. The gas has 3 translational and 2 rotational degrees of freedom, so the total number of degrees of freedom is f = 5, of which 2 are rotational.

$\text{Fraction} = \frac{\text{rotational degrees of freedom}}{\text{total degrees of freedom}} = \frac{2}{5}$

Hence the fraction of the total energy associated with rotational motion is 2/5.

The volume of the gas was decreased by a factor of 4 at constant temperature

Reason — The mean free path is inversely proportional to the number density of the gas,

$\lambda = \frac{1}{\sqrt2\pi \text n \text d^2}, \qquad \text n = \frac{\text N}{\text V}$

If the volume is decreased by a factor of 4 at constant temperature, the same number of molecules N is confined in one-fourth of the volume, so the number density n becomes 4 times and the mean free path becomes one-fourth of its earlier value. Decreasing the number of molecules or the pressure would increase the mean free path, and raising the temperature at constant pressure would also increase it.

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

Explanation

Assertion (A) is correct: At a constant temperature, when the volume of a given mass of gas is increased, the molecules have to travel a larger distance before striking the walls and the number of collisions per unit area per second falls. Hence the pressure of the gas decreases.

Reason (R) is also correct: Boyle's law states that for a fixed amount of gas at constant temperature, the pressure of a gas is inversely proportional to its volume, $\text P \propto \frac{1}{\text V}$, or P₁V₁ = P₂V₂.

Since P is inversely proportional to V, an increase in V must be accompanied by a decrease in P. 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: According to the postulates of the kinetic theory, gas molecules move in continuous random motion in all directions, and their speeds vary over a wide range. This is why the root-mean-square speed has to be defined as an average over all the molecules.

Reason (R) is false: The molecules of a gas do not all have the same kinetic energy at a given temperature. Their speeds, and hence their kinetic energies, are distributed over a wide range, and the molecules keep exchanging energy among themselves at every collision. What the kinetic theory fixes is only the average kinetic energy per molecule, $\overline{\text E} = \frac{3}{2}\text{kT}$, which depends on the temperature alone.

The Reason would have been correct had it stated that the molecules have the same average kinetic energy at a given temperature.

Therefore, assertion is true but reason is false.

Note: The printed answer key marks option 2, treating the Reason as true. As worded, the Reason asserts that individual molecules have equal kinetic energies, which contradicts the distribution of molecular speeds that the Assertion itself describes.*

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

Explanation

Assertion (A) is correct: A gas has neither a definite volume nor a definite shape. It covers the entire volume of the containing vessel, which is why, when the cork of an ammonia bottle is opened, the smell of ammonia spreads in the whole room.

Reason (R) is also correct: According to the postulates of the kinetic theory, gas molecules do not exert attractive or repulsive forces on each other, except during collisions.

Since the molecules hardly attract one another and possess large kinetic energy, nothing holds them together and they continue to move throughout the entire available space. 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: Raising the temperature of a gas increases the average kinetic energy of its molecules, and hence the speed of the molecules increases.

Reason (R) is also correct: From the kinetic theory,

$\text v_{rms} = \sqrt{\frac{3\text{RT}}{\text M}} \quad \Rightarrow \quad \text v_{rms} \propto \sqrt{\text T}$

that is, the root-mean-square speed is directly proportional to the square-root of the absolute temperature.

Since v_rms varies as $\sqrt{\text T}$, a rise in temperature necessarily raises the root-mean-square speed. 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: One of the postulates of the kinetic theory is that the collisions between the gas molecules, and between the molecules and the walls of the container, are perfectly elastic.

Reason (R) is also correct: In a perfectly elastic collision the total kinetic energy of the colliding bodies is conserved, so no kinetic energy is lost in the collision.

It is precisely because the collisions involve no loss of kinetic energy that the total kinetic energy of the gas remains constant as long as the temperature is constant, though energy may be transferred between the molecules. 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: According to the kinetic theory, the average kinetic energy of a gas molecule is $\overline{\text E} = \frac{3}{2}\text{kT}$, so it is directly proportional to the absolute temperature of the gas.

Reason (R) is false: The relation $\overline{\text E} = \frac{3}{2}\text{kT}$ contains only the temperature. The average kinetic energy of the molecules depends on the temperature alone and not on the pressure. In fact, different gases at the same temperature have the same average kinetic energy per molecule whatever their pressure may be.

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: Since $\overline{\text E} = \frac{3}{2}\text{kT}$, putting T = 0 gives $\overline{\text E}$ = 0. Absolute zero is that temperature at which the motion of all the molecules of the gas stops, so their kinetic energy becomes zero.

Reason (R) is also correct: Temperature is a measure of the average kinetic energy of the molecules of a gas, since $\text T = \frac{2\overline{\text E}}{3\text k}$. This is the kinetic interpretation of temperature.

As the temperature is itself the measure of the average kinetic energy, the vanishing of the temperature means the vanishing of the kinetic 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: By the pressure law, at constant volume $\text P \propto \text T$, so the pressure of an ideal gas increases with a rise in temperature when the volume is held constant.

Reason (R) is also correct: From the kinetic theory $\text P = \frac{2}{3}\text E$, where E is the translational kinetic energy of the gas per unit volume. Hence the pressure of a gas is proportional to the average kinetic energy of its molecules.

A rise in temperature increases the average kinetic energy of the molecules, so they strike the walls more frequently and with greater momentum, and the pressure rises. 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 Boyle's law, for a given amount of gas at constant temperature $\text P \propto \frac{1}{\text V}$, so the pressure decreases when the volume increases.

Reason (R) is also correct: When the volume of the vessel is increased at constant temperature, the molecules have to cover a greater distance between two successive collisions with the walls, so the number of collisions per unit area of the walls per second decreases.

Since the pressure of a gas arises from the momentum transferred to the walls by these collisions, fewer collisions per unit area mean a smaller pressure. 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: Boyle's law is strictly true only for a perfect gas. Real gases obey it only approximately, and they deviate from it appreciably under extreme conditions, so it is not valid for real gases under all conditions of temperature and pressure.

Reason (R) is correct: At high pressures the actual volume of the molecules is no longer negligible, and at low temperatures the intermolecular forces are no longer negligible. Hence the behaviour of real gases deviates from ideal gas behaviour at high pressures and low temperatures.

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: When a gas is heated at constant pressure it expands, so its volume increases with a rise in temperature.

Reason (R) is also correct: Charles' law states that the volume of a given mass of gas at constant pressure is directly proportional to its absolute temperature, $\text V \propto \text T$, or $\frac{\text V}{\text T}$ = a constant.

Since the volume is directly proportional to the absolute temperature, a rise in temperature must be accompanied by an increase in volume. 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: From Charles' law, at constant pressure $\frac{\text V_1}{\text T_1} = \frac{\text V_2}{\text T_2}$. Putting T₂ = 2T₁ gives V₂ = 2V₁, so the volume also doubles.

Reason (R) is also correct: Charles' law states that the volume of a gas at constant pressure is directly proportional to its absolute temperature.

The doubling of the volume follows directly from this proportionality, provided the temperature used is the absolute 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: By the pressure law, at constant volume $\text P \propto \text T$, so heating a gas in a closed vessel of fixed volume raises its pressure.

Reason (R) is also correct: A rise in temperature increases the average kinetic energy of the molecules, and hence their mean speed. This increases both the momentum transferred in each collision with the walls and the number of collisions per second.

Since the pressure of a gas is the force per unit area produced by these collisions, more frequent and more forceful collisions raise the pressure. 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 the pressure law, at constant volume the pressure of an ideal gas is directly proportional to its absolute temperature, so the pressure decreases as the temperature decreases.

Reason (R) is also correct: Since $\overline{\text E} = \frac{3}{2}\text{kT}$, a fall in temperature lowers the average kinetic energy of the molecules and hence their mean speed, so they strike the walls less often and with smaller momentum.

The pressure being the result of these collisions, the weaker and less frequent collisions produce a smaller pressure. 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 diatomic molecule has five degrees of freedom at room temperature, three with respect to translation and two with respect to rotation.

Reason (R) is also correct: A diatomic molecule can move bodily in three independent directions and can rotate about two of the three coordinate axes, its moment of inertia about the axis joining the two atoms being negligible. The molecule can also vibrate along the line joining the two atoms, but the vibrational motion does not usually occur at ordinary temperatures.

Adding the three translational and the two rotational motions gives exactly five degrees of freedom. 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 monoatomic gas molecule, such as He or Ar, consists of a single atom and has three degrees of freedom, all translational.

Reason (R) is also correct: The translational motion of a single atom can take place in any direction in space and can be resolved along the three coordinate axes, giving three independent motions. A monoatomic molecule can rotate also, but its moment of inertia is so small that the kinetic energy of rotation is insignificant, and it has no vibrational motion.

Since only the three translational motions are available, the molecule has exactly three degrees of freedom. 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 there is no internal potential energy and the total internal energy is entirely the kinetic energy of the molecules. For μ moles of a monoatomic gas $\text U = \frac{3}{2}\mu \text{RT}$, which depends only on the temperature.

Reason (R) is also correct: By the law of equipartition of energy, each degree of freedom contributes an average energy $\frac{1}{2}$ kT per molecule, so for f degrees of freedom the internal energy of 1 mole is $\text U = \frac{1}{2}\text{fRT}$.

As every one of these contributions contains the temperature and nothing else, the internal energy can depend only on 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 but reason is not the correct explanation of assertion.

Explanation

Assertion (A) is correct: The rotational degrees of freedom of a diatomic molecule do contribute to the internal energy of the gas, each contributing $\frac{1}{2}$ kT per molecule.

Reason (R) is also correct: The law of equipartition of energy states that in a system in equilibrium at absolute temperature T, each degree of freedom contributes an equal average energy $\frac{1}{2}$ kT.

However, the two rotational degrees of freedom of a diatomic molecule are already active at ordinary room temperature, and it is the vibrational motion that is excited only at high temperatures. The Reason therefore does not explain the Assertion as it is worded.

Therefore, both assertion and reason are true but reason is not 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 μ moles of an ideal gas having f degrees of freedom, $\text U = \frac{1}{2}\mu \text{fRT}$. For a monoatomic gas f = 3 and for a diatomic gas f = 5, so at the same temperature $\text U_{diatomic} = \frac{5}{2}\mu \text{RT}$ is greater than $\text U_{monoatomic} = \frac{3}{2}\mu \text{RT}$.

Reason (R) is also correct: A monoatomic molecule has three degrees of freedom, all translational, while a diatomic molecule has five, three translational and two rotational.

Since each degree of freedom contributes an equal share of energy, the gas with the greater number of degrees of freedom has the greater internal energy at the same 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: A diatomic molecule can vibrate along the line joining the two atoms, but this vibrational motion does not usually occur at ordinary temperatures. Hence the vibrational modes make no contribution to the internal energy of a diatomic gas at room temperature, and its energy corresponds to only five degrees of freedom.

Reason (R) is also correct: The vibrational energy levels of a diatomic molecule are not excited at low temperatures, and become excited only when the temperature is raised sufficiently.

Since the vibrational motion is not excited at room temperature, it can contribute nothing to the internal energy there. The Reason therefore explains the Assertion.

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

Given,

• Initial temperature, T₁ = 127°C = 127 + 273 = 400 K
• Final volume, V₂ = 1.5 V₁
• The pressure remains constant

By Charles' law, at constant pressure $\frac{\text V}{\text T}$ = a constant, so

$\frac{\text V_1}{\text T_1} = \frac{\text V_2}{\text T_2} \quad \Rightarrow \quad \text T_2 = \text T_1 \times \frac{\text V_2}{\text V_1}$

$\text T_2 = 400 \times \frac{1.5\text V_1}{\text V_1} = 600\ \text K$

$\text t_2 = 600 - 273 = 327^\circ\text C$

Hence, the new temperature of the gas is 327°C.

Given,

• Initial temperature, T₁ = −173°C = −173 + 273 = 100 K
• Initial pressure, P₁ = 1 atmosphere
• Final pressure, P₂ = 2 atmospheres
• The volume remains constant

By the pressure law, at constant volume $\frac{\text P}{\text T}$ = a constant, so

$\frac{\text P_1}{\text T_1} = \frac{\text P_2}{\text T_2} \quad \Rightarrow \quad \text T_2 = \text T_1 \times \frac{\text P_2}{\text P_1}$

$\text T_2 = 100 \times \frac{2}{1} = 200\ \text K$

$\text t_2 = 200 - 273 = -73^\circ\text C$

Hence, the gas should be heated to −73°C.

Given,

• Ratio of the vapour densities, ρ₁ : ρ₂ = 8 : 9
• Both the gases are at the same temperature

From the kinetic theory, $\text P = \frac{1}{3}\rho\overline{\text v^2}$, so at the same temperature and pressure

$\frac{\text v_{1rms}}{\text v_{2rms}} = \sqrt{\frac{\text M_2}{\text M_1}} = \sqrt{\frac{\rho_2}{\rho_1}}$

Substituting the values,

$\frac{\text v_{1rms}}{\text v_{2rms}} = \sqrt{\frac{9}{8}} = \frac{3}{2\sqrt2}$

Hence, the root-mean-square speeds of the molecules of the two gases are in the ratio $3 : 2\sqrt2$.

From the kinetic theory, the pressure of a gas is

$\text P = \frac{1}{3}\text d\overline{\text v^2}$

where d is the density of the gas. At a constant temperature the mean-square speed $\overline{\text v^2}$ of the molecules remains constant, so

$\frac{\text P}{\text d} = \frac{1}{3}\overline{\text v^2} = \text{a constant}$

Hence, at a constant temperature the pressure of an ideal gas is directly proportional to its density, that is, $\frac{\text P}{\text d}$ = constant.

For a gas containing n molecules, the gas equation in terms of Boltzmann's constant is

$\text{PV} = \text{nkT}$

Therefore the number of molecules of the gas is

$\text n = \frac{\text{PV}}{\text{kT}}$

where P is the pressure, V the volume, T the absolute temperature and k the Boltzmann constant.

According to the kinetic theory, the average kinetic energy of one molecule of a gas at absolute temperature T is

$\overline{\text E} = \frac{3}{2}\text{kT}$

where k is Boltzmann's constant.

Since $\frac{3}{2}$ k is a constant, $\overline{\text E} \propto \text T$.

Hence, in an ideal gas the average kinetic energy of one molecule is directly proportional to the absolute temperature of the gas. This relation contains no term for the mass of the molecule, which means that different gases at the same temperature have the same average kinetic energy per molecule.

The molecules of ammonia do not travel in straight lines across the room. As they move, they continuously collide with the molecules of air and with one another, and after each collision the magnitudes and the directions of their speeds change.

Because of these frequent collisions, an ammonia molecule follows a short zig-zag path, its mean free path being very small. Hence, although the root-mean-square speed of the molecules is very large, they are not able to advance speedily in one particular direction.

Hence, the smell of ammonia takes time in reaching the other corner of the room.

According to the kinetic theory, the average kinetic energy of a molecule of a gas at absolute temperature T is

$\text E = \frac{3}{2}\text{kT}$

where k is Boltzmann's constant, k = R/N.

Since $\frac{3}{2}$ k is a constant,

$\text E \propto \text T$

Hence, the average kinetic energy of a gas molecule is directly proportional to the absolute temperature of the gas. This is known as the kinetic interpretation of temperature.

