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Multiple Choice Questions

One and More than One Correct

1. $C_v$ and $C_p$ denote the molar specific heat capacities of a gas at constant volume and constant pressure, respectively. Then [JEE 2009 P-I]

a. $C_p - C_v$ is larger for a diatomic ideal gas than for a monatomic ideal gas.

b. $C_p + C_v$ is larger for a diatomic ideal gas than for a monatomic ideal gas.

c. $C_p / C_v$ is larger for a diatomic ideal gas than for a monatomic ideal gas.

d. $C_p \cdot C_v$ is larger for a diatomic ideal gas than for a monatomic ideal gas.

Answer: (b, d)

Solution:

For Monatomic gas $C_v = \frac{3}{2}R$, $C_p = \frac{5}{2}R$

For diatomic gas $C_v = \frac{5}{2}R$, $C_p = \frac{7}{2}R$

2. A real gas behaves like an ideal gas if its [JEE 2010 P-I]

a. pressure and temperature are both high

b. pressure and temperature are both low

c. pressure is high and temperature is low

d. pressure is low and temperature is high

Answer: (d)

3. A mixture of 2 moles of helium gas (atomic mass $= 4\text{ amu}$) and 1 mole of argon gas (atomic mass $= 40\text{ amu}$) is kept at $300\text{ K}$ in a container. The ratio of the r m s speeds $\left(\frac{v_{\text{rms}}(\text{helium})}{v_{\text{rms}}(\text{argon})}\right)$ is: [JEE 2012 P-I]

a. 0.32

b. 0.45

c. 2.24

d. 3.16

Answer: (d)

Solution:

$$v_{\text{rms}} = \sqrt{\frac{3RT}{M}}$$ $$\Rightarrow \frac{v_{\text{rms}}(\text{He})}{v_{\text{rms}}(\text{Ar})} = \sqrt{\frac{M_{\text{Ar}}}{M_{\text{He}}}} = \sqrt{\frac{40}{4}} = \sqrt{10}$$ $$\Rightarrow \frac{v_{\text{rms}}(\text{He})}{v_{\text{rms}}(\text{Ar})} \approx 3.16$$

4. Two non-reactive monatomic ideal gases have their atomic masses in the ratio $2 : 3$. The ratio of their partial pressures, when enclosed in a vessel kept at a constant temperature, is $4 : 3$. The ratio of their densities is: [JEE Adv. 2013 P-I]

a. 1 : 4

b. 1 : 2

c. 6 : 9

d. 8 : 9

Answer: (d)

Solution:

$$m = \frac{PV}{RT} M$$ $$\Rightarrow P M = \rho R T \implies \rho \propto P M$$ $$\Rightarrow \frac{\rho_1}{\rho_2} = \left(\frac{P_1}{P_2}\right) \left(\frac{M_1}{M_2}\right) = \left(\frac{4}{3}\right) \left(\frac{2}{3}\right) = \frac{8}{9}$$

Here $\rho_1$ and $\rho_2$ are the densities of gases in the vessel containing the mixture.

5. One mole of diatomic ideal gas undergoes a cyclic process ABC as shown in figure. The process BC is adiabatic. The temperatures at A, B and C are $400\text{ K}$, $800\text{ K}$ and $600\text{ K}$ respectively. Choose the correct statement: [JEE Main 2014]

a. The change in internal energy in the process AB is –350 R.

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

c. The change in internal energy in whole cyclic process is 250 R.

d. The change in internal energy in the process CA is 700 R.

Answer: (b)

Solution:

AB: $\Delta U_{\text{AB}} = n C_v \Delta T = 1 \cdot \frac{5}{2} R (800 - 400) = 1000R$

CA: $\Delta U_{\text{CA}} = n C_v \Delta T = 1 \cdot \frac{5}{2} R (400 - 600) = -500R$

$$\Rightarrow \Delta U_{\text{cycle}} = 0$$ $$\Delta U_{\text{AB}} + \Delta U_{\text{BC}} + \Delta U_{\text{CA}} = 0$$ $$\Rightarrow 1000R + \Delta U_{\text{BC}} - 500R = 0$$ $$\Rightarrow \Delta U_{\text{BC}} = -500R$$

6. A ring of mass M and radius R is rotating with angular speed $\omega$ about a fixed vertical axis passing through its centre O with two point masses each of mass $\frac{M}{8}$ at rest at O. These masses can move radially outwards along two massless rods fixed on the ring as shown in the figure. At some instant the angular speed of the system is $\frac{8}{9}\omega$ and one of the masses is at a distance of $\frac{3}{5}R$ from O. At this instant the distance of the other mass from O is [JEE Adv. 2015 P-I]

a. $\frac{2}{3}R$

b. $\frac{1}{3}R$

c. $\frac{3}{5}R$

d. $\frac{4}{5}R$

Answer: (d)

