Kinetic Theory — Competition Zone

At what temperature will the rms speed of oxygen molecules become just sufficient for escaping from the Earth's atmosphere? (Given mass of oxygen molecule (m) = 2.76 × 10^-26 kg, Boltzmann's constant k = 1.38 × 10^-23 JK^-1.)

1. 5.016 × 10⁴ K
2. 8.360 × 10⁴ K
3. 2.508 × 10⁴ K
4. 1.254 × 10⁴ K

Answer

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,

$\text v_{rms} = \text v_e \quad \Rightarrow \quad \sqrt{\dfrac{3\text{kT}}{\text m}} = \text v_e$

Squaring both sides,

$\dfrac{3\text{kT}}{\text m} = \text v_e^2 \quad \Rightarrow \quad \text T = \dfrac{\text m\text v_e^2}{3\text k}$

Substituting the values,

$\text T = \dfrac{(2.76 \times 10^{-26}) \times (11.2 \times 10^3)^2}{3 \times (1.38 \times 10^{-23})} \\[1em] = \dfrac{(2.76 \times 10^{-26}) \times (1.2544 \times 10^8)}{4.14 \times 10^{-23}} \\[1em] = \dfrac{3.462 \times 10^{-18}}{4.14 \times 10^{-23}}$

$\text T = 8.360 \times 10^4\ \text K$

A diatomic gas with rigid molecules does 10 J of work when expanded at constant pressure. What would be the heat energy absorbed by the gas, in this process?

1. 25 J
2. 35 J
3. 30 J
4. 40 J

Answer

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

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

$\Delta \text Q = \text n\text C_p \Delta \text T = \text n\left(\dfrac{7}{2}\text R\right)\Delta \text T = \dfrac{7}{2}(\text{nR}\Delta \text T)$

Substituting nRΔT = 10 J,

$\Delta \text Q = \dfrac{7}{2} \times 10 \\[1em] = 35\ \text J$

Two moles of helium gas are mixed with three moles of hydrogen molecules (taken to be rigid). What is the molar specific heat of mixture at constant volume? [R = 8.3 J/mol-K]

1. 19.7 J/mol-K
2. 15.7 J/mol-K
3. 17.4 J/mol-K
4. 21.6 J/mol-K

Answer

17.4 J/mol-K

Reason — Given,

• Helium (monoatomic), n₁ = 2 mol, $\text C_{v_1} = \dfrac{3}{2}\text R$
• Hydrogen (diatomic, rigid), n₂ = 3 mol, $\text C_{v_2} = \dfrac{5}{2}\text R$
• 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,

$5(\text C_v)_{mix} = 2 \times \dfrac{3}{2}\text R + 3 \times \dfrac{5}{2}\text R \\[1em] = 3\text R + 7.5\text R = 10.5\text R$

$(\text C_v)_{mix} = \dfrac{10.5\text R}{5} = \dfrac{10.5 \times 8.3}{5}$

$(\text C_v)_{mix} = 17.4\ \text{J/mol-K}$

One mole of an ideal gas passes through a process, where pressure and volume obey the relation

$\text P = \text P_0\left[1 - \dfrac{1}{2}\left(\dfrac{\text V_0}{\text V}\right)^2\right]$

Here P₀ and V₀ are constants. The change in the temperature of the gas, if its volume changes from V₀ to 2V₀, will be :

1. $\dfrac{1}{2}\dfrac{\text P_0 \text V_0}{\text R}$

2. $\dfrac{1}{4}\dfrac{\text P_0 \text V_0}{\text R}$

3. $\dfrac{3}{4}\dfrac{\text P_0 \text V_0}{\text R}$

4. $\dfrac{5}{4}\dfrac{\text P_0 \text V_0}{\text R}$

Answer

$\dfrac{5}{4}\dfrac{\text P_0 \text V_0}{\text R}$

Reason — Given,

• Number of moles = 1
• $\text P = \text P_0\left[1 - \dfrac{1}{2}\left(\dfrac{\text V_0}{\text V}\right)^2\right]$
• The volume changes from V₀ to 2V₀