The kinetic energy of a gas at the absolute zero is zero.

The average kinetic energy of a molecule of a gas is $\text E = \frac{3}{2}\text{kT}$, that is, E ∝ T. Putting T = 0 gives E = 0.

The same result follows from $\text v_{rms} = \sqrt{\frac{3\text{RT}}{\text M}}$, which becomes zero at T = 0. Hence absolute zero is that temperature at which the motion of all the molecules of the gas stops.

Hydrogen will leak out more rapidly.

The rate at which a gas escapes through a fine hole is proportional to the root-mean-square speed of its molecules, and

$\text v_{rms} = \sqrt{\frac{3\text{RT}}{\text M}} \quad \Rightarrow \quad \text v_{rms} \propto \frac{1}{\sqrt{\text M}}$

Both the gases are in the same box at the same temperature. The molecular weight of hydrogen (2) is much less than that of oxygen (32), so the hydrogen molecules have the larger root-mean-square speed and reach the hole more often.

The molar internal energy of an ideal gas having f degrees of freedom is $\text U_m = \frac{1}{2}\text{fRT}$, where R is the universal gas constant and T the absolute temperature.

Monoatomic gas : f = 3, so

$\text U_m = \frac{3}{2}\text{RT}\ \text{J mol}^{-1}$

Diatomic gas : f = 5 at ordinary temperatures, so

$\text U_m = \frac{5}{2}\text{RT}\ \text{J mol}^{-1}$

For μ moles of the gas the total internal energy is $\text U = \mu \text U_m$ measured in joule.

The degrees of freedom of the flight of the mosquito are 3.

The degrees of freedom of a particle indicate the number of independent motions which the particle can undergo. A mosquito flying in a room moves bodily and its translational motion can take place in any direction in space, so it can be resolved along the three coordinate axes. Hence it has three independent motions, all translational.

Yes, the number of degrees of freedom increases on raising the temperature of the gas.

At ordinary temperatures the vibrational motion of the atoms within a molecule does not occur, so a diatomic molecule has only five degrees of freedom, three translational and two rotational.

On raising the temperature of the gas the vibrational motion of the atoms of the molecule is also promoted, and the molecule then possesses vibrational degrees of freedom in addition. Hence the number of degrees of freedom increases.

From the kinetic theory, the pressure of a gas is

$\text P = \frac{1}{3}\rho\overline{\text v^2} = \frac{2}{3}\left(\frac{1}{2}\rho\overline{\text v^2}\right)$

$\text P = \frac{2}{3}\text E$

where $\text E \left(= \frac{1}{2}\rho\overline{\text v^2}\right)$ is the translational kinetic energy of the gas per unit volume.

Hence, the pressure of a gas is equal to two-thirds of its translational kinetic energy per unit volume.

On the basis of the kinetic theory, the pressure of an ideal gas is

$\text P = \frac{1}{3}\frac{\text{mn}}{\text V}\text v_{rms}^2$

where

• m = mass of one molecule of the gas
• n = total number of molecules of the gas
• V = volume of the gas
• v_rms = root-mean-square speed of the molecules

From the gas equation PV = μRT, we have $\text R = \frac{\text{PV}}{\mu \text T}$. Therefore

$$\begin{aligned}\text{unit of R} = \frac{\text{unit of pressure} \times \text{unit of volume}}{\text{mole} \times \text{unit of absolute temperature}} \\[1em] = \frac{\text{newton metre}^{-2} \times \text{metre}^3}{\text{mole K}} = \frac{\text{joule}}{\text{mole K}}\end{aligned}$$

Hence, the SI unit of the gas constant is J/(mol-K) or J mol^-1 K^-1.

The Boltzmann constant k is the gas constant per molecule. It is the ratio of the universal gas constant R to Avogadro's number N,

$\text k = \frac{\text R}{\text N} = \frac{8.31\ \text{J mol}^{-1}\text K^{-1}}{6.02 \times 10^{23}\ \text{mol}^{-1}} = 1.38 \times 10^{-23}\ \text{J K}^{-1}$

With this constant the gas equation for n molecules of a gas takes the form PV = nkT, and the average kinetic energy of one molecule is $\frac{3}{2}$ kT.

The unit of the Boltzmann constant is J/K (joule per kelvin).

Since k = R/N, its unit is the unit of R, that is J mol^-1 K^-1, divided by Avogadro's number, which is a pure number per mole. Hence the unit of k is J K^-1.

The Boltzmann constant is the gas constant per molecule, so

$\text k = \frac{\text R}{\text N}$

where R is the universal gas constant and N is Avogadro's number, that is, the number of molecules in 1 mole of a gas.

On the basis of the kinetic theory, the pressure of an ideal gas is

$\text P = \frac{1}{3}\rho\overline{\text v^2}$

where ρ is the density of the gas and $\overline{\text v^2}$ is the mean-square speed of its molecules.

The volume of one mole of a gas at normal temperature and pressure is 22.4 L, that is, 22.4 × 10^-3 m³.

This quantity of the gas contains 6.02 × 10²³ molecules, which is Avogadro's number.

A molecule of a monoatomic gas has f = 3 degrees of freedom. The ratio of the two specific heats is

$\gamma = \frac{\text C_p}{\text C_v} = 1 + \frac{2}{\text f} = 1 + \frac{2}{3} = \frac{5}{3}$

Hence, for a monoatomic gas $\frac{\text C_p}{\text C_v} = \frac{5}{3} = 1.67$.

Given,

• Initial temperature, T₁ = 27°C = 27 + 273 = 300 K
• Final mean kinetic energy, $\overline{\text E}_2 = \frac{1}{3}\overline{\text E}_1$

The mean kinetic energy of a molecule is $\overline{\text E} = \frac{3}{2}\text{kT}$, so $\overline{\text E} \propto \text T$. Therefore

$\frac{\overline{\text E}_2}{\overline{\text E}_1} = \frac{\text T_2}{\text T_1} \quad \Rightarrow \quad \text T_2 = \text T_1 \times \frac{\overline{\text E}_2}{\overline{\text E}_1}$

$\text T_2 = 300 \times \frac{1}{3} = 100\ \text K$

Hence, at 100 K the mean kinetic energy of the molecules will be one-third of its value at 27°C.

The speed of sound in a gas is

$\text v = \sqrt{\frac{\gamma \text P}{\rho}}$

and the root-mean-square speed of the molecules of the gas is

$\text v_{rms} = \sqrt{\frac{3\text P}{\rho}}$

Dividing the first by the second,

$\frac{\text v}{\text v_{rms}} = \sqrt{\frac{\gamma}{3}} \quad \Rightarrow \quad \text v = \sqrt{\frac{\gamma}{3}}\ \text v_{rms}$

Hence, the speed of sound in a gas is $\sqrt{\frac{\gamma}{3}}$ times the root-mean-square speed of its molecules, where γ is the ratio of the two specific heats of the gas.

Given,

• V_A = 2V_B
• T_A = 2T_B
• P_A = 2P_B

For n molecules of a gas, PV = nkT, so

$\text n = \frac{\text{PV}}{\text{kT}}$

Taking the ratio for the two vessels,

$\frac{\text n_{\text A}}{\text n_{\text B}} = \left(\frac{\text P_{\text A}}{\text P_{\text B}}\right)\left(\frac{\text V_{\text A}}{\text V_{\text B}}\right)\left(\frac{\text T_{\text B}}{\text T_{\text A}}\right)$

Substituting the values,

$\frac{\text n_{\text A}}{\text n_{\text B}} = 2 \times 2 \times \frac{1}{2} = \frac{2}{1}$

Hence, the ratio of the number of gas molecules in A and B is 2 : 1.

Given,

• T₁ = 150 K, v_{1 rms} = v
• T₂ = 600 K

From the kinetic theory, $\text v_{rms} \propto \sqrt{\text T}$. Therefore

$\frac{\text v_{2rms}}{\text v_{1rms}} = \sqrt{\frac{\text T_2}{\text T_1}} = \sqrt{\frac{600}{150}} = \sqrt4 = 2$

$\text v_{2rms} = 2\text v$

Hence, the root-mean-square speed of the molecules at 600 K is 2v.

Given,

• Molecular weight of hydrogen, M_H = 2
• Molecular weight of oxygen, M_O = 32
• Both the gases are at the same temperature

The ratio of the rms speeds of the molecules of two different gases at the same temperature is inversely proportional to the square-root of their molecular weights,

$$\begin{aligned}\frac{(\text v_{rms})_{\text H}}{(\text v_{rms})_{\text O}} = \sqrt{\frac{\text M_{\text O}}{\text M_{\text H}}} \\[1em] = \sqrt{\frac{32}{2}} = \sqrt{16} = \frac{4}{1}\end{aligned}$$

Hence, the ratio of the rms speeds of hydrogen molecules to that of oxygen molecules is 4 : 1.

Given,

• Initial temperature, T₁ = 0°C = 273 K
• Final temperature, T₂ = 273°C = 273 + 273 = 546 K

The mean kinetic energy of a molecule is $\overline{\text E} = \frac{3}{2}\text{kT}$, so $\overline{\text E} \propto \text T$. Therefore

$\frac{\overline{\text E}_2}{\overline{\text E}_1} = \frac{\text T_2}{\text T_1} = \frac{546}{273} = \frac{2}{1}$

Hence, the mean kinetic energy of the molecules will be increased in the ratio 2 : 1, that is, it will become twice.

Given,

• The number of molecules in the box is increased from n to 2n, the volume and the temperature remaining unchanged

(i) Pressure of the gas : For n molecules, PV = nkT, so at constant V and T,

$\text P \propto \text n$

Hence the pressure of the gas will be doubled.

(ii) Total kinetic energy of the gas : The average kinetic energy of one molecule is $\frac{3}{2}$ kT, which depends only on the temperature. The total kinetic energy of the gas is $\text n \times \frac{3}{2}\text{kT}$, which is proportional to n.

Hence the total kinetic energy of the gas will also be doubled.

(iii) Root-mean-square speed : Since $\text v_{rms} = \sqrt{\frac{3\text{RT}}{\text M}}$, the rms speed depends only upon the temperature of the gas and its molecular weight, and not on the number of molecules.

Hence the root-mean-square speed of the molecules will remain the same.

From the ideal gas equation PV = μRT, we have

$\text R = \frac{\text{PV}}{\mu \text T}$

Therefore the unit of R is

$\text{unit of R} = \frac{\text{unit of pressure} \times \text{unit of volume}}{\text{mole} \times \text{unit of absolute temperature}}$

$= \frac{\text{newton metre}^{-2} \times \text{metre}^3}{\text{mole K}} = \frac{\text{newton metre}}{\text{mole K}}$

$= \frac{\text{joule}}{\text{mole K}}$

Hence, the unit of R is J mol^-1 K^-1, that is, J/(mol-K).

Since a difference of 1 K is equal to a difference of 1°C, the unit of R can also be written as J mol^-1 °C^-1.

Given,

• Number of moles, μ = 1 mol
• Normal pressure, P = 1.013 × 10⁵ N m^-2
• Normal temperature, T = 0°C = 273 K
• Volume of 1 mole of a gas at N.T.P., V = 22.4 L = 22.4 × 10^-3 m³

According to Avogadro's hypothesis, at normal temperature and normal pressure the volume of 1 mole of any gas is 22.4 litre.

From the ideal gas equation PV = μRT,

$\text R = \frac{\text{PV}}{\mu \text T}$

Substituting the values,

$$\begin{aligned}\text R = \frac{(1.013 \times 10^5\ \text N\text m^{-2}) \times (22.4 \times 10^{-3}\ \text m^3)}{1\ \text{mol} \times 273\ \text K} \\[1em] = \frac{2269.12}{273}\ \text{N m mol}^{-1}\text K^{-1}\end{aligned}$$

$\text R = 8.31\ \text{J mol}^{-1}\text K^{-1}$

Hence, the value of the universal gas constant R is 8.31 J/(mol-K).

From the kinetic theory, the pressure of a gas is

$\text P = \frac{1}{3}\rho\overline{\text v^2}$

For 1 gram-molecule of the gas, if M is the molecular weight and V the volume, then the density is $\rho = \frac{\text M}{\text V}$. Hence

$\text P = \frac{1}{3}\frac{\text M}{\text V}\overline{\text v^2} \quad \Rightarrow \quad \overline{\text v^2} = \frac{3\text{PV}}{\text M}$

The average kinetic energy of 1 gram-molecule of the gas is

$\frac{1}{2}\text M\overline{\text v^2} = \frac{1}{2}\text M\left(\frac{3\text P}{\rho}\right) = \frac{1}{2}\text M\left(\frac{3\text{RT}}{\rho \text V}\right) \qquad [\because\ \text{PV} = \text{RT}]$

Since M = ρV,

$\frac{1}{2}\text M\overline{\text v^2} = \frac{3}{2}\text{RT}$

There are N (Avogadro's number) molecules in 1 gram-molecule of the gas. Hence the average kinetic energy of one molecule is

$\overline{\text E} = \frac{(3/2)\text{RT}}{\text N} = \frac{3}{2}\left(\frac{\text R}{\text N}\right)\text T = \frac{3}{2}\text{kT}$

where k (= R/N) is Boltzmann's constant.

Since $\frac{3}{2}$ k is a constant,

$\overline{\text E} \propto \text T$

Hence, the average kinetic energy of a molecule of an ideal gas is directly proportional to the absolute temperature of the gas.

Given,

• Atomic weight of helium, M_He = 4
• Molecular weight of hydrogen, M_H = 2
• Both the gases are in the same vessel at 30°C

The rms velocity of the molecules of a gas is

$\text v_{rms} = \sqrt{\frac{3\text{RT}}{\text M}}$

At the same temperature, the ratio of the rms velocities of the molecules of two different gases is inversely proportional to the square-root of their molecular weights,

$\frac{(\text v_{rms})_{\text{He}}}{(\text v_{rms})_{\text H}} = \sqrt{\frac{\text M_{\text H}}{\text M_{\text{He}}}}$

Substituting the values,

$\frac{(\text v_{rms})_{\text{He}}}{(\text v_{rms})_{\text H}} = \sqrt{\frac{2}{4}} = \frac{1}{\sqrt2}$

Hence, the root-mean-square velocities of the molecules of helium and hydrogen are in the ratio $1 : \sqrt2$.

Given,

• Mass of a krypton molecule, m_Kr = 2.25 m_H
• Equal masses of the two gases are enclosed in the vessel
• The temperature is constant

The rms velocity of the molecules of a gas is

$\text v_{rms} = \sqrt{\frac{3\text{kT}}{\text m}} \quad \Rightarrow \quad \text v_{rms} \propto \frac{1}{\sqrt{\text m}}$

where m is the mass of one molecule. Hence, at the same temperature,

$\frac{(\text v_{rms})_{\text{Kr}}}{(\text v_{rms})_{\text H}} = \sqrt{\frac{\text m_{\text H}}{\text m_{\text{Kr}}}}$

Substituting m_Kr = 2.25 m_H,

$$\begin{aligned}\frac{(\text v_{rms})_{\text{Kr}}}{(\text v_{rms})_{\text H}} = \sqrt{\frac{\text m_{\text H}}{2.25\text m_{\text H}}} \\[1em] = \sqrt{\frac{1}{2.25}} = \frac{1}{1.5} = \frac{2}{3}\end{aligned}$$

Hence, the root-mean-square velocities of the molecules of krypton and hydrogen are in the ratio 2 : 3.