Solution:

Using conservation of angular momentum

$$M R^2 \omega = \left( M R^2 + \frac{M}{8} \left(\frac{3}{5}R\right)^2 + \frac{M}{8} x^2 \right) \frac{8}{9}\omega$$ $$\Rightarrow x = \frac{4}{5}R$$

7. A container of fixed volume has a mixture of one mole of hydrogen and one mole of helium in equilibrium at temperature T. Assuming the gases are ideal, the correct statement(s) is(are) [JEE Adv. 2015 P-I]

a. The average energy per mole of the gas mixture is 2RT

b. The ratio of speed of sound in the gas mixture to that in helium gas is $\sqrt{6/5}$

c. The ratio of the rms speed of helium atoms to that of hydrogen molecules is 1/2

d. The ratio of the rms speed of helium atoms to that of hydrogen molecules is $1/\sqrt{2}$

Answer: (a, b, d)

Solution:

Average energy per mole

$$U = n_1 C_{v1} T + n_2 C_{v2} T = 1 \cdot \frac{5}{2} R T + 1 \cdot \frac{3}{2} R T = 4 R T$$ $$C_{v,\text{mix}} T = \frac{4RT}{2} = 2RT$$ $$C_{v,\text{mix}} = 2R$$ $$\frac{\gamma_{\text{mix}}}{\gamma_{\text{He}}} = \sqrt{\frac{C_{p,\text{mix}}}{C_{v,\text{mix}}} \times \frac{C_{v,\text{He}}}{C_{p,\text{He}}}} = \sqrt{\frac{3R}{2R} \times \frac{3/2}{5/2}} = \sqrt{\frac{6}{5}}$$ $$\frac{v_{\text{rms}}(\text{He})}{v_{\text{rms}}(\text{H}_2)} = \sqrt{\frac{M_{\text{H}_2}}{M_{\text{He}}}} = \sqrt{\frac{2}{4}} = \frac{1}{\sqrt{2}}$$

8. Consider an ideal gas confined in an isolated closed chamber. As the gas undergoes an adiabatic expansion, the average time of collision between molecules increases as $V^q$, where V is the volume of the gas. The value of $q$ is $\left( \text{where } \gamma = \frac{C_p}{C_v} \right)$: [JEE Main 2015]

a. $\frac{\gamma + 1}{2}$

b. $\frac{\gamma - 1}{2}$

c. $\frac{\gamma + 1}{2}$

d. $\frac{\gamma - 1}{2}$

Answer: (c)

Solution:

Average time taken between collisions

$$t \propto \frac{\lambda}{v_{\text{rms}}}$$

$\lambda = \text{Mean free path} = \frac{1}{\sqrt{2} \pi d^2 (N/V)}$

$N = \text{Number of molecules}$, $V = \text{volume}$

$$\therefore t \propto \frac{V}{\sqrt{2} \pi d^2 N \sqrt{\frac{3RT}{M}}} \propto \frac{V}{\sqrt{T}}$$

For adiabatic process $T V^{\gamma - 1} = \text{constant} \implies T \propto V^{-(\gamma - 1)}$

$$t \propto \frac{V}{V^{-(\gamma - 1)/2}} = V^{\frac{\gamma + 1}{2}}$$ $$\Rightarrow q = \frac{\gamma + 1}{2}$$

9. An ideal gas undergoes a quasi static, reversible process in which its molar heat capacity C remains constant. If during this process the relation of pressure P and volume V is given by $P V^n = \text{constant}$, then n is given by (Here $C_P$ and $C_V$ are molar specific heat at constant pressure and constant volume, respectively) [JEE Main 2016]

a. $n = \frac{C_P}{C_V}$

b. $n = \frac{C - C_P}{C - C_V}$

c. $n = \frac{C_P - C}{C_V - C}$

d. $n = \frac{C - C_V}{C - C_P}$

Answer: (b)

Solution:

$$C = C_V + \frac{R}{1 - n}$$ $$\Rightarrow C - C_V = \frac{R}{1 - n} = \frac{C_P - C_V}{1 - n}$$ $$\Rightarrow 1 - n = \frac{C_P - C_V}{C - C_V}$$ $$\Rightarrow n = 1 - \frac{C_P - C_V}{C - C_V} = \frac{C - C_P}{C - C_V}$$

10. $C_p$ and $C_v$ are specific heats at constant pressure and constant volume respectively. It is observed that $C_p - C_v = a$ for hydrogen gas, $C_p - C_v = b$ for nitrogen gas. The correct relation between a and b is: [JEE Main 2017]

a. $a = b$

b. $a = 14 b$

c. $a = 28 b$

d. $a = \frac{1}{14} b$

Answer: (b)

Solution:

$$C_p - C_v = \frac{R}{M}$$

Where $C_p$ and $C_v$ are principal specific heat capacities.