For 1 mole of an ideal gas, PV = RT, so

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

Substituting the given value of P,

$\text T = \dfrac{\text P_0 \text V}{\text R}\left[1 - \dfrac{1}{2}\left(\dfrac{\text V_0}{\text V}\right)^2\right]$

When the volume is V₀ :

$\text T_1 = \dfrac{\text P_0 \text V_0}{\text R}\left[1 - \dfrac{1}{2}\left(\dfrac{\text V_0}{\text V_0}\right)^2\right] = \dfrac{\text P_0 \text V_0}{\text R}\left[1 - \dfrac{1}{2}\right] \\[1em] = \dfrac{\text P_0 \text V_0}{2\text R}$

When the volume is 2V₀ :

$\text T_2 = \dfrac{\text P_0 (2\text V_0)}{\text R}\left[1 - \dfrac{1}{2}\left(\dfrac{\text V_0}{2\text V_0}\right)^2\right] = \dfrac{2\text P_0 \text V_0}{\text R}\left[1 - \dfrac{1}{8}\right] \\[1em] = \dfrac{2\text P_0 \text V_0}{\text R} \times \dfrac{7}{8} = \dfrac{7}{4}\dfrac{\text P_0 \text V_0}{\text R}$

Change in temperature :

$\Delta \text T = \text T_2 - \text T_1 = \dfrac{\text P_0 \text V_0}{\text R}\left[\dfrac{7}{4} - \dfrac{1}{2}\right] \\[1em] = \dfrac{5}{4}\dfrac{\text P_0 \text V_0}{\text R}$

A cylinder with fixed capacity of 67.2 L contains helium gas at S.T.P. The amount of heat needed to raise the temperature of the gas by 20°C is : [Take, R = 8.31 J mol^-1 K^-1]

1. 700 J
2. 748 J
3. 374 J
4. 350 J

Answer

748 J

Reason — Given,

• Capacity of the cylinder, V = 67.2 L (fixed)
• The gas is helium (monoatomic) at S.T.P., so $\text C_v = \dfrac{3}{2}\text R$
• 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

$\text n = \dfrac{67.2\ \text L}{22.4\ \text L} = 3\ \text{mol}$

The capacity of the cylinder is fixed, so the heat is supplied at constant volume,

$\text Q = \text n\text C_v\Delta \text T = \text n\left(\dfrac{3}{2}\text R\right)\Delta \text T$

Substituting the values,

$\text Q = 3 \times \dfrac{3}{2} \times 8.31 \times 20 \\[1em] = 90 \times 8.31 = 747.9\ \text J$

$\text Q \approx 748\ \text J$

The temperature, at which the root-mean-square velocity of hydrogen molecules equals their escape velocity from the earth, is closest to : [Boltzmann constant K_B = 1.38 × 10^-23 J/K, Avogadro number N_A = 6.02 × 10²⁶ /kg. Radius of earth = 6.4 × 10⁶ m, g = 10 m/s²]

1. 10⁴ K
2. 650 K
3. 3 × 10⁵ K
4. 800 K

Answer

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

$\text m = \dfrac{2}{6.02 \times 10^{26}}\ \text{kg} = 3.32 \times 10^{-27}\ \text{kg}$

The rms velocity of the molecules and the escape velocity from the earth are

$\text v_{rms} = \sqrt{\dfrac{3\text K_\text B \text T}{\text m}} \quad \text{and} \quad \text v_e = \sqrt{2\text{gR}_e}$

Equating the two and squaring,

$\dfrac{3\text K_\text B \text T}{\text m} = 2\text{gR}_e \quad \Rightarrow \quad \text T = \dfrac{2\text{gR}_e \text m}{3\text K_\text B}$

Substituting the values,

$\text T = \dfrac{2 \times 10 \times (6.4 \times 10^6) \times (3.32 \times 10^{-27})}{3 \times (1.38 \times 10^{-23})} \\[1em] = \dfrac{4.25 \times 10^{-19}}{4.14 \times 10^{-23}}$

$\text T \approx 10^4\ \text K$

An ideal gas occupies a volume of 2m³ at a pressure of 3 × 10⁶ Pa. The energy of the gas is :