The masses of the two gases taken do not affect the result, since the rms velocity depends only on the temperature and the mass of a molecule.

Both will have an equal number of molecules.

From the kinetic theory, for two gases occupying equal volumes V,

$\text{PV} = \frac{1}{3}\text m_1 \text n_1 \overline{\text v_1^2} \quad \text{and} \quad \text{PV} = \frac{1}{3}\text m_2 \text n_2 \overline{\text v_2^2}$

Since the two samples are at the same pressure and occupy the same volume,

$\text m_1 \text n_1 \overline{\text v_1^2} = \text m_2 \text n_2 \overline{\text v_2^2}$

Both the gases are at the same temperature, so their average kinetic energies of translation per molecule are equal,

$\frac{1}{2}\text m_1 \overline{\text v_1^2} = \frac{1}{2}\text m_2 \overline{\text v_2^2}$

Applying this result in the previous expression,

$\text n_1 = \text n_2$

Hence, 1 cm³ of hydrogen and 1 cm³ of oxygen at N.T.P. contain an equal number of molecules. This is Avogadro's law, which states that equal volumes of all gases under the same conditions of temperature and pressure contain an equal number of molecules.

The sensation of heat depends upon the quantity of heat absorbed by the skin, that is, upon the heat-density of the surroundings, and not upon the kinetic temperature alone.

As we go up in the atmosphere, the number of air molecules per unit volume goes on decreasing. Hence the quantity of heat per unit volume also goes on decreasing.

The kinetic temperature is the measure of the translational kinetic energy per molecule. In the upper part of the atmosphere the kinetic energy per molecule is quite large, so the kinetic temperature is high, but the molecules are so few that the heat-density is very low.

Hence, in the upper part of the atmosphere one feels severe cold although the kinetic temperature there is of the order of 1000 K.

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

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

so the graph of PV against P must be a straight line parallel to the P-axis for an ideal gas.

In the given graph, the line for the gas C is parallel to the P-axis, showing that the product PV keeps the same value at all pressures.

For the gas B the product PV increases with pressure, and for the gas A it decreases with pressure. Hence these two gases do not obey Boyle's law at all pressures and are real gases.

Hence, the gas C is ideal, because it obeys Boyle's law at all pressures.

Difference between real gases and an ideal gas :

Ideal gas	Real gas
It strictly obeys Boyle's law, Charles' law and the law of pressure under all conditions of temperature and pressure.	It obeys these laws only approximately, and the deviations are different for different gases.
Its molecules are infinitesimally small, so their actual volume is negligible.	The actual volume of its molecules is not always negligible compared with the volume of the gas.
There is no force of attraction between its molecules.	Its molecules attract one another, which is why a real gas can be liquefied.
Its pressure coefficient and volume coefficient are exactly equal to each other.	The two coefficients are not exactly equal.

Conditions under which a real gas behaves like an ideal gas :

At low pressure the volume of the gas is large, so the actual volume of the molecules is negligible in comparison with the volume of the gas.

At high temperature the molecules have large kinetic energy, so the effect of the intermolecular force on their motion is negligible.

Hence, at low pressure and high temperature the behaviour of a real gas is approximately ideal.

The two chief properties of the molecules of an ideal gas are that their volume is zero and that there is no mutual force between them.

At low pressure the volume of the gas is large, so the volume of the molecules is negligible in comparison with the volume of the gas.

At high temperature the molecules have large kinetic energy, so the effect of the intermolecular force on their motion is negligible.

Hence, at low pressure and high temperature the real gases obey the gas equation $\frac{\text{PV}}{\text{RT}}$ more correctly, that is, their behaviour is approximately ideal.

The mean free path of the molecules of a gas is

$\lambda = \frac{1}{\sqrt2\pi \text n \text d^2}$

where n is the number of molecules per unit volume and d is the diameter of a molecule. For N molecules of an ideal gas, PV = NkT gives

$\text n = \frac{\text N}{\text V} = \frac{\text P}{\text{kT}}$

Substituting this value of n,

$\lambda = \frac{\text{kT}}{\sqrt2\pi \text d^2 \text P}$

Dependence on temperature : $\lambda \propto \text T$, that is, the mean free path is directly proportional to the absolute temperature of the gas.

Dependence on pressure : $\lambda \propto \frac{1}{\text P}$, that is, the mean free path is inversely proportional to the pressure of the gas.

Avogadro's number : The Avogadro number is the number of molecules in 1 mole of a substance. It is denoted by N_A and its value is 6.0221 × 10²³.

Thus 1 mole of a substance may also be defined as the mass of the substance containing 6.0221 × 10²³ molecules.

Why it has the same value for all substances : Let 1 mole of a substance A have an absolute mass M_A in gram and let a molecule of A have an absolute mass m_A. A ₆C¹² atom has an absolute mass m_C and 1 mole of ₆C¹² has a mass of exactly 12 g. Since the mass of 1 mole is proportional to the mass of its molecule,

$\frac{\text M_{\text A}}{12} = \frac{\text m_{\text A}}{\text m_{\text C}} \quad \Rightarrow \quad \frac{\text M_{\text A}}{\text m_{\text A}} = \frac{12}{\text m_{\text C}}$

The ratio $\frac{\text M_{\text A}}{\text m_{\text A}}$ is the number of molecules in 1 mole of the substance A, which is the Avogadro number N_A. The right-hand side contains only the mass of a carbon-12 atom and does not depend on the substance A at all.

Hence, the Avogadro number is the same for all substances. It is essentially a counting unit fixed by convention, just as a dozen is.

The degrees of freedom of a particle indicate the number of independent motions which the particle can undergo, or the number of independent methods of exchanging energy.

For a gas molecule these independent motions may be translational, rotational or vibrational.

Monoatomic gas : The molecule consists of a single atom, and its translational motion can be resolved along the three coordinate axes. Hence it has three degrees of freedom, all translational.

Diatomic gas : The molecule is made up of two atoms joined rigidly. It can move bodily in three independent directions and can rotate about two of the three coordinate axes, its moment of inertia about the axis joining the two atoms being negligible. Hence it has five degrees of freedom, three translational and two rotational.

Polyatomic gas : The molecule can rotate about any of the three coordinate axes, so it has six degrees of freedom, three translational and three rotational.

(i) The collisions of gas molecules with the walls of the container

The molecules of a gas keep on colliding with the walls of the vessel and thus transfer momentum to them. A steady force therefore acts perpendicular to the walls, and the magnitude of this force per unit area of the walls is called the pressure of the gas.

(ii) It decreases by half

By Boyle's law, at constant temperature P₁V₁ = P₂V₂. Putting V₂ = 2V₁,

$\text P_2 = \frac{\text P_1 \text V_1}{2\text V_1} = \frac{\text P_1}{2}$

Hence, on doubling the volume the pressure falls to half its original value.

(iii) It increases

Charles' law states that at constant pressure the volume of a gas is directly proportional to its absolute temperature, V ∝ T. Hence, when the temperature is increased at constant pressure, the volume of the gas increases.

(iv) Gay-Lussac's law

Gay-Lussac's law, also called the pressure law, states that for a fixed mass of gas at constant volume the pressure is directly proportional to the absolute temperature, P ∝ T.

(v) Energy is distributed equally among all available degrees of freedom

According to the law of equipartition of energy, in a system in equilibrium at absolute temperature T, the average energy associated with each degree of freedom is $\frac{1}{2}$ kT, where k is Boltzmann's constant. The energy is therefore shared equally among all the available degrees of freedom.

(i) The average distance a molecule travels before colliding with another molecule

Between two successive collisions a molecule moves in a straight line with a constant speed, so it follows short zig-zag paths of different lengths. The mean of these free paths is called the mean free path.

(ii) It decreases

The mean free path is $\lambda = \frac{\text{kT}}{\sqrt2\pi \text d^2 \text P}$, so at constant temperature $\lambda \propto \frac{1}{\text P}$. On increasing the pressure the molecules are packed more closely together, so a molecule collides with another after covering a shorter distance.

(iii) Increasing temperature increases molecular speed

The root-mean-square speed of the molecules is $\text v_{rms} = \sqrt{\frac{3\text{RT}}{\text M}}$, that is, $\text v_{rms} \propto \sqrt{\text T}$. Hence a rise in temperature increases the speed of the molecules and also their collision frequency.

(iv) It increases

From the ideal gas law PV = nRT, at constant volume P ∝ T. Hence, when the temperature is increased and the volume is kept constant, the pressure of the gas increases.

(v) The temperature, pressure, and size of gas molecules

Since $\lambda = \frac{\text{kT}}{\sqrt2\pi \text d^2 \text P}$, the mean free path depends on the absolute temperature T, on the pressure P, and on the diameter d of the gas molecules.

(i) Pressure increases

From the kinetic theory, $\text P = \frac{1}{3}\rho \text v_{rms}^2$, so the pressure is proportional to the square of the rms speed. An increase in the rms speed makes the collisions of the molecules with the walls both more frequent and more forceful, and hence the pressure increases.

(ii) 3

The molecule of a monoatomic gas consists of a single atom whose translational motion can be resolved along the three coordinate axes. Hence it has three degrees of freedom, all translational.

(iii) $\frac{1}{2}$ kT

According to the law of equipartition of energy, the average energy associated with each degree of freedom of a particle in a system in equilibrium at absolute temperature T is $\frac{1}{2}$ kT, where k is Boltzmann's constant.

(iv) Less than translational energy

A diatomic molecule has three translational and two rotational degrees of freedom at room temperature. By the law of equipartition of energy, the energy in the translational modes is $3 \times \frac{1}{2}\text{kT} = \frac{3}{2}\text{kT}$, while the energy in the rotational modes is $2 \times \frac{1}{2}\text{kT} = \text{kT}$. Since $\text{kT} \lt \frac{3}{2}\text{kT}$, the rotational contribution is less than the translational contribution.

Note: The printed answer key marks option 2, "Equal to translational energy". A diatomic molecule has only two rotational degrees of freedom against three translational ones, so by the law of equipartition of energy the rotational contribution is kT against a translational contribution of $\frac{3}{2}$ kT. Option 3 is therefore the correct choice.*

(v) Vibrational

At ordinary temperatures the vibrational motion of a diatomic molecule does not occur. At high temperatures this motion is promoted, so vibrational degrees of freedom become available and contribute to the internal energy.

(i) Formula for the pressure of an ideal gas :

From the kinetic theory, the pressure of an ideal gas is

$\text P = \frac{1}{3}\frac{\text{mn}}{\text V}\text v_{rms}^2$

where

• m = mass of one molecule of the gas
• n = total number of molecules of the gas
• V = volume of the gas
• v_rms = root-mean-square speed of the molecules

(ii) Explanation of Boyle's law :

Boyle's law states that the product of the pressure and the volume of a given mass of gas at constant temperature is constant. From the kinetic theory,

$\text P = \frac{1}{3}\frac{\text{mn}}{\text V}\overline{\text v^2} \quad \Rightarrow \quad \text{PV} = \frac{1}{3}\text{mn}\overline{\text v^2}$

Here mn is the mass of the gas, which is constant for a given mass of gas. If the temperature remains constant, the mean-square speed $\overline{\text v^2}$ of the molecules also remains constant, because $\overline{\text v^2} \propto \text T$. Therefore the whole of the right-hand side is constant,

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

This is Boyle's law.

(iii) Given,

• Both the vessels are of the same volume and are filled with the same gas at the same temperature
• Ratio of the pressures, P₁ : P₂ = 1 : 2

The root-mean-square speed of the molecules of a gas is

$\text v_{rms} = \sqrt{\frac{3\text{RT}}{\text M}} \quad \Rightarrow \quad \text v_{rms} \propto \sqrt{\text T}$

so it depends only upon the absolute temperature of the gas and its molecular weight, and not upon the pressure. Since the same gas is at the same temperature in both the vessels,

$\frac{(\text v_{rms})_1}{(\text v_{rms})_2} = \frac{1}{1}$

Hence, the ratio of the root-mean-square speeds of the molecules in the two vessels is 1 : 1.

Ideal gas : In practice, gases do not obey Boyle's law, Charles' law and the law of pressure strictly under all conditions of temperature and pressure. Experiments show that as the pressure is decreased, different gases become more and more similar in behaviour and more and more obedient to the gas laws. So we may imagine a gas whose properties are similar to the properties of a real gas at infinitely low pressure. Such an imaginary gas is called a 'perfect gas' or an 'ideal gas'.

The following properties are imagined in a perfect gas :

(i) It strictly obeys Boyle's law, Charles' law and the law of pressure under all conditions of temperature and pressure.

(ii) Its pressure coefficient and volume coefficient are exactly equal to each other.

(iii) Its molecules are infinitesimally small.

(iv) There is no force of attraction between its molecules. Hence a perfect gas cannot be converted into the liquid or the solid state.

Conditions under which a gas can be assumed to be ideal :

At low pressure the volume of the gas is large, so the actual volume of the molecules is negligible in comparison with the volume of the gas.

At high temperature the molecules have large kinetic energy, so the effect of the intermolecular force on their motion is negligible.

Hence, at low pressure and high temperature a gas can be assumed to be an ideal gas. In practice, the gases which are difficult to liquefy, such as oxygen, nitrogen, hydrogen and helium, can be considered as perfect.

Gas constant for one gram-molecule : For 1 gram-molecule (1 mole) of a gas the equation of state is

$\text{PV} = \text{RT} \quad \Rightarrow \quad \text R = \frac{\text{PV}}{\text T}$

According to Avogadro's hypothesis, at normal temperature (0°C or 273 K) and normal pressure (1 atmosphere or 1.013 × 10⁵ N m^-2) the volume of 1 mole of a gas is 22.4 litre, that is, 22.4 × 10^-3 m³.

Substituting the values,

$$\begin{aligned}\text R = \frac{(1.013 \times 10^5\ \text N\text m^{-2}) \times (22.4 \times 10^{-3}\ \text m^3)}{1\ \text{mol} \times 273\ \text K} \\[1em] = 8.31\ \text{N m mol}^{-1}\text K^{-1}\end{aligned}$$

$\text R = 8.31\ \text{J mol}^{-1}\text K^{-1}$

Hence, the gas constant for one gram-molecule of a gas is 8.31 J mol^-1 K^-1. As it has the same value for all gases, it is called the universal gas constant.

Establishment of the equation PV = RT :

The volume, pressure and temperature of a given mass of a gas are inter-related. Suppose that initially the pressure, volume and absolute temperature of a given mass of a gas are P₁, V₁ and T₁, and that they finally change to P₂, V₂ and T₂. This change may be divided into two parts.