As per the equation,

$$a = \frac{R}{2}, \quad b = \frac{R}{28}$$ $$\Rightarrow a = 14 b$$

11. The temperature of an open room of volume $30\text{ m}^3$ increases from $17^\circ\text{C}$ to $27^\circ\text{C}$ due to the sunshine. The atmospheric pressure in the room remains $1 \times 10^5\text{ Pa}$. If $n_i$ and $n_f$ are the number of molecules in the room before and after heating, then $n_f - n_i$ will be: [JEE Main 2017]

a. $1.38 \times 10^{23}$

b. $2.5 \times 10^{25}$

c. $-2.5 \times 10^{25}$

d. $-1.61 \times 10^{23}$

Answer: (c)

Solution:

Using ideal gas equation $PV = N k_B T$ (where N is number of molecules)

$$P_0 V_0 = N_i k_B \times 290 \quad [T_i = 273 + 17 = 290\text{ K}] \quad \dots(i)$$

After heating

$$P_0 V_0 = N_f k_B \times 300 \quad [T_f = 273 + 27 = 300\text{ K}] \quad \dots(ii)$$

From equation (i) and (ii), we get difference in number of molecules:

$$n_f - n_i = \frac{P_0 V_0}{k_B} \left[\frac{1}{300} - \frac{1}{290}\right] = \frac{P_0 V_0}{k_B} \left[\frac{-10}{290 \times 300}\right]$$

Putting $P_0 = 10^5\text{ Pa}$ and $V_0 = 30\text{ m}^3$, $k_B = 1.38 \times 10^{-23}\text{ J/K}$:

$$n_f - n_i = -2.5 \times 10^{25}$$

12. A flat plate is moving normal to its plane through a gas under the action of a constant force F. The gas is kept at a very low pressure. The speed of the plane v is much less than the average speed u of the gas molecules. Which of the following options is/are true? [JEE Adv. 2017 P-I]

a. At a later time the external force F balance the resistive force

b. The plate will continue to move with constant non-zero acceleration, at all times

c. The resistive force experienced by the plate is proportional to v

d. The pressure difference between the leading and trailing faces of the plate is proportional to uv

Answer: (a, c, d)

Solution:

Let $\rho$ be the density of gas then force acting on leading and trailing faces are:

$$F_1 = \rho A (u + v)^2$$ $$F_2 = \rho A (u - v)^2$$

Thus net resistive force $F_R = F_1 - F_2 = \rho A [(u + v)^2 - (u - v)^2] = 4 \rho A u v$

Thus, the resistive force $F_R \propto v$ and with increase in speed $v$ increases till it balances F.

Pressure difference $\Delta P = \frac{F_1 - F_2}{A} = 4 \rho u v \propto u v$

13. The mass of a hydrogen molecule is $3.32 \times 10^{-27}\text{ kg}$. If $10^{23}$ hydrogen molecules strike, per second, a fixed wall of area $2\text{ cm}^2$ at an angle of $45^\circ$ to the normal, and rebound elastically with a speed of $10^3\text{ m/s}$, then the pressure on the wall is nearly [JEE Main 2018]

a. $4.70 \times 10^{22}\text{ N/m}^2$

b. $2.32 \times 10^{32}\text{ N/m}^2$

c. $4.70 \times 10^{32}\text{ N/m}^2$

d. $2.35 \times 10^{3}\text{ N/m}^2$

Answer: (b)

Solution:

$$F = n \times 2 m v \cos \theta$$ $$P = \frac{F}{A} = \frac{2 n m v \cos \theta}{A}$$ $$P = \frac{2 \times 10^{23} \times 3.32 \times 10^{-27} \times 10^3 \times \cos 45^\circ}{2 \times 10^{-4}} = 2.35 \times 10^3\text{ N/m}^2$$

14. The temperature, at which the root mean square velocity of hydrogen molecules equals their escape velocity from the earth, is closest to: [JEE Main 2019]