1. 6 × 10⁴ J
2. 10⁸ J
3. 9 × 10⁶ J
4. 3 × 10² J

Answer

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

$\text U = \dfrac{3}{2}\text{nRT}$

But for n moles PV = nRT, so nRT may be replaced by PV,

$\text U = \dfrac{3}{2}\text{PV}$

Substituting the values,

$\text U = \dfrac{3}{2} \times (3 \times 10^6) \times 2 \\[1em] = 9 \times 10^6\ \text J$

A gas mixture consists of 3 moles of oxygen and 5 moles of argon at temperature T. Considering only translational and rotational modes, the total internal energy of the system is :

1. 12 RT
2. 15 RT
3. 20 RT
4. 4 RT

Answer

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

$\text U = \dfrac{1}{2}\text{nfRT}$

For oxygen :

$\text U_{\text{O}_2} = \dfrac{1}{2} \times 3 \times 5 \times \text{RT} = 7.5\ \text{RT}$

For argon :

$\text U_{\text{Ar}} = \dfrac{1}{2} \times 5 \times 3 \times \text{RT} = 7.5\ \text{RT}$

Total internal energy of the system :

$\text U = \text U_{\text{O}_2} + \text U_{\text{Ar}} = 7.5\ \text{RT} + 7.5\ \text{RT} \\[1em] = 15\ \text{RT}$

The volume occupied by the molecules contained in 4.5 kg water at STP, if the intermolecular forces vanish away is :

1. 5.6 × 10^-3 m³
2. 5.6 m³
3. 5.6 × 10⁶ m³
4. 5.6 × 10³ m³

Answer

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

$\text n = \dfrac{\text m}{\text M} = \dfrac{4.5 \times 10^3}{18} = 250\ \text{mol}$

From the ideal gas equation PV = nRT,

$\text V = \dfrac{\text{nRT}}{\text P}$

Substituting the values,

$\text V = \dfrac{250 \times 8.3 \times 273}{1.01 \times 10^5} \\[1em] = \dfrac{5.665 \times 10^5}{1.01 \times 10^5}$

$\text V = 5.6\ \text m^3$

P-T diagram of an ideal gas having three different densities ρ₁, ρ₂, ρ₃ (in three different cases) is shown in the figure. Which of the following is correct?

1. ρ₂ < ρ₃
2. ρ₁ > ρ₂
3. ρ₁ < ρ₂
4. ρ₁ = ρ₂ = ρ₃

Answer

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,

$\text{PV} = \text{nRT} = \dfrac{\text m}{\text M}\text{RT}$

$\text P = \left(\dfrac{\text m}{\text V}\right)\dfrac{\text{RT}}{\text M} \quad \Rightarrow \quad \text P = \dfrac{\rho \text{RT}}{\text M}$

Hence, on a P-T diagram the slope of the straight line is

$\dfrac{\text P}{\text T} = \dfrac{\rho \text R}{\text M} \quad \Rightarrow \quad \text{slope} \propto \rho$

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.

The graph represents the T-V curves of an ideal gas (where T is the temperature and V is the volume) at all three pressures P₁, P₂ and P₃ compared with those of Charles' Law represented as dotted lines. The correct relation is;

1. P₃ > P₂ > P₁
2. P₁ > P₃ > P₂
3. P₂ > P₁ > P₃
4. P₁ > P₂ > P₃

Answer

P₁ > P₂ > P₃

Reason — Refer to the T-V curves given in the question.

For n moles of an ideal gas, PV = nRT, so

$\text T = \left(\dfrac{\text P}{\text{nR}}\right)\text V$

Comparing this with the equation of a straight line y = mx, the slope of the T-V curve is

$\text{slope} = \dfrac{\text P}{\text{nR}} \quad \Rightarrow \quad \text{slope} \propto \text P$

In the given graph the curve drawn for P₁ has the greatest slope and that for P₃ the least. Therefore

$\text P_1 \gt \text P_2 \gt \text P_3$

An oxygen cylinder of volume 30 L has 18.20 moles of oxygen. After some oxygen is withdrawn from the cylinder, its gauge pressure drops to 11 atmospheric pressure at temperature 27°C. The mass of the oxygen withdrawn from the cylinder is nearly equal to: [Given, R = $\dfrac{100}{12}$ J mol^-1K^-1 and molecular mass of O₂ = 32, 1 atm pressure = 1.01 × 10⁵ N/m²]