(1) Change of pressure at constant temperature : Let the temperature be kept constant at T₁ and the pressure be changed from P₁ to P₂. If the volume changes from V₁ to V', then by Boyle's law

$\text P_2 \text V' = \text P_1 \text V_1 \quad \Rightarrow \quad \text V' = \frac{\text P_1 \text V_1}{\text P_2} \qquad \ldots(\text i)$

(2) Change of temperature at constant pressure : Now let the pressure be kept constant at P₂ and the absolute temperature be changed from T₁ to T₂. If the volume changes from V' to V₂, then by Charles' law

$\frac{\text V'}{\text T_1} = \frac{\text V_2}{\text T_2} \quad \Rightarrow \quad \text V' = \frac{\text T_1 \text V_2}{\text T_2} \qquad \ldots(\text{ii})$

From equations (i) and (ii),

$\frac{\text P_1 \text V_1}{\text P_2} = \frac{\text T_1 \text V_2}{\text T_2} \quad \Rightarrow \quad \frac{\text P_1 \text V_1}{\text T_1} = \frac{\text P_2 \text V_2}{\text T_2}$

Hence, for a given mass of a gas the value of $\frac{\text{PV}}{\text T}$ remains constant. If this constant is taken as R for 1 gram-molecule of the gas, then

$\frac{\text{PV}}{\text T} = \text R \quad \Rightarrow \quad \text{PV} = \text{RT}$

Since the gas laws are strictly true only for an ideal gas, this equation is also strictly true only for an ideal gas, and it is therefore called the ideal gas equation. For μ moles of a gas it takes the form PV = μRT.

Dimensional formula for R : From the gas equation,

$\text R = \frac{\text{PV}}{\text T}$

The dimensions of pressure are [M L^-1 T^-2], those of volume are [L³] and those of temperature are [θ]. Therefore

$[\text R] = \frac{[\text M \text L^{-1}\text T^{-2}][\text L^3]}{[\theta]} = \frac{[\text M \text L^2 \text T^{-2}]}{[\theta]}$

$[\text R] = [\text M \text L^2 \text T^{-2}\theta^{-1}]$

Hence, the dimensional formula for R is [M L² T^-2 θ^-1], the mole being dimensionless.

Expression for the pressure of a gas :

Consider an ideal gas contained in a cubical vessel whose walls are perfectly elastic. Let l be the length of each side of the cube, m the mass of each molecule and n the total number of molecules in the gas. The vessel is placed so that its three edges lie along the rectangular coordinate axes.

Let v be the velocity of a molecule and let v_x, v_y and v_z be its components along the X-, Y- and Z-axes, so that

$\text v^2 = \text v_x^2 + \text v_y^2 + \text v_z^2$

Consider the motion of this molecule along the X-axis. When it collides with the face A, the collision being perfectly elastic, its velocity changes from v_x to −v_x. Hence the change in its momentum in one collision is

$\Delta \text p = -\text{mv}_x - (\text{mv}_x) = -2\text{mv}_x$

By the law of conservation of momentum, the momentum transferred to the wall A in one collision is + 2 mv_x.

After rebounding from A the molecule collides with the opposite wall and returns, so it covers a distance 2l between two successive collisions with A. Since its velocity is v_x, it takes $\frac{2\text l}{\text v_x}$ second between two collisions with A, that is, it collides with A $\frac{\text v_x}{2\text l}$ times per second.

Therefore the momentum transferred to the wall A in 1 second is

$= 2\text{mv}_x \times \frac{\text v_x}{2\text l} = \frac{\text m}{\text l}\text v_x^2$

According to Newton's law of motion, the rate of change of momentum is equal to the force exerted. Since there are n molecules, the total force exerted on the wall A is

$\frac{\text m}{\text l}\left(\text v_{x_1}^2 + \text v_{x_2}^2 + \ldots + \text v_{x_n}^2\right)$

The area of the wall is l², so the pressure on the wall A is

$$\begin{aligned}\text P = \frac{\text m}{\text l^3}\left(\text v_{x_1}^2 + \text v_{x_2}^2 + \ldots + \text v_{x_n}^2\right) \\[1em] = \frac{\text{mn}}{\text l^3}\left(\frac{\text v_{x_1}^2 + \text v_{x_2}^2 + \ldots + \text v_{x_n}^2}{\text n}\right)\end{aligned}$$

But l³ = V, the volume of the cube, and the bracketed quantity is the average value $\overline{\text v_x^2}$ of v_x² for all the n molecules. Therefore

$\text P = \frac{\text{mn}}{\text V}\overline{\text v_x^2} \qquad \ldots(\text i)$

Since v² = v_x² + v_y² + v_z² holds for every molecule, it also holds for the mean values,

$\overline{\text v^2} = \overline{\text v_x^2} + \overline{\text v_y^2} + \overline{\text v_z^2} \qquad \ldots(\text{ii})$

The directions of motion of the molecules are quite at random, that is, all directions are equally possible. Hence

$\overline{\text v_x^2} = \overline{\text v_y^2} = \overline{\text v_z^2}$

From equation (ii),

$\overline{\text v^2} = 3\overline{\text v_x^2} \quad \Rightarrow \quad \overline{\text v_x^2} = \frac{1}{3}\overline{\text v^2}$

Substituting this value of $\overline{\text v_x^2}$ in equation (i),

$\text P = \frac{1}{3}\frac{\text{mn}}{\text V}\overline{\text v^2} \qquad \ldots(\text{iii})$

This is the expression for the pressure of the gas, $\overline{\text v^2}$ being the mean-square velocity of the molecules. Since mn is the mass of the whole gas and V its volume, mn/V is the density ρ of the gas, and the above expression can be written as

$\text P = \frac{1}{3}\rho\overline{\text v^2}$

Relation between pressure and kinetic energy per unit volume :

The expression obtained above may be written as

$\text P = \frac{1}{3}\rho\overline{\text v^2} = \frac{2}{3}\left(\frac{1}{2}\rho\overline{\text v^2}\right)$

Here $\frac{1}{2}\rho\overline{\text v^2}$ is the translational kinetic energy of the gas per unit volume, since ρ is the mass of the gas per unit volume. Denoting it by E,

$\text P = \frac{2}{3}\text E$

Hence, the pressure of a gas is equal to two-thirds of its translational kinetic energy per unit volume.

Kinetic interpretation of temperature :

Suppose 1 mole of a gas is at an absolute temperature T and has a volume V. Let the mass of a molecule of the gas be m and the mean-square speed of its molecules be $\overline{\text v^2}$. There are N (Avogadro's number) molecules in 1 mole. According to the kinetic theory, the pressure of the gas is

$\text P = \frac{1}{3}\frac{\text{mN}}{\text V}\overline{\text v^2} \quad \Rightarrow \quad \text{PV} = \frac{1}{3}\text{mN}\overline{\text v^2}$

But mN, the mass of one molecule multiplied by the number of molecules in 1 mole, is the mass M of 1 mole of the gas. Therefore

$\text{PV} = \frac{1}{3}\text M\overline{\text v^2}$

The ideal gas equation for 1 mole is PV = RT. Hence

$\text{RT} = \frac{1}{3}\text M\overline{\text v^2} \quad \Rightarrow \quad \overline{\text v^2} = \frac{3\text{RT}}{\text M} \qquad \ldots(\text i)$

Taking the square-root,

$\text v_{rms} = \sqrt{\overline{\text v^2}} = \sqrt{\frac{3\text{RT}}{\text M}} \qquad \ldots(\text{ii})$

Since 3R/M is a constant for a given gas,

$\text v_{rms} \propto \sqrt{\text T}$

Hence, the root-mean-square velocity of the molecules of a gas is directly proportional to the square-root of the absolute temperature of the gas. This is the kinetic interpretation of temperature: the faster the motion of the molecules, the higher is the temperature of the gas.

It is also clear from this relation that if the absolute temperature becomes zero, the motion of the molecules ceases (v_rms = 0). Hence absolute zero is that temperature at which the motion of all the molecules of the gas stops. Since v_rms can never be negative, no temperature is possible below the absolute zero.

Mean kinetic energy of a molecule :

The kinetic energy of 1 mole of the gas is

$$\begin{aligned}\frac{1}{2}\text M\overline{\text v^2} = \frac{1}{2}\text M\left(\frac{3\text{RT}}{\text M}\right) \qquad [\text{from eq. (i)}] \\[1em] = \frac{3}{2}\text{RT}\end{aligned}$$

There are N molecules in 1 mole of the gas. Therefore the average kinetic energy of one molecule is

$\overline{\text E} = \frac{(3/2)\text{RT}}{\text N} = \frac{3}{2}\left(\frac{\text R}{\text N}\right)\text T$

The ratio R/N is a constant which is called Boltzmann's constant and is denoted by k. Hence

$\overline{\text E} = \frac{3}{2}\text{kT}$

Hence, the mean kinetic energy of a molecule of a gas is $\frac{3}{2}$ kT.

This expression does not contain the mass of the molecule, which means that different gases at the same temperature have the same average kinetic energy per molecule.

Proof : According to the kinetic theory, the pressure of a gas is

$\text P = \frac{1}{3}\frac{\text M}{\text V}\overline{\text v^2} \quad \Rightarrow \quad \text{PV} = \frac{1}{3}\text M\overline{\text v^2}$

where M is the molecular weight of the gas, that is, the mass of 1 mole, and V is the volume of 1 mole.

The ideal gas equation for 1 mole of the gas is

$\text{PV} = \text{RT}$

Comparing the two expressions,

$\frac{1}{3}\text M\overline{\text v^2} = \text{RT}$

$\overline{\text v^2} = \frac{3\text{RT}}{\text M}$

Since $\overline{\text v^2} = \text v_{rms}^2$,

$\text v_{rms}^2 = \frac{3\text{RT}}{\text M} \quad \Rightarrow \quad \frac{\text v_{rms}^2}{\text T} = \frac{3\text R}{\text M}$

Hence, $\frac{\text v_{rms}^2}{\text T} = \frac{3\text R}{\text M}$ is proved.

Change in the mean kinetic energy of oxygen :

Given,

• Initial temperature, T₁ = 0°C = 273 K
• Final temperature, T₂ = 273°C = 273 + 273 = 546 K

The mean kinetic energy of a molecule is $\overline{\text E} = \frac{3}{2}\text{kT}$, so $\overline{\text E} \propto \text T$. Therefore

$\frac{\overline{\text E}_2}{\overline{\text E}_1} = \frac{\text T_2}{\text T_1} = \frac{546}{273} = 2$

$\overline{\text E}_2 = 2\overline{\text E}_1$

Hence, when the temperature of oxygen changes from 0°C to 273°C, its mean kinetic energy becomes twice.

Degree of freedom : The degrees of freedom of a particle indicate the number of independent motions which the particle can undergo, or the number of independent methods of exchanging energy.

Monoatomic gas : The molecule of a monoatomic gas, such as He or Ar, consists of a single atom. Its translational motion can take place in any direction in space and can therefore be resolved along the three coordinate axes, so that it has three independent motions.

Hence, a monoatomic gas has three degrees of freedom, all translational.

A monoatomic molecule can rotate also, but it has such a small moment of inertia that its kinetic energy of rotation is insignificant. Therefore it possesses only kinetic energy of translation.

Diatomic gas : The molecule of a diatomic gas, such as H₂ or O₂, is made up of two atoms joined rigidly to one another through a bond. It cannot only move bodily, but can also rotate about any of the three coordinate axes. However, its moment of inertia about the axis joining the two atoms is negligible, so it can have only two rotational motions.

Hence, a diatomic molecule has five degrees of freedom, three with respect to translation and two with respect to rotation.

The diatomic molecule can vibrate also along the line joining the two atoms, but the vibrational motion does not usually occur at ordinary temperatures.

The number of degrees of freedom can also be obtained from the molar specific heat at constant volume, since $\text C_v = \frac{1}{2}\text{fR}$, that is,

$\text f = \frac{2\text C_v}{\text R}$

For a monoatomic gas $\text C_v = \frac{3}{2}\text R$, giving f = 3, and for a diatomic gas $\text C_v = \frac{5}{2}\text R$, giving f = 5.

Postulates of the kinetic theory of gases :

(i) Gas particles are in constant, random motion : Gas molecules move in continuous, random motion in all directions. They travel in straight lines until they collide with other gas molecules or with the walls of the container.

(ii) Negligible volume of gas molecules : The actual volume of the individual gas molecules is extremely small compared to the total volume of the container. Most of the space in a gas is empty, which allows gas molecules to be treated as point particles with no volume of their own.

(iii) No intermolecular forces : Gas molecules do not exert attractive or repulsive forces on each other, except during collisions. Between collisions the gas particles move independently of one another.

(iv) Elastic collisions : Collisions between the gas molecules, and between the molecules and the walls of the container, are perfectly elastic, so that no kinetic energy is lost in the collisions.

(v) Average kinetic energy is proportional to temperature : The average kinetic energy of the gas molecules is directly proportional to the absolute temperature of the gas.

(vi) Large number of molecules : A gas contains a very large number of molecules, which makes a statistical treatment of the properties of the gas possible.

(vii) Pressure is due to collisions : The pressure exerted by a gas on the walls of its container is due to the collisions of the gas molecules with the walls.

Expression for the pressure of a gas :

Consider an ideal gas contained in a cubical vessel whose walls are perfectly elastic. Let l be the length of each side of the cube, m the mass of each molecule and n the total number of molecules in the gas. The vessel is placed so that its three edges lie along the rectangular coordinate axes.

Let v be the velocity of a molecule and let v_x, v_y and v_z be its components along the X-, Y- and Z-axes, so that

$\text v^2 = \text v_x^2 + \text v_y^2 + \text v_z^2$

Consider the motion of this molecule along the X-axis. When it collides with the face A, the collision being perfectly elastic, its velocity changes from v_x to −v_x. Hence the change in its momentum in one collision is

$\Delta \text p = -\text{mv}_x - (\text{mv}_x) = -2\text{mv}_x$

By the law of conservation of momentum, the momentum transferred to the wall A in one collision is + 2 mv_x.

After rebounding from A the molecule collides with the opposite wall and returns, so it covers a distance 2l between two successive collisions with A. Since its velocity is v_x, it takes $\frac{2\text l}{\text v_x}$ second between two collisions with A, that is, it collides with A $\frac{\text v_x}{2\text l}$ times per second.

Therefore the momentum transferred to the wall A in 1 second is

$= 2\text{mv}_x \times \frac{\text v_x}{2\text l} = \frac{\text m}{\text l}\text v_x^2$

According to Newton's law of motion, the rate of change of momentum is equal to the force exerted. Since there are n molecules, the total force exerted on the wall A is

$\frac{\text m}{\text l}\left(\text v_{x_1}^2 + \text v_{x_2}^2 + \ldots + \text v_{x_n}^2\right)$

The area of the wall is l², so the pressure on the wall A is

$$\begin{aligned}\text P = \frac{\text m}{\text l^3}\left(\text v_{x_1}^2 + \text v_{x_2}^2 + \ldots + \text v_{x_n}^2\right) \\[1em] = \frac{\text{mn}}{\text l^3}\left(\frac{\text v_{x_1}^2 + \text v_{x_2}^2 + \ldots + \text v_{x_n}^2}{\text n}\right)\end{aligned}$$

But l³ = V, the volume of the cube, and the bracketed quantity is the average value $\overline{\text v_x^2}$ of v_x² for all the n molecules. Therefore

$\text P = \frac{\text{mn}}{\text V}\overline{\text v_x^2} \qquad \ldots(\text i)$

Since v² = v_x² + v_y² + v_z² holds for every molecule, it also holds for the mean values,

$\overline{\text v^2} = \overline{\text v_x^2} + \overline{\text v_y^2} + \overline{\text v_z^2} \qquad \ldots(\text{ii})$

The directions of motion of the molecules are quite at random, that is, all directions are equally possible. Hence

$\overline{\text v_x^2} = \overline{\text v_y^2} = \overline{\text v_z^2}$

From equation (ii),

$\overline{\text v^2} = 3\overline{\text v_x^2} \quad \Rightarrow \quad \overline{\text v_x^2} = \frac{1}{3}\overline{\text v^2}$

Substituting this value of $\overline{\text v_x^2}$ in equation (i),

$\text P = \frac{1}{3}\frac{\text{mn}}{\text V}\overline{\text v^2} \qquad \ldots(\text{iii})$

This is the expression for the pressure of the gas, $\overline{\text v^2}$ being the mean-square velocity of the molecules. Since mn is the mass of the whole gas and V its volume, mn/V is the density ρ of the gas, and the above expression can be written as

$\text P = \frac{1}{3}\rho\overline{\text v^2}$

Hence, the pressure of a gas on the basis of the kinetic theory is $\text P = \frac{1}{3}\frac{\text{mn}}{\text V}\overline{\text v^2} = \frac{1}{3}\rho\overline{\text v^2}$.