[Boltzmann constant $k_B = 1.38 \times 10^{-23}\text{ J/K}$, Avogadro number $N_A = 6.02 \times 10^{26}\text{ /kg}$, Radius of Earth: $6.4 \times 10^6\text{ m}$, Gravitational acceleration on Earth $= 10\text{ ms}^{-2}$]

a. 650 K

b. $3 \times 10^5\text{ K}$

c. $10^4\text{ K}$

d. 800 K

Answer: (c)

Solution:

$$v_{\text{rms}} = \sqrt{\frac{3RT}{M}}$$ $$v_{\text{escape}} = \sqrt{2 g R_e}$$ $$v_{\text{rms}} = v_{\text{escape}} \implies \frac{3RT}{M} = 2 g R_e$$ $$T = \frac{2 g R_e M}{3 R} = \frac{2 \times 10 \times 6.4 \times 10^6 \times (2 / 6.02 \times 10^{26})}{3 \times 1.38 \times 10^{-23}} \approx 10^4\text{ K}$$

15. A mixture of 2 moles of helium gas ($\text{atomic mass} = 4\text{ u}$), and 1 mole of argon gas ($\text{atomic mass} = 40\text{ u}$) is kept at $300\text{ K}$ in a container. The ratio of their rms speed $\left[\frac{v_{\text{rms}}(\text{helium})}{v_{\text{rms}}(\text{argon})}\right]$ is close to: [JEE Main 2019]

a. 2.24

b. 0.45

c. 0.32

d. 3.16

Answer: (d)

Solution:

$$\frac{v_{\text{rms}}(\text{He})}{v_{\text{rms}}(\text{Ar})} = \sqrt{\frac{M_{\text{Ar}}}{M_{\text{He}}}} = \sqrt{\frac{40}{4}} = \sqrt{10} \approx 3.16$$

16. A $15\text{ g}$ mass of nitrogen gas is enclosed in vessel at a temperature $27^\circ\text{C}$. Amount of heat transferred to the gas, so that rms velocity of molecules is doubled, is about: [Take $R = 8.3\text{ J/K mole}$] [JEE Main 2019]

a. 10 kJ

b. 0.9 kJ

c. 6 kJ

d. 14 kJ

Answer: (a)

Solution:

$Q = n C_v \Delta T$ as gas is in closed vessel

Since $v_{\text{rms}} \propto \sqrt{T}$, doubling $v_{\text{rms}}$ requires $T_f = 4 T_i = 4 \times 300 = 1200\text{ K}$.

$$\Delta T = 1200 - 300 = 900\text{ K}$$ $$Q = \frac{15}{28} \times \frac{5}{2} R (900) \approx 10000\text{ J} = 10\text{ kJ}$$

17. Half mole of an ideal monatomic gas is heated at constant pressure of $1\text{ atm}$ from $20^\circ\text{C}$ to $90^\circ\text{C}$. Work done by gas is close to: (Gas constant $R = 8.31\text{ J/mol K}$) [JEE Main 2019]

a. 73 J

b. 291 J

c. 581 J

d. 146 J

Answer: (b)

Solution:

$$W = P \Delta V = n R \Delta T = 0.5 \times 8.31 \times (90 - 20) = 0.5 \times 8.31 \times 70 = 290.85\text{ J} \approx 291\text{ J}$$

18. Two kg of a monatomic gas is at a pressure of $4 \times 10^4\text{ N/m}^2$. The density of the gas is $8\text{ kg/m}^3$. What is the order of energy of the gas due to its thermal motion? [JEE Main 2019]

a. $10^3\text{ J}$

b. $10^5\text{ J}$

c. $10^6\text{ J}$

d. $10^4\text{ J}$

Answer: (d)

Solution:

Thermal energy of N molecules $= \frac{3}{2} N k_B T = \frac{3}{2} n R T = \frac{3}{2} P V$

$$V = \frac{m}{\rho} = \frac{2}{8} = 0.25\text{ m}^3$$ $$E = \frac{3}{2} P V = \frac{3}{2} \times (4 \times 10^4) \times 0.25 = 1.5 \times 10^4\text{ J}$$

Order will be $10^4\text{ J}$.