1. 0.125 kg
2. 0.144 kg
3. 0.116 kg
4. 0.156 kg

Answer

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
• $\text R = \dfrac{100}{12}$ 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

$\text P_f = \text P_g + \text P_{atm} = 11 + 1 = 12\ \text{atm} \\[1em] = 12 \times 1.01 \times 10^5\ \text N\text m^{-2}$

From the gas equation P_fV = n_fRT, the number of moles left in the cylinder is

$\text n_f = \dfrac{\text P_f \text V}{\text{RT}}$

Substituting the values,

$\text n_f = \dfrac{(12 \times 1.01 \times 10^5) \times (30 \times 10^{-3})}{\dfrac{100}{12} \times 300} \\[1em] = \dfrac{3.636 \times 10^4}{2.5 \times 10^3} = 14.544\ \text{mol}$

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

$\text m = 3.656 \times \dfrac{32}{1000}\ \text{kg} \\[1em] = 0.116\ \text{kg}$

A container has two chambers of volumes V₁ = 2 L and V₂ = 3 L separated by a partition made of a thermal insulator. The chambers contain n₁ = 5 and n₂ = 4 moles of ideal gas at pressures P₁ = 1 atm and P₂ = 2 atm, respectively. When the partition is removed, the mixture attains an equilibrium pressure of:

1. 1.3 atm
2. 1.6 atm
3. 1.4 atm
4. 1.8 atm

Answer

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 $\text U = \dfrac{1}{2}\text{nfRT}$. Taking the same f for the gases,

$\dfrac{\text n_1 \text f\text{RT}_1}{2} + \dfrac{\text n_2 \text f\text{RT}_2}{2} = \dfrac{(\text n_1 + \text n_2)\text f\text{RT}}{2}$

Using nRT = PV for each gas and (n₁ + n₂)RT = P_mix(V₁ + V₂) for the mixture, and cancelling the common factor $\dfrac{\text f}{2}$,

$\text P_1 \text V_1 + \text P_2 \text V_2 = \text P_{mix}(\text V_1 + \text V_2)$

Substituting the values,

$(1 \times 2) + (2 \times 3) = \text P_{mix} \times (2 + 3) \\[1em] 2 + 6 = 5\text P_{mix}$

$\text P_{mix} = \dfrac{8}{5} = 1.6\ \text{atm}$

γ_A is the specific heat ratio of monoatomic gas A having 3 translational degrees of freedom. γ_B is the specific heat ratio of polyatomic gas B having 3 translational, 3 rotational degrees of freedom and 1 vibrational mode. If $\dfrac{γ_\text A}{γ_\text B} = \left(1 + \dfrac{1}{\text n}\right)$, then the value of n is ............... .

Answer

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

$\gamma = 1 + \dfrac{2}{\text f} = \dfrac{\text f + 2}{\text f}$

For gas A : The gas is monoatomic with only three translational degrees of freedom, so

$\text f_{\text A} = 3 \quad \Rightarrow \quad \gamma_{\text A} = \dfrac{3 + 2}{3} = \dfrac{5}{3}$

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

$\text f_{\text B} = 3 + 3 + (2 \times 1) = 8$

$\gamma_{\text B} = \dfrac{8 + 2}{8} = \dfrac{10}{8}$

Ratio of the specific heat ratios :

$\dfrac{\gamma_{\text A}}{\gamma_{\text B}} = \dfrac{\dfrac{5}{3}}{\dfrac{10}{8}} = \dfrac{5}{3} \times \dfrac{8}{10} \\[1em] = \dfrac{40}{30} = \dfrac{4}{3}$

Comparing with the given relation,

$1 + \dfrac{1}{\text n} = \dfrac{4}{3} \quad \Rightarrow \quad \dfrac{1}{\text n} = \dfrac{4}{3} - 1 = \dfrac{1}{3}$

$\text n = 3$

Hence, the value of n is 3.