Postulates of the kinetic theory of gases :

(i) Gas particles are in constant, random motion : Gas molecules move in continuous, random motion in all directions. They travel in straight lines until they collide with other gas molecules or with the walls of the container.

(ii) Negligible volume of gas molecules : The actual volume of the individual gas molecules is extremely small compared to the total volume of the container. Most of the space in a gas is empty, which allows gas molecules to be treated as point particles with no volume of their own.

(iii) No intermolecular forces : Gas molecules do not exert attractive or repulsive forces on each other, except during collisions. Between collisions the gas particles move independently of one another.

(iv) Elastic collisions : Collisions between the gas molecules, and between the molecules and the walls of the container, are perfectly elastic, so that no kinetic energy is lost in the collisions.

(v) Average kinetic energy is proportional to temperature : The average kinetic energy of the gas molecules is directly proportional to the absolute temperature of the gas.

(vi) Large number of molecules : A gas contains a very large number of molecules, which makes a statistical treatment of the properties of the gas possible.

(vii) Pressure is due to collisions : The pressure exerted by a gas on the walls of its container is due to the collisions of the gas molecules with the walls.

Expression for the pressure of a gas :

Consider an ideal gas contained in a cubical vessel whose walls are perfectly elastic. Let l be the length of each side of the cube, m the mass of each molecule and n the total number of molecules in the gas. The vessel is placed so that its three edges lie along the rectangular coordinate axes.

Let v be the velocity of a molecule and let v_x, v_y and v_z be its components along the X-, Y- and Z-axes, so that

$\text v^2 = \text v_x^2 + \text v_y^2 + \text v_z^2$

Consider the motion of this molecule along the X-axis. When it collides with the face A, the collision being perfectly elastic, its velocity changes from v_x to −v_x. Hence the change in its momentum in one collision is

$\Delta \text p = -\text{mv}_x - (\text{mv}_x) = -2\text{mv}_x$

By the law of conservation of momentum, the momentum transferred to the wall A in one collision is + 2 mv_x.

After rebounding from A the molecule collides with the opposite wall and returns, so it covers a distance 2l between two successive collisions with A. Since its velocity is v_x, it takes $\frac{2\text l}{\text v_x}$ second between two collisions with A, that is, it collides with A $\frac{\text v_x}{2\text l}$ times per second.

Therefore the momentum transferred to the wall A in 1 second is

$= 2\text{mv}_x \times \frac{\text v_x}{2\text l} = \frac{\text m}{\text l}\text v_x^2$

According to Newton's law of motion, the rate of change of momentum is equal to the force exerted. Since there are n molecules, the total force exerted on the wall A is

$\frac{\text m}{\text l}\left(\text v_{x_1}^2 + \text v_{x_2}^2 + \ldots + \text v_{x_n}^2\right)$

The area of the wall is l², so the pressure on the wall A is

$$\begin{aligned}\text P = \frac{\text m}{\text l^3}\left(\text v_{x_1}^2 + \text v_{x_2}^2 + \ldots + \text v_{x_n}^2\right) \\[1em] = \frac{\text{mn}}{\text l^3}\left(\frac{\text v_{x_1}^2 + \text v_{x_2}^2 + \ldots + \text v_{x_n}^2}{\text n}\right)\end{aligned}$$

But l³ = V, the volume of the cube, and the bracketed quantity is the average value $\overline{\text v_x^2}$ of v_x² for all the n molecules. Therefore

$\text P = \frac{\text{mn}}{\text V}\overline{\text v_x^2} \qquad \ldots(\text i)$

Since v² = v_x² + v_y² + v_z² holds for every molecule, it also holds for the mean values,

$\overline{\text v^2} = \overline{\text v_x^2} + \overline{\text v_y^2} + \overline{\text v_z^2} \qquad \ldots(\text{ii})$

The directions of motion of the molecules are quite at random, that is, all directions are equally possible. Hence

$\overline{\text v_x^2} = \overline{\text v_y^2} = \overline{\text v_z^2}$

From equation (ii),

$\overline{\text v^2} = 3\overline{\text v_x^2} \quad \Rightarrow \quad \overline{\text v_x^2} = \frac{1}{3}\overline{\text v^2}$

Substituting this value of $\overline{\text v_x^2}$ in equation (i),

$\text P = \frac{1}{3}\frac{\text{mn}}{\text V}\overline{\text v^2} \qquad \ldots(\text{iii})$

This is the expression for the pressure of the gas, $\overline{\text v^2}$ being the mean-square velocity of the molecules. Since mn is the mass of the whole gas and V its volume, mn/V is the density ρ of the gas, and the above expression can be written as

$\text P = \frac{1}{3}\rho\overline{\text v^2}$

Pressure in terms of the translational kinetic energy per unit volume :

The above expression may be rewritten as

$\text P = \frac{1}{3}\rho\overline{\text v^2} = \frac{2}{3}\left(\frac{1}{2}\rho\overline{\text v^2}\right)$

Since ρ is the mass of the gas per unit volume, the quantity $\frac{1}{2}\rho\overline{\text v^2}$ is the translational kinetic energy of the gas per unit volume. Denoting it by E,

$\text P = \frac{2}{3}\text E$

Hence, the pressure P exerted by the gas on the walls of the container is $\frac{2}{3}$ times the translational kinetic energy per unit volume.

Formula for the pressure of an ideal gas : On the basis of the kinetic theory, the pressure of an ideal gas is

$\text P = \frac{1}{3}\frac{\text{mn}}{\text V}\overline{\text v^2}$

where

• m = mass of one molecule of the gas
• n = total number of molecules of the gas
• V = volume of the gas
• $\overline{\text v^2}$ = mean-square velocity of the molecules

Relation between the kinetic energy of the molecules and the temperature of the gas :

Consider 1 mole of the gas. It contains N (Avogadro's number) molecules, so putting n = N in the above formula,

$\text P = \frac{1}{3}\frac{\text{mN}}{\text V}\overline{\text v^2} \quad \Rightarrow \quad \text{PV} = \frac{1}{3}\text{mN}\overline{\text v^2}$

But mN is the mass M of 1 mole of the gas, so

$\text{PV} = \frac{1}{3}\text M\overline{\text v^2}$

The ideal gas equation for 1 mole is PV = RT. Comparing the two,

$\frac{1}{3}\text M\overline{\text v^2} = \text{RT} \quad \Rightarrow \quad \overline{\text v^2} = \frac{3\text{RT}}{\text M}$

The kinetic energy of 1 mole of the gas is therefore

$\frac{1}{2}\text M\overline{\text v^2} = \frac{1}{2}\text M\left(\frac{3\text{RT}}{\text M}\right) = \frac{3}{2}\text{RT}$

Dividing by the N molecules present in 1 mole, the average kinetic energy of one molecule is

$\overline{\text E} = \frac{(3/2)\text{RT}}{\text N} = \frac{3}{2}\left(\frac{\text R}{\text N}\right)\text T = \frac{3}{2}\text{kT}$

where k (= R/N) is Boltzmann's constant. Since $\frac{3}{2}$ k is a constant,

$\overline{\text E} \propto \text T$

Hence, the average kinetic energy of a molecule of a gas is directly proportional to the absolute temperature of the gas. As this expression contains no term for the mass of the molecule, different gases at the same temperature have the same average kinetic energy per molecule.

Law of equipartition of energy : According to this law, in a classical system of particles which is in equilibrium at absolute temperature T, the average internal (kinetic) energy per particle associated with each degree of freedom is $\frac{1}{2}$ kT, where k is Boltzmann's constant. If the particle has f degrees of freedom, its average kinetic energy would be $\frac{1}{2}$ f kT.

Proof that $\gamma = 1 + \frac{2}{\text f}$ :

Consider 1 mole of an ideal gas at absolute temperature T. It has N_A molecules, and since there are no intermolecular forces in an ideal gas there is no internal potential energy. Hence the internal energy of 1 mole of the gas is entirely kinetic,

$\text U = \text N_\text A \times \frac{1}{2}\text{fkT} = \frac{1}{2}\text{fRT} \qquad [\because\ \text k = \text R/\text N_\text A]$

Differentiating with respect to T,

$\frac{\text{dU}}{\text{dT}} = \frac{1}{2}\text{fR}$

Molar specific heat at constant volume : Suppose the gas is heated at constant volume until its temperature rises through dT. The heat given is dQ = C_v dT, and no external work is done as the volume remains constant. Therefore, by the first law of thermodynamics, dU = dQ − dW gives

$\text{dU} = \text C_v\text{dT} \quad \Rightarrow \quad \text C_v = \frac{\text{dU}}{\text{dT}}$

Substituting the value of dU/dT obtained above,

$\text C_v = \frac{1}{2}\text{fR}$

Molar specific heat at constant pressure : By Mayer's formula, C_p − C_v = R, so

$\text C_p = \text C_v + \text R = \frac{1}{2}\text{fR} + \text R = \left(\frac{\text f}{2} + 1\right)\text R$

Ratio of the two specific heats :

$\gamma = \frac{\text C_p}{\text C_v} = \frac{\left(\frac{\text f}{2} + 1\right)\text R}{\frac{\text f}{2}\text R} = \frac{\frac{\text f + 2}{2}}{\frac{\text f}{2}} = \frac{\text f + 2}{\text f}$

$\gamma = 1 + \frac{2}{\text f}$

Hence, for an ideal gas $\gamma = 1 + \frac{2}{\text f}$ is proved.

It follows that γ decreases with an increase in the number of degrees of freedom. For a monoatomic gas f = 3 gives γ = 1.67, for a diatomic gas f = 5 gives γ = 1.40, and for a triatomic gas f = 6 gives γ = 1.33.

(a) Boyle's law : According to this law, the product of the pressure and the volume of a given mass of gas at constant temperature is constant.

From the kinetic theory, the pressure of a given mass of an ideal gas is

$\text P = \frac{1}{3}\frac{\text{mn}}{\text V}\overline{\text v^2}$

where m is the mass of a molecule, n the total number of molecules, V the volume of the gas and $\overline{\text v^2}$ the mean-square speed of the molecules. Thus

$\text{PV} = \frac{1}{3}\text{mn}\overline{\text v^2}$

Here mn is the mass of the gas, which is constant for a given mass of gas. If the temperature remains constant, the mean-square speed $\overline{\text v^2}$ of the molecules also remains constant, since the average kinetic energy of a molecule depends only on the temperature. Therefore

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

This is Boyle's law.

(b) Charles' law : According to this law, the volume of a given mass of gas at constant pressure is directly proportional to the absolute temperature of the gas.

From the kinetic theory,

$\text P = \frac{1}{3}\frac{\text{mn}}{\text V}\overline{\text v^2}$

$\text V = \frac{2\text n}{3\text P}\left(\frac{1}{2}\text m\overline{\text v^2}\right)$

Since the average kinetic energy of one molecule is $\frac{1}{2}\text m\overline{\text v^2} = \frac{3}{2}\text{kT}$,

$\text V = \frac{2\text n}{3\text P}\left(\frac{3}{2}\text{kT}\right) = \frac{\text n}{\text P}\text{kT}$

If the pressure P is constant, then for a given mass of the gas n and k are also constant, so

$\text V \propto \text T$

This is Charles' law.

Given,

• Number of molecules, n = 4 × 10²⁴
• Temperature, T = 50 K
• Pressure, P = 0.03 atmospheric = 0.03 × 1.01 × 10⁵ = 3.03 × 10³ N/m²
• Boltzmann constant, k = 1.38 × 10^-23 J/K

For n molecules of a gas, the gas equation is

$\text{PV} = \text{nkT} \quad \Rightarrow \quad \text V = \frac{\text{nkT}}{\text P}$

Substituting the values,

$$\begin{aligned}\text V = \frac{(4 \times 10^{24}) \times (1.38 \times 10^{-23}) \times 50}{3.03 \times 10^3} \\[1em] = \frac{2.76 \times 10^3}{3.03 \times 10^3}\end{aligned}$$

$\text V = 0.91\ \text m^3$

Hence, the volume of the vessel is 0.91 m³.

Given,

• Gas constant, R = 8.3 × 10³ J/(k mol-K)
• Avogadro's number, N = 6.03 × 10²⁶/(k mol)

The Boltzmann constant is the gas constant per molecule,

$\text k = \frac{\text R}{\text N}$

Substituting the values,

$$\begin{aligned}\text k = \frac{8.3 \times 10^3\ \text{J (k mol)}^{-1}\text K^{-1}}{6.03 \times 10^{26}\ (\text{k mol})^{-1}} \\[1em] = 1.376 \times 10^{-23}\ \text{J/K}\end{aligned}$$

Hence, the value of the Boltzmann constant is 1.376 × 10^-23 J/K.

Given,

• Pressure, P = 1.2 × 10⁷ N/m²
• Temperature, T = 127°C = 127 + 273 = 400 K
• Volume, V = 2.0 litre = 2.0 × 10^-3 m³
• Boltzmann constant, k = 1.38 × 10^-23 J/K

For n molecules of a gas, the gas equation is

$\text{PV} = \text{nkT} \quad \Rightarrow \quad \text n = \frac{\text{PV}}{\text{kT}}$

Substituting the values,

$$\begin{aligned}\text n = \frac{(1.2 \times 10^7) \times (2.0 \times 10^{-3})}{(1.38 \times 10^{-23}) \times 400} \\[1em] = \frac{2.4 \times 10^4}{5.52 \times 10^{-21}}\end{aligned}$$

$\text n = 4.35 \times 10^{24}$

Hence, the number of molecules in the gas is 4.35 × 10²⁴.

Given,

• Residual pressure, P = 10^-11 N/m²
• Temperature, T = 27°C = 27 + 273 = 300 K
• Boltzmann constant, k = 1.38 × 10^-23 J/K

For n molecules of a gas, PV = nkT. Hence the number of molecules per unit volume is

$\frac{\text n}{\text V} = \frac{\text P}{\text{kT}}$

Substituting the values,

$$\begin{aligned}\frac{\text n}{\text V} = \frac{10^{-11}}{(1.38 \times 10^{-23}) \times 300} \\[1em] = \frac{10^{-11}}{4.14 \times 10^{-21}} \\[1em] = 2.415 \times 10^9\ \text{per m}^3\end{aligned}$$

Since 1 m³ = 10⁶ cm³, the number of molecules per cm³ is

$$\begin{aligned}= (2.415 \times 10^9) \times 10^{-6} \\[1em] = 2415\end{aligned}$$

Hence, 2415 molecules per cm³ still remain in the vessel.