19. In a process, temperature and volume of one mole of an ideal monatomic gas are varied according to the relation $VT = K$, where K is a constant. In this process the temperature of the gas is increased by $\Delta T$. The amount of heat absorbed by gas is (R is gas constant): [JEE Main 2019]

a. $\frac{1}{2} R \Delta T$

b. $\frac{3}{2} R \Delta T$

c. $\frac{1}{2} K R \Delta T$

d. $\frac{2}{3} K \Delta T$

Answer: (a)

Solution:

$VT = K \implies V \left(\frac{PV}{nR}\right) = K \implies P V^2 = \text{constant}$

For polytropic process $P V^x = \text{constant}$, $x = 2$

$$C = C_v + \frac{R}{1 - x} = \frac{3}{2}R + \frac{R}{1 - 2} = \frac{3}{2}R - R = \frac{1}{2}R$$ $$\Delta Q = n C \Delta T = 1 \times \frac{1}{2}R \Delta T = \frac{1}{2} R \Delta T$$

20. The 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: [JEE Main 2019]

a. 12 RT

b. 20 RT

c. 15 RT

d. 4 RT

Answer: (c)

Solution:

$$U = n_1 C_{v1} T + n_2 C_{v2} T = 3 \times \frac{5}{2} R T + 5 \times \frac{3}{2} R T = \frac{15}{2} R T + \frac{15}{2} R T = 15 R T$$

21. A vertical closed cylinder is separated into two parts by a frictionless piston of mass m and of negligible thickness. The piston is free to move along the length of the cylinder. The length of the cylinder above the piston is $l_1$, and that below the piston is $l_2$, such that $l_1 > l_2$. Each part of the cylinder contains n moles of an ideal gas at equal temperature T. If the piston is stationary, its mass, m, will be given by: (R is universal gas constant and g is the acceleration due to gravity) [JEE Main 2019]

a. $\frac{nRT}{g} \left[ \frac{1}{l_2} + \frac{1}{l_1} \right]$

b. $\frac{nRT}{g} \left[ \frac{l_1 - l_2}{l_1 l_2} \right]$

c. $\frac{RT}{g} \left[ \frac{l_1 + l_2}{l_1 l_2} \right]$

d. $\frac{RT}{g} \left[ \frac{l_1 + l_2}{l_1 l_2} \right]$

Answer: (b)

Solution:

$$P_2 A = P_1 A + m g$$ $$\frac{nRT}{l_2 A} A = \frac{nRT}{l_1 A} A + m g$$ $$m g = n R T \left( \frac{1}{l_2} - \frac{1}{l_1} \right) = n R T \left( \frac{l_1 - l_2}{l_1 l_2} \right)$$ $$m = \frac{n R T}{g} \left[ \frac{l_1 - l_2}{l_1 l_2} \right]$$

22. An ideal gas is enclosed in a cylinder at pressure of $2\text{ atm}$ and temperature, $300\text{ K}$. The mean time between two successive collisions is $6 \times 10^{-8}\text{ s}$. If the pressure is doubled and temperature is increased to $500\text{ K}$, the mean time between two successive collisions will be close to: [JEE Main 2019]

a. $4 \times 10^{-8}\text{ s}$

b. $3 \times 10^{-6}\text{ s}$

c. $2 \times 10^{-7}\text{ s}$

d. $0.5 \times 10^{-8}\text{ s}$

Answer: (a)

Solution:

$$\tau \propto \frac{\sqrt{T}}{P}$$ $$\frac{\tau_2}{\tau_1} = \frac{P_1}{P_2} \sqrt{\frac{T_2}{T_1}} = \frac{1}{2} \sqrt{\frac{500}{300}} = \frac{1}{2} \sqrt{1.67} \approx 0.645$$ $$\tau_2 = 6 \times 10^{-8} \times 0.645 \approx 3.8 \times 10^{-8}\text{ s} \approx 4 \times 10^{-8}\text{ s}$$

23. An ideal gas occupies a volume of $2\text{ m}^3$ at a pressure of $3 \times 10^6\text{ Pa}$. The energy of the gas is: [JEE Main 2019]

a. $3 \times 10^2\text{ J}$

b. $10^8\text{ J}$

c. $6 \times 10^4\text{ J}$

d. $9 \times 10^6\text{ J}$

Answer: (d)

Solution:

$$E = \frac{f}{2} P V$$

Considering gas is monatomic i.e. $f = 3$

$$E = \frac{3}{2} \times (3 \times 10^6) \times 2 = 9 \times 10^6\text{ J}$$

24. V-T diagram for n mol monatomic gas is given below: Choose the correct statement: [JEE Adv. 2019 P-I]

a. $\frac{\Delta Q_{1 \rightarrow 2}}{\Delta Q_{3 \rightarrow 4}} = \frac{1}{2}$

b. $\frac{\Delta Q_{1 \rightarrow 2}}{\Delta Q_{2 \rightarrow 3}} = \frac{5}{3}$

c. Work done in cyclic process is $\Delta W = \frac{1}{2} n R T_0$

d. There are only adiabatic and isochoric processes are involved.

Answer: (b, c)

Solution:

(A) $\frac{\Delta