Given,

• Density of the gas, ρ = 1.2 kg/m³
• Normal pressure, P = 1.0 × 10⁵ N/m²
• Temperature, T₁ = 0°C = 273 K

(i) Root-mean-square speed at 0°C :

From the kinetic theory, $\text P = \frac{1}{3}\rho\overline{\text v^2}$, so

$\text v_{rms} = \sqrt{\overline{\text v^2}} = \sqrt{\frac{3\text P}{\rho}}$

Substituting the values,

$$\begin{aligned}\text v_{rms} = \sqrt{\frac{3 \times (1.0 \times 10^5)}{1.2}} = \sqrt{2.5 \times 10^5} \\[1em] = 500\ \text{m/s}\end{aligned}$$

(ii) Temperature at which the speed becomes three times :

The rms speed is proportional to the square-root of the absolute temperature,

$\frac{\text v_2}{\text v_1} = \sqrt{\frac{\text T_2}{\text T_1}}$

Here v₂ = 3v₁, so

$3 = \sqrt{\frac{\text T_2}{273}} \quad \Rightarrow \quad \frac{\text T_2}{273} = 9$

$\text T_2 = 9 \times 273 = 2457\ \text K$

$\text t_2 = 2457 - 273 = 2184^\circ\text C$

Hence, the rms speed of the molecules at 0°C is 500 m/s, and the speed becomes three times at 2184°C.

Given,

• rms speed of oxygen molecules, (v_rms)_O = 150 m/s
• Molecular weight of oxygen, M_O = 32
• Molecular weight of hydrogen, M_H = 2
• Both the gases are at the same temperature

At the same temperature, the ratio of the rms speeds of the molecules of two different gases is inversely proportional to the square-root of their molecular weights,

$\frac{(\text v_{rms})_{\text H}}{(\text v_{rms})_{\text O}} = \sqrt{\frac{\text M_{\text O}}{\text M_{\text H}}}$

Substituting the values,

$\frac{(\text v_{rms})_{\text H}}{150} = \sqrt{\frac{32}{2}} = \sqrt{16} = 4$

$$\begin{aligned}(\text v_{rms})_{\text H} = 150 \times 4 \\[1em] = 600\ \text{m/s}\end{aligned}$$

Hence, the root-mean-square speed of hydrogen molecules at the same temperature is 600 m/s.

Given,

• Temperature, T = 27°C = 27 + 273 = 300 K
• Boltzmann constant, k = 1.38 × 10^-23 J/K

According to the kinetic theory, the mean kinetic energy of a gas molecule at absolute temperature T is

$\overline{\text E} = \frac{3}{2}\text{kT}$

Substituting the values,

$$\begin{aligned}\overline{\text E} = \frac{3}{2} \times (1.38 \times 10^{-23}) \times 300 \\[1em] = 6.21 \times 10^{-21}\ \text J\end{aligned}$$

Hence, the mean kinetic energy of a gas molecule at 27°C is 6.21 × 10^-21 J.

Given,

• Average kinetic energy of a molecule, $\overline{\text E}$ = 4.2 × 10^-19 J
• Boltzmann constant, k = 1.4 × 10^-23 J/K

The average kinetic energy of a molecule of any gas at absolute temperature T is

$\overline{\text E} = \frac{3}{2}\text{kT} \quad \Rightarrow \quad \text T = \frac{2\overline{\text E}}{3\text k}$

Substituting the values,

$$\begin{aligned}\text T = \frac{2 \times (4.2 \times 10^{-19})}{3 \times (1.4 \times 10^{-23})} \\[1em] = \frac{8.4 \times 10^{-19}}{4.2 \times 10^{-23}}\end{aligned}$$

$\text T = 2 \times 10^4\ \text K$

Hence, the required temperature is 2 × 10⁴ K.

Given,

• Initial temperature, T₁ = 27°C = 27 + 273 = 300 K
• Kinetic energy at T₁, E₁ = 3400 J
• Final temperature, T₂ = 327°C = 327 + 273 = 600 K

The kinetic energy of 1 gram-molecule of a gas is $\text E = \frac{3}{2}\text{RT}$, so E ∝ T. Therefore

$\frac{\text E_2}{\text E_1} = \frac{\text T_2}{\text T_1} \quad \Rightarrow \quad \text E_2 = \text E_1 \times \frac{\text T_2}{\text T_1}$

Substituting the values,

$$\begin{aligned}\text E_2 = 3400 \times \frac{600}{300} \\[1em] = 3400 \times 2 = 6800\ \text J\end{aligned}$$

Hence, the kinetic energy of 1 g-molecule of the gas at 327°C is 6800 J.

Given,

• Average kinetic energy at 27°C, $\overline{\text E}_1$ = 9.3 × 10^-21 J
• Initial temperature, T₁ = 27°C = 27 + 273 = 300 K
• Final temperature, T₂ = 227°C = 227 + 273 = 500 K
• Mass of a hydrogen molecule, m = 3.1 × 10^-27 kg

(i) Average kinetic energy at 227°C :

The average kinetic energy of a molecule is $\overline{\text E} = \frac{3}{2}\text{kT}$, so $\overline{\text E} \propto \text T$,

$\frac{\overline{\text E}_2}{\overline{\text E}_1} = \frac{\text T_2}{\text T_1} \quad \Rightarrow \quad \overline{\text E}_2 = \overline{\text E}_1 \times \frac{\text T_2}{\text T_1}$

Substituting the values,

$$\begin{aligned}\overline{\text E}_2 = (9.3 \times 10^{-21}) \times \frac{500}{300} \\[1em] = 15.5 \times 10^{-21}\ \text J = 1.55 \times 10^{-20}\ \text J\end{aligned}$$

(ii) Root-mean-square speed at 27°C :

The average kinetic energy of a molecule is also $\overline{\text E} = \frac{1}{2}\text m\overline{\text v^2}$, so

$\text v_{rms} = \sqrt{\overline{\text v^2}} = \sqrt{\frac{2\overline{\text E}_1}{\text m}}$

Substituting the values,

$$\begin{aligned}\text v_{rms} = \sqrt{\frac{2 \times (9.3 \times 10^{-21})}{3.1 \times 10^{-27}}} = \sqrt{6 \times 10^6} \\[1em] = 2.45 \times 10^3\ \text{m/s}\end{aligned}$$

Hence, the average kinetic energy at 227°C is 1.55 × 10^-20 J and the root-mean-square speed of a hydrogen molecule at 27°C is 2.45 × 10³ m/s.

Given,

• Degrees of freedom of a molecule of the gas, f = 5

For an ideal gas the molar specific heats are

$\text C_v = \frac{1}{2}\text{fR} \quad \text{and} \quad \text C_p = \left(\frac{\text f}{2} + 1\right)\text R$

Therefore the ratio of the two specific heats is

$\gamma = \frac{\text C_p}{\text C_v} = 1 + \frac{2}{\text f}$

Substituting f = 5,

$$\begin{aligned}\gamma = 1 + \frac{2}{5} = \frac{7}{5} \\[1em] = 1.4\end{aligned}$$

Hence, $\frac{\text C_p}{\text C_v}$ of the gas is 1.4.

Given,

• Volume of the gas, V = 1.00 cm³ = 1.00 × 10^-6 m³
• At N.T.P. the volume of 1 mole of a gas is 22.4 L = 22.4 × 10³ cm³
• Avogadro's number, N = 6.02 × 10²³/mol

(i) Number of molecules : Since 22.4 × 10³ cm³ of the gas contains 6.02 × 10²³ molecules, the number of molecules in 1.00 cm³ is

$$\begin{aligned}\text n = \frac{6.02 \times 10^{23}}{22.4 \times 10^3} \\[1em] = 2.68 \times 10^{19}\end{aligned}$$

(ii) Volume occupied by one molecule : In an ideal gas the volume of the molecules themselves is negligible, so the whole of the 1.00 cm³ is available to the 2.68 × 10¹⁹ molecules. Hence the average volume available to one molecule is

$$\begin{aligned}= \frac{1.00 \times 10^{-6}\ \text m^3}{2.68 \times 10^{19}} \\[1em] = 37.3 \times 10^{-27}\ \text m^3\end{aligned}$$

(iii) Average distance between two molecules : If this space is assumed to be cubical, the mean distance between two molecules is the side of the cube,

$$\begin{aligned}\text d = (37.3 \times 10^{-27})^{\frac{1}{3}} \\[1em] = 3.34 \times 10^{-9}\ \text m\end{aligned}$$

Hence, 1.00 cm³ of the gas at N.T.P. contains 2.68 × 10¹⁹ molecules, the volume available to one molecule is 37.3 × 10^-27 m³, and the average distance between two molecules is 3.34 × 10^-9 m.

Given,

• Temperature, T = 273 K
• Pressure, P = 1.60 × 10^-3 N/m²
• Boltzmann constant, k = 1.38 × 10^-23 J/K

(i) Number of molecules in unit volume : For n molecules of a gas, PV = nkT, so

$\frac{\text n}{\text V} = \frac{\text P}{\text{kT}}$

Substituting the values,

$$\begin{aligned}\frac{\text n}{\text V} = \frac{1.60 \times 10^{-3}}{(1.38 \times 10^{-23}) \times 273} \\[1em] = \frac{1.60 \times 10^{-3}}{3.767 \times 10^{-21}} \\[1em] = 4.25 \times 10^{17}\ \text{per m}^3\end{aligned}$$

(ii) Average distance between the molecules : The average volume available to one molecule is

$= \frac{1}{4.25 \times 10^{17}} = 2.353 \times 10^{-18}\ \text m^3$

If this space is assumed to be cubical, the average distance between two molecules is the side of the cube,

$$\begin{aligned}\text d = (2.353 \times 10^{-18})^{\frac{1}{3}} \\[1em] = 1.33 \times 10^{-6}\ \text m\end{aligned}$$

Hence, the number of molecules in unit volume is 4.25 × 10¹⁷ per m³ and the average distance between the molecules is 1.33 × 10^-6 m.

Given,

• Volume of the flask, V = 1 × 10^-3 m³
• Number of oxygen molecules, n = 3.0 × 10¹²
• Mass of one molecule of oxygen, m = 5.3 × 10^-26 kg
• Root-mean-square speed, v_rms = 400 m/s

From the kinetic theory, the pressure of the gas is

$\text P = \frac{1}{3}\frac{\text{mn}}{\text V}\text v_{rms}^2$

Substituting the values,

$$\begin{aligned}\text P = \frac{1}{3} \times \frac{(5.3 \times 10^{-26}) \times (3.0 \times 10^{12})}{1 \times 10^{-3}} \times (400)^2 \\[1em] = \frac{1}{3} \times (1.59 \times 10^{-10}) \times (1.6 \times 10^5) \\[1em] = \frac{2.544 \times 10^{-5}}{3}\end{aligned}$$

$\text P = 8.48 \times 10^{-6}\ \text N/\text m^2$

Hence, the pressure of the oxygen gas in the flask is 8.48 × 10^-6 N/m².

Given,

• Mass of the smoke particle, m = 5 × 10^-17 kg
• Normal temperature, T = 273 K
• Boltzmann constant, k = 1.38 × 10^-23 J/K

A smoke particle performing Brownian motion behaves like a very heavy molecule of the gas, so its mean kinetic energy is also $\frac{3}{2}$ kT,

$\frac{1}{2}\text m\text v_{rms}^2 = \frac{3}{2}\text{kT} \quad \Rightarrow \quad \text v_{rms} = \sqrt{\frac{3\text{kT}}{\text m}}$

Substituting the values,

$$\begin{aligned}\text v_{rms} = \sqrt{\frac{3 \times (1.38 \times 10^{-23}) \times 273}{5 \times 10^{-17}}} \\[1em] = \sqrt{\frac{1.13 \times 10^{-20}}{5 \times 10^{-17}}} = \sqrt{2.26 \times 10^{-4}}\end{aligned}$$

$\text v_{rms} = 1.5 \times 10^{-2}\ \text{m/s}$

Hence, the root-mean-square speed of the smoke particle is 1.5 × 10^-2 m/s.

Given,

• rms speed of oxygen molecules at 0°C, (v_rms)_O = 460 m/s
• Temperature of oxygen, T_O = 0°C = 273 K
• Temperature of helium, T_He = 40°C = 40 + 273 = 313 K
• Molecular weight of oxygen, M_O = 32 g/mol
• Molecular weight of helium, M_He = 4 g/mol

The rms speed of the molecules of a gas is

$\text v_{rms} = \sqrt{\frac{3\text{RT}}{\text M}}$

Taking the ratio for the two gases,

$\frac{(\text v_{rms})_{\text{He}}}{(\text v_{rms})_{\text O}} = \sqrt{\frac{\text T_{\text{He}}}{\text T_{\text O}} \times \frac{\text M_{\text O}}{\text M_{\text{He}}}}$

Substituting the values,

$$\begin{aligned}\frac{(\text v_{rms})_{\text{He}}}{460} = \sqrt{\frac{313}{273} \times \frac{32}{4}} = \sqrt{1.1465 \times 8} \\[1em] = \sqrt{9.172} = 3.028\end{aligned}$$

$$\begin{aligned}(\text v_{rms})_{\text{He}} = 460 \times 3.028 \\[1em] = 1394\ \text{m/s}\end{aligned}$$

Hence, the root-mean-square speed of the helium molecules at 40°C is 1394 m/s.

Given,

• Mean kinetic energy at 27°C, E₁ = 600 J
• Initial temperature, T₁ = 27°C = 27 + 273 = 300 K
• Final temperature, T₂ = 127°C = 127 + 273 = 400 K

The mean kinetic energy of a gas is $\text E = \frac{3}{2}\mu \text{RT}$, so E ∝ T for a given quantity of the gas. Therefore

$\frac{\text E_2}{\text E_1} = \frac{\text T_2}{\text T_1} \quad \Rightarrow \quad \text E_2 = \text E_1 \times \frac{\text T_2}{\text T_1}$

Substituting the values,

$$\begin{aligned}\text E_2 = 600 \times \frac{400}{300} \\[1em] = 800\ \text J\end{aligned}$$

Hence, the mean kinetic energy of 1 kg-mol of nitrogen at 127°C is 800 J.

Given,

• Number of moles, μ = 1 mol
• Temperature, T = 0°C = 273 K
• Gas constant, R = 8.31 J/(mol-K)

The kinetic energy of translation of 1 mole of a gas at absolute temperature T is

$\text E = \frac{3}{2}\text{RT}$

Substituting the values,

$$\begin{aligned}\text E = \frac{3}{2} \times 8.31 \times 273 \\[1em] = \frac{3 \times 2268.63}{2}\end{aligned}$$

$\text E = 3.4 \times 10^3\ \text J$

Hence, the kinetic energy of translation of 1 mole of the gas at 0°C is 3.4 × 10³ J.

Given,

• Temperature, T = 127°C = 127 + 273 = 400 K
• Boltzmann constant, k = 1.38 × 10^-23 J/(molecule-K)

Oxygen is a diatomic gas, so a molecule of oxygen has f = 5 degrees of freedom, three translational and two rotational. By the law of equipartition of energy, the total kinetic energy of one molecule is

$\text E = \frac{1}{2}\text{fkT} = \frac{5}{2}\text{kT}$

Substituting the values,

$$\begin{aligned}\text E = \frac{5}{2} \times (1.38 \times 10^{-23}) \times 400 \\[1em] = \frac{5 \times 5.52 \times 10^{-21}}{2}\end{aligned}$$

$\text E = 1.38 \times 10^{-20}\ \text J$

Hence, the total kinetic energy of an O₂ molecule at 127°C is 1.38 × 10^-20 J.

Given,

• Volume of the bulb, V = 500 cm³ = 500 × 10^-6 m³ = 5 × 10^-4 m³
• Pressure, h = 10^-3 mm of mercury = 10^-6 m of mercury
• Temperature, T = 27°C = 27 + 273 = 300 K
• Density of mercury, ρ = 13.6 × 10³ kg/m³
• g = 10 m/s²
• Boltzmann constant, k = 1.38 × 10^-23 J/K

Pressure inside the bulb :

$$\begin{aligned}\text P = \text h\rho \text g = (10^{-6}) \times (13.6 \times 10^3) \times 10 \\[1em] = 0.136\ \text N/\text m^2\end{aligned}$$

Number of molecules : For n molecules of a gas, PV = nkT, so

$\text n = \frac{\text{PV}}{\text{kT}}$

Substituting the values,

$$\begin{aligned}\text n = \frac{0.136 \times (5 \times 10^{-4})}{(1.38 \times 10^{-23}) \times 300} \\[1em] = \frac{6.8 \times 10^{-5}}{4.14 \times 10^{-21}}\end{aligned}$$

$\text n = 16.425 \times 10^{15}$

Hence, the number of molecules of air inside the bulb is 16.425 × 10¹⁵.

Given,

• Molecular weight of oxygen, M = 32 g/mol = 32 × 10^-3 kg/mol
• Temperature, T = 0°C = 273 K
• Pressure, P = 1 atmosphere = 1.01 × 10⁵ N/m²

At 0°C and 1 atmospheric pressure the volume of 1 gram-molecule of a gas is 22.4 L, that is, 2.24 × 10^-2 m³. Hence the density of oxygen is

$$\begin{aligned}\rho = \frac{\text M}{\text V} = \frac{32 \times 10^{-3}}{2.24 \times 10^{-2}} \\[1em] = 1.4286\ \text{kg/m}^3\end{aligned}$$

From the kinetic theory,

$\text v_{rms} = \sqrt{\frac{3\text P}{\rho}}$

Substituting the values,

$\text v_{rms} = \sqrt{\frac{3 \times (1.01 \times 10^5)}{1.4286}} = \sqrt{2.121 \times 10^5}$

$\text v_{rms} = 460.5\ \text{m/s}$

Hence, the root-mean-square speed of the oxygen molecules is 460.5 m/s.

Given,

• rms speed of helium atoms, v_rms = 1300 m/s
• Normal pressure, P = 1.01 × 10⁵ N/m²
• Normal temperature, T = 0°C = 273 K
• Atomic weight of helium, M = 4
• Avogadro's number, N = 6.02 × 10²³/mol

(i) Density of helium at 0°C :

From the kinetic theory, $\text v_{rms} = \sqrt{\frac{3\text P}{\rho}}$. Squaring both sides,

$\text v_{rms}^2 = \frac{3\text P}{\rho} \quad \Rightarrow \quad \rho = \frac{3\text P}{\text v_{rms}^2}$

Substituting the values,

$\rho = \frac{3 \times (1.01 \times 10^5)}{(1300)^2} = \frac{3.03 \times 10^5}{1.69 \times 10^6}$

$\rho = 0.179\ \text{kg/m}^3$

(ii) Mass of a helium atom :

The mass of one atom is the mass of 1 mole divided by Avogadro's number,

$$\begin{aligned}\text m = \frac{\text M}{\text N_\text A} = \frac{4\ \text g}{6.02 \times 10^{23}} \\[1em] = 6.64 \times 10^{-24}\ \text g\end{aligned}$$

Hence, the density of helium at 0°C is 0.179 kg/m³ and the mass of a helium atom is 6.64 × 10^-24 g.

Given,

• Density of hydrogen at N.T.P., ρ_H = 0.09 kg/m³
• Normal pressure, P = 1.01 × 10⁵ N/m²
• Oxygen is 16 times heavier than hydrogen

Root-mean-square speed of hydrogen molecules : From the kinetic theory,

$\text v_{rms} = \sqrt{\frac{3\text P}{\rho}}$

Substituting the values,

$(\text v_{rms})_{\text H} = \sqrt{\frac{3 \times (1.01 \times 10^5)}{0.09}} = \sqrt{3.367 \times 10^6}$

$(\text v_{rms})_{\text H} = 1.84 \times 10^3\ \text{m/s}$

Root-mean-square speed of oxygen molecules : At the same temperature the rms speed is inversely proportional to the square-root of the molecular weight,

$$\begin{aligned}\frac{(\text v_{rms})_{\text O}}{(\text v_{rms})_{\text H}} = \sqrt{\frac{\text M_{\text H}}{\text M_{\text O}}} \\[1em] = \sqrt{\frac{1}{16}} = \frac{1}{4}\end{aligned}$$

$$\begin{aligned}(\text v_{rms})_{\text O} = \frac{1.84 \times 10^3}{4} \\[1em] = 4.6 \times 10^2\ \text{m/s}\end{aligned}$$

Hence, the rms speed of hydrogen molecules is 1.84 × 10³ m/s and that of oxygen molecules is 4.6 × 10² m/s.

Given,

• Initial temperature, T₁ = 40 K
• Final rms speed, v₂ = 3v₁

The rms speed of the molecules of a gas is proportional to the square-root of the absolute temperature,

$\frac{\text v_2}{\text v_1} = \sqrt{\frac{\text T_2}{\text T_1}}$

Substituting v₂ = 3v₁,

$3 = \sqrt{\frac{\text T_2}{40}} \quad \Rightarrow \quad \frac{\text T_2}{40} = 9$

$\text T_2 = 9 \times 40 = 360\ \text K$

Hence, at 360 K the rms speed of the molecules will be thrice the rms speed at 40 K.

Given,

• Molecular weight of oxygen, M_O = 32
• Molecular weight of nitrogen, M_N = 28
• Temperature of nitrogen, T_N = 0°C = 273 K

The rms speed of the molecules of a gas is

$\text v_{rms} = \sqrt{\frac{3\text{RT}}{\text M}}$

The two speeds are to be equal, so

$\sqrt{\frac{3\text{RT}_{\text O}}{\text M_{\text O}}} = \sqrt{\frac{3\text{RT}_{\text N}}{\text M_{\text N}}}$

Squaring both sides and cancelling 3R,

$\frac{\text T_{\text O}}{\text M_{\text O}} = \frac{\text T_{\text N}}{\text M_{\text N}} \quad \Rightarrow \quad \text T_{\text O} = \text T_{\text N} \times \frac{\text M_{\text O}}{\text M_{\text N}}$

Substituting the values,

$$\begin{aligned}\text T_{\text O} = 273 \times \frac{32}{28} \\[1em] = 312\ \text K\end{aligned}$$

$\text t_{\text O} = 312 - 273 = 39^\circ\text C$

Hence, at 39°C the effective speed of an oxygen molecule will be equal to the effective speed of a nitrogen molecule at 0°C.

Given,

• Number of hydrogen molecules, n = 6 × 10²¹
• Volume of the vessel, V = 200 cm³
• Initial temperature, T₁ = 27°C = 27 + 273 = 300 K
• Initial pressure, P₁ = 10⁵ N/m²
• Final temperature, T₂ = 47°C = 47 + 273 = 320 K

(i) Number of molecules per unit volume : The vessel is closed, so neither the number of molecules nor the volume changes on heating. Hence

$\frac{\text n}{\text V} = \frac{6 \times 10^{21}}{200\ \text{cm}^3} = 3 \times 10^{19}\ \text{per cm}^3$

The number of molecules per unit volume will remain unchanged, that is, the ratio is 1 : 1.

(ii) Pressure of the gas : For n molecules, PV = nkT. Since n and V are constant, P ∝ T, so

$\frac{\text P_1}{\text P_2} = \frac{\text T_1}{\text T_2} = \frac{300}{320} = \frac{15}{16}$

The pressure will change in the ratio 15 : 16.

(iii) Average kinetic energy of hydrogen : The average kinetic energy of a molecule is $\frac{3}{2}$ kT, so E ∝ T,

$\frac{\text E_1}{\text E_2} = \frac{\text T_1}{\text T_2} = \frac{300}{320} = \frac{15}{16}$

The average kinetic energy will also change in the ratio 15 : 16.

Given,

• Molecular weight of hydrogen, M = 2 g/mol = 2 × 10^-3 kg/mol
• Temperature, T = 27°C = 27 + 273 = 300 K
• Gas constant, R = 8.31 J/(mol-K)

(i) Kinetic energy of one gram-molecule :

$$\begin{aligned}\text E = \frac{3}{2}\text{RT} = \frac{3}{2} \times 8.31 \times 300 \\[1em] = 3.74 \times 10^3\ \text J\end{aligned}$$

(ii) Kinetic energy of one gram of the gas : Since 1 gram-molecule of hydrogen has a mass of 2 g, the kinetic energy of 1 g of the gas is

$$\begin{aligned}\text E' = \frac{3}{2}\frac{\text{RT}}{\text M} = \frac{3.74 \times 10^3}{2} \\[1em] = 1.87 \times 10^3\ \text J\end{aligned}$$

(iii) Root-mean-square speed of the molecules :

$$\begin{aligned}\text v_{rms} = \sqrt{\frac{3\text{RT}}{\text M}} = \sqrt{\frac{3 \times 8.31 \times 300}{2 \times 10^{-3}}} \\[1em] = \sqrt{3.7395 \times 10^6}\end{aligned}$$

$\text v_{rms} = 1.93 \times 10^3\ \text{m/s}$

Hence, the kinetic energy of one gram-molecule is 3.74 × 10³ J, that of one gram of the gas is 1.87 × 10³ J, and the root-mean-square speed of the molecules is 1.93 × 10³ m/s.

Given,

• Energy of the first excited state above the ground state, E = 10.2 eV
• 1 eV = 1.6 × 10^-19 J
• Boltzmann constant, k = 1.38 × 10^-23 J/K

Converting the energy into joule,

$$\begin{aligned}\text E = 10.2 \times (1.6 \times 10^{-19}) \\[1em] = 1.632 \times 10^{-18}\ \text J\end{aligned}$$

The hydrogen atom can go from the ground state to the first excited state when its average kinetic energy becomes equal to this energy. The average kinetic energy of an atom at absolute temperature T is $\frac{3}{2}$ kT, so

$\frac{3}{2}\text{kT} = \text E \quad \Rightarrow \quad \text T = \frac{2\text E}{3\text k}$

Substituting the values,

$$\begin{aligned}\text T = \frac{2 \times (1.632 \times 10^{-18})}{3 \times (1.38 \times 10^{-23})} \\[1em] = \frac{3.264 \times 10^{-18}}{4.14 \times 10^{-23}}\end{aligned}$$

$\text T = 7.88 \times 10^4\ \text K$

Hence, at 7.88 × 10⁴ K the hydrogen atom will go from its ground state to the first excited state.

Given,

• Temperature, T = 127°C = 127 + 273 = 400 K
• Boltzmann constant, k_B = 1.38 × 10^-23 J K^-1

Average translatory kinetic energy : A molecule has three translational degrees of freedom, so

$\text E_{trans} = \frac{3}{2}\text k_\text B \text T$

Substituting the values,

$$\begin{aligned}\text E_{trans} = \frac{3}{2} \times (1.38 \times 10^{-23}) \times 400 \\[1em] = 8.28 \times 10^{-21}\ \text J\end{aligned}$$

Total kinetic energy : Oxygen is a diatomic gas, so a molecule of oxygen has f = 5 degrees of freedom, three translational and two rotational. By the law of equipartition of energy,

$\text E_{total} = \frac{1}{2}\text{fk}<em>\text B \text T = \frac{5}{2}\text k</em>\text B \text T$

Substituting the values,

$$\begin{aligned}\text E_{total} = \frac{5}{2} \times (1.38 \times 10^{-23}) \times 400 \\[1em] = 1.38 \times 10^{-20}\ \text J\end{aligned}$$

Hence, the average translatory kinetic energy of an oxygen molecule at 127°C is 8.28 × 10^-21 J and its total kinetic energy is 1.38 × 10^-20 J.

Note: The unit of the Boltzmann constant is quoted in the question as J/mol-K but its unit is J K^-1.

Given,

• Mass of oxygen, m = 4.0 g
• Molecular weight of oxygen, M = 32 g/mol
• Temperature, T = 27°C = 27 + 273 = 300 K
• Number of degrees of freedom, f = 5
• Gas constant, R = 2.0 cal/mol-K

The number of moles of oxygen is

$\mu = \frac{\text m}{\text M} = \frac{4.0\ \text g}{32\ \text{g/mol}} = 0.125\ \text{mol}$

The total internal energy of μ moles of an ideal gas having f degrees of freedom is

$\text U = \frac{1}{2}\text f\mu \text{RT}$

Substituting the values,

$$\begin{aligned}\text U = \frac{1}{2} \times 5 \times 0.125 \times 2.0 \times 300 \\[1em] = \frac{5 \times 0.125 \times 600}{2}\end{aligned}$$

$\text U = 187.5\ \text{cal}$

Hence, the total internal energy of 4.0 g of oxygen at 27°C is 187.5 cal.

Given,

• Initial temperature, T₁ = −73°C = −73 + 273 = 200 K

(i) When the root-mean-square speed is doubled :

The rms speed is proportional to the square-root of the absolute temperature,

$\frac{\text v_2}{\text v_1} = \sqrt{\frac{\text T_2}{\text T_1}}$

Putting v₂ = 2v₁,

$2 = \sqrt{\frac{\text T_2}{200}} \quad \Rightarrow \quad \frac{\text T_2}{200} = 4$

$\text T_2 = 4 \times 200 = 800\ \text K$

$\text t_2 = 800 - 273 = 527^\circ\text C$

(ii) When the average kinetic energy is doubled :

The average kinetic energy of a molecule is $\frac{3}{2}$ kT, so E ∝ T,

$\frac{\text E_2}{\text E_1} = \frac{\text T_2}{\text T_1} \quad \Rightarrow \quad 2 = \frac{\text T_2}{200}$

$\text T_2 = 2 \times 200 = 400\ \text K$

$\text t_2 = 400 - 273 = 127^\circ\text C$

Hence, the gas should be heated to 527°C for the rms speed to be doubled and to 127°C for the average kinetic energy to be doubled.

8.360 × 10⁴ K

Reason — Given,

• Mass of an oxygen molecule, m = 2.76 × 10^-26 kg
• Boltzmann constant, k = 1.38 × 10^-23 J K^-1
• Escape velocity from the Earth, v_e = 11.2 km/s = 11.2 × 10³ m/s

The rms speed of the molecules is just sufficient for escaping when it becomes equal to the escape velocity,

$\text v_{rms} = \text v_e \quad \Rightarrow \quad \sqrt{\frac{3\text{kT}}{\text m}} = \text v_e$

Squaring both sides,

$\frac{3\text{kT}}{\text m} = \text v_e^2 \quad \Rightarrow \quad \text T = \frac{\text m\text v_e^2}{3\text k}$

Substituting the values,

$$\begin{aligned}\text T = \frac{(2.76 \times 10^{-26}) \times (11.2 \times 10^3)^2}{3 \times (1.38 \times 10^{-23})} \\[1em] = \frac{(2.76 \times 10^{-26}) \times (1.2544 \times 10^8)}{4.14 \times 10^{-23}} \\[1em] = \frac{3.462 \times 10^{-18}}{4.14 \times 10^{-23}}\end{aligned}$$

$\text T = 8.360 \times 10^4\ \text K$

35 J

Reason — Given,

• Work done by the gas at constant pressure, W = 10 J
• The gas is diatomic with rigid molecules, so f = 5 and $\text C_p = \frac{7}{2}\text R$

The work done by a gas expanding at constant pressure is

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

For n moles of an ideal gas, PV = nRT, so at constant pressure PΔV = nRΔT. Therefore

$\text{nR}\Delta \text T = \text W = 10\ \text J$

The heat absorbed at constant pressure is

$\Delta \text Q = \text n\text C_p \Delta \text T = \text n\left(\frac{7}{2}\text R\right)\Delta \text T = \frac{7}{2}(\text{nR}\Delta \text T)$

Substituting nRΔT = 10 J,

$$\begin{aligned}\Delta \text Q = \frac{7}{2} \times 10 \\[1em] = 35\ \text J\end{aligned}$$

17.4 J/mol-K

Reason — Given,

• Helium (monoatomic), n₁ = 2 mol, $\text C_{v_1} = \frac{3}{2}\text R$
• Hydrogen (diatomic, rigid), n₂ = 3 mol, $\text C_{v_2} = \frac{5}{2}\text R$
• R = 8.3 J/mol-K

The heat required to raise the temperature of the mixture through ΔT is the sum of the heats required for the two gases,

$\text n(\text C_v)_{mix}\Delta \text T = \text n_1 \text C_{v_1}\Delta \text T + \text n_2 \text C_{v_2}\Delta \text T$

$\text n(\text C_v)_{mix} = \text n_1 \text C_{v_1} + \text n_2 \text C_{v_2}$

where n = n₁ + n₂ = 2 + 3 = 5 mol.

Substituting the values,

$$\begin{aligned}5(\text C_v)_{mix} = 2 \times \frac{3}{2}\text R + 3 \times \frac{5}{2}\text R \\[1em] = 3\text R + 7.5\text R = 10.5\text R\end{aligned}$$

$(\text C_v)_{mix} = \frac{10.5\text R}{5} = \frac{10.5 \times 8.3}{5}$

$(\text C_v)_{mix} = 17.4\ \text{J/mol-K}$

$\frac{5}{4}\frac{\text P_0 \text V_0}{\text R}$

Reason — Given,

• Number of moles = 1
• $\text P = \text P_0\left[1 - \frac{1}{2}\left(\frac{\text V_0}{\text V}\right)^2\right]$
• The volume changes from V₀ to 2V₀

For 1 mole of an ideal gas, PV = RT, so

$\text T = \frac{\text{PV}}{\text R}$

Substituting the given value of P,

$\text T = \frac{\text P_0 \text V}{\text R}\left[1 - \frac{1}{2}\left(\frac{\text V_0}{\text V}\right)^2\right]$

When the volume is V₀ :

$$\begin{aligned}\text T_1 = \frac{\text P_0 \text V_0}{\text R}\left[1 - \frac{1}{2}\left(\frac{\text V_0}{\text V_0}\right)^2\right] = \frac{\text P_0 \text V_0}{\text R}\left[1 - \frac{1}{2}\right] \\[1em] = \frac{\text P_0 \text V_0}{2\text R}\end{aligned}$$

When the volume is 2V₀ :

$$\begin{aligned}\text T_2 = \frac{\text P_0 (2\text V_0)}{\text R}\left[1 - \frac{1}{2}\left(\frac{\text V_0}{2\text V_0}\right)^2\right] = \frac{2\text P_0 \text V_0}{\text R}\left[1 - \frac{1}{8}\right] \\[1em] = \frac{2\text P_0 \text V_0}{\text R} \times \frac{7}{8} = \frac{7}{4}\frac{\text P_0 \text V_0}{\text R}\end{aligned}$$

Change in temperature :

$$\begin{aligned}\Delta \text T = \text T_2 - \text T_1 = \frac{\text P_0 \text V_0}{\text R}\left[\frac{7}{4} - \frac{1}{2}\right] \\[1em] = \frac{5}{4}\frac{\text P_0 \text V_0}{\text R}\end{aligned}$$

748 J

Reason — Given,

• Capacity of the cylinder, V = 67.2 L (fixed)
• The gas is helium (monoatomic) at S.T.P., so $\text C_v = \frac{3}{2}\text R$
• Rise in temperature, ΔT = 20°C = 20 K
• R = 8.31 J mol^-1 K^-1

At S.T.P. the volume of 1 mole of any ideal gas is 22.4 L. Hence the number of moles of helium in the cylinder is

$\text n = \frac{67.2\ \text L}{22.4\ \text L} = 3\ \text{mol}$

The capacity of the cylinder is fixed, so the heat is supplied at constant volume,

$\text Q = \text n\text C_v\Delta \text T = \text n\left(\frac{3}{2}\text R\right)\Delta \text T$

Substituting the values,

$$\begin{aligned}\text Q = 3 \times \frac{3}{2} \times 8.31 \times 20 \\[1em] = 90 \times 8.31 = 747.9\ \text J\end{aligned}$$

$\text Q \approx 748\ \text J$

10⁴ K

Reason — Given,

• Boltzmann constant, K_B = 1.38 × 10^-23 J/K
• Avogadro number, N_A = 6.02 × 10²⁶/kg
• Radius of the earth, R_e = 6.4 × 10⁶ m
• g = 10 m/s²
• Molecular mass of hydrogen = 2

The mass of one hydrogen molecule is

$\text m = \frac{2}{6.02 \times 10^{26}}\ \text{kg} = 3.32 \times 10^{-27}\ \text{kg}$

The rms velocity of the molecules and the escape velocity from the earth are

$\text v_{rms} = \sqrt{\frac{3\text K_\text B \text T}{\text m}} \quad \text{and} \quad \text v_e = \sqrt{2\text{gR}_e}$

Equating the two and squaring,

$\frac{3\text K_\text B \text T}{\text m} = 2\text{gR}_e \quad \Rightarrow \quad \text T = \frac{2\text{gR}_e \text m}{3\text K_\text B}$

Substituting the values,

$$\begin{aligned}\text T = \frac{2 \times 10 \times (6.4 \times 10^6) \times (3.32 \times 10^{-27})}{3 \times (1.38 \times 10^{-23})} \\[1em] = \frac{4.25 \times 10^{-19}}{4.14 \times 10^{-23}}\end{aligned}$$

$\text T \approx 10^4\ \text K$

9 × 10⁶ J

Reason — Given,

• Volume, V = 2 m³
• Pressure, P = 3 × 10⁶ Pa

The internal energy of n moles of an ideal gas at absolute temperature T is

$\text U = \frac{3}{2}\text{nRT}$

But for n moles PV = nRT, so nRT may be replaced by PV,

$\text U = \frac{3}{2}\text{PV}$

Substituting the values,

$$\begin{aligned}\text U = \frac{3}{2} \times (3 \times 10^6) \times 2 \\[1em] = 9 \times 10^6\ \text J\end{aligned}$$

15 RT

Reason — Given,

• Oxygen : 3 moles, diatomic, so considering translational and rotational modes f = 5
• Argon : 5 moles, monoatomic, so f = 3
• Temperature = T

The internal energy of n moles of a gas having f degrees of freedom is

$\text U = \frac{1}{2}\text{nfRT}$

For oxygen :

$\text U_{\text{O}_2} = \frac{1}{2} \times 3 \times 5 \times \text{RT} = 7.5\ \text{RT}$

For argon :

$\text U_{\text{Ar}} = \frac{1}{2} \times 5 \times 3 \times \text{RT} = 7.5\ \text{RT}$

Total internal energy of the system :

$$\begin{aligned}\text U = \text U_{\text{O}_2} + \text U_{\text{Ar}} = 7.5\ \text{RT} + 7.5\ \text{RT} \\[1em] = 15\ \text{RT}\end{aligned}$$

5.6 m³

Reason — Given,

• Mass of water, m = 4.5 kg = 4.5 × 10³ g
• Molecular mass of water, M = 18
• At STP, P = 1.01 × 10⁵ N/m² and T = 273 K
• R = 8.3 J mol^-1 K^-1

If the intermolecular forces vanish away, the water molecules no longer hold together and the whole of the water exists in the vapour phase, that is, it behaves like an ideal gas.

The number of moles of water is

$\text n = \frac{\text m}{\text M} = \frac{4.5 \times 10^3}{18} = 250\ \text{mol}$

From the ideal gas equation PV = nRT,

$\text V = \frac{\text{nRT}}{\text P}$

Substituting the values,

$$\begin{aligned}\text V = \frac{250 \times 8.3 \times 273}{1.01 \times 10^5} \\[1em] = \frac{5.665 \times 10^5}{1.01 \times 10^5}\end{aligned}$$

$\text V = 5.6\ \text m^3$

ρ₁ > ρ₂

Reason — Refer to the P-T diagram given in the question.

For n moles of an ideal gas of mass m and molecular mass M,

$\text{PV} = \text{nRT} = \frac{\text m}{\text M}\text{RT}$

$\text P = \left(\frac{\text m}{\text V}\right)\frac{\text{RT}}{\text M} \quad \Rightarrow \quad \text P = \frac{\rho \text{RT}}{\text M}$

Hence, on a P-T diagram the slope of the straight line is

$\frac{\text P}{\text T} = \frac{\rho \text R}{\text M} \quad \Rightarrow \quad \text{slope} \propto \rho$

In the given figure the line for ρ₁ has the greatest slope and that for ρ₃ the least. Therefore

$\rho_1 \gt \rho_2 \gt \rho_3$

Of the four options, only ρ₁ > ρ₂ agrees with this result.

P₁ > P₂ > P₃

Reason — Refer to the T-V curves given in the question.

For n moles of an ideal gas, PV = nRT, so

$\text T = \left(\frac{\text P}{\text{nR}}\right)\text V$

Comparing this with the equation of a straight line y = mx, the slope of the T-V curve is

$\text{slope} = \frac{\text P}{\text{nR}} \quad \Rightarrow \quad \text{slope} \propto \text P$

In the given graph the curve drawn for P₁ has the greatest slope and that for P₃ the least. Therefore

$\text P_1 \gt \text P_2 \gt \text P_3$

0.116 kg

Reason — Given,

• Volume of the cylinder, V = 30 L = 30 × 10^-3 m³
• Initial number of moles, n_i = 18.20 mol
• Final gauge pressure, P_g = 11 atm
• Temperature, T = 27°C = 27 + 273 = 300 K
• $\text R = \frac{100}{12}$ J mol^-1 K^-1
• Molecular mass of O₂ = 32, 1 atm = 1.01 × 10⁵ N/m²

The gauge pressure is the pressure above the atmospheric pressure, so the final absolute pressure is

$$\begin{aligned}\text P_f = \text P_g + \text P_{atm} = 11 + 1 = 12\ \text{atm} \\[1em] = 12 \times 1.01 \times 10^5\ \text N\text m^{-2}\end{aligned}$$

From the gas equation P_fV = n_fRT, the number of moles left in the cylinder is

$\text n_f = \frac{\text P_f \text V}{\text{RT}}$

Substituting the values,

$$\begin{aligned}\text n_f = \frac{(12 \times 1.01 \times 10^5) \times (30 \times 10^{-3})}{\frac{100}{12} \times 300} \\[1em] = \frac{3.636 \times 10^4}{2.5 \times 10^3} = 14.544\ \text{mol}\end{aligned}$$

Therefore the number of moles withdrawn is

$\text n_i - \text n_f = 18.20 - 14.544 = 3.656\ \text{mol}$

The mass of oxygen withdrawn is

$$\begin{aligned}\text m = 3.656 \times \frac{32}{1000}\ \text{kg} \\[1em] = 0.116\ \text{kg}\end{aligned}$$

1.6 atm

Reason — Given,

• V₁ = 2 L, n₁ = 5 mol, P₁ = 1 atm
• V₂ = 3 L, n₂ = 4 mol, P₂ = 2 atm
• The partition is a thermal insulator

The chambers are separated by a thermal insulator, so no energy is lost to the surroundings. When the partition is removed, the total energy of the mixture is equal to the sum of the energies of the two gases,

$\text E_1 + \text E_2 = \text E_{mix}$

The internal energy of n moles of a gas having f degrees of freedom is $\text U = \frac{1}{2}\text{nfRT}$. Taking the same f for the gases,

$\frac{\text n_1 \text f\text{RT}_1}{2} + \frac{\text n_2 \text f\text{RT}_2}{2} = \frac{(\text n_1 + \text n_2)\text f\text{RT}}{2}$

Using nRT = PV for each gas and (n₁ + n₂)RT = P_mix(V₁ + V₂) for the mixture, and cancelling the common factor $\frac{\text f}{2}$,

$\text P_1 \text V_1 + \text P_2 \text V_2 = \text P_{mix}(\text V_1 + \text V_2)$

Substituting the values,

$$\begin{aligned}(1 \times 2) + (2 \times 3) = \text P_{mix} \times (2 + 3) \\[1em] 2 + 6 = 5\text P_{mix}\end{aligned}$$

$\text P_{mix} = \frac{8}{5} = 1.6\ \text{atm}$

3

Given,

• Gas A is monoatomic with 3 translational degrees of freedom
• Gas B is polyatomic with 3 translational and 3 rotational degrees of freedom and 1 vibrational mode

For an ideal gas the ratio of the two specific heats is

$\gamma = 1 + \frac{2}{\text f} = \frac{\text f + 2}{\text f}$

For gas A : The gas is monoatomic with only three translational degrees of freedom, so

$\text f_{\text A} = 3 \quad \Rightarrow \quad \gamma_{\text A} = \frac{3 + 2}{3} = \frac{5}{3}$

For gas B : A vibrational mode carries two terms of energy, one kinetic and one potential, so it contributes 2 to the number of degrees of freedom. Hence

$\text f_{\text B} = 3 + 3 + (2 \times 1) = 8$

$\gamma_{\text B} = \frac{8 + 2}{8} = \frac{10}{8}$

Ratio of the specific heat ratios :

$$\begin{aligned}\frac{\gamma_{\text A}}{\gamma_{\text B}} = \frac{\frac{5}{3}}{\frac{10}{8}} = \frac{5}{3} \times \frac{8}{10} \\[1em] = \frac{40}{30} = \frac{4}{3}\end{aligned}$$

Comparing with the given relation,

$1 + \frac{1}{\text n} = \frac{4}{3} \quad \Rightarrow \quad \frac{1}{\text n} = \frac{4}{3} - 1 = \frac{1}{3}$

$\text n = 3$

Hence, the value of n is 3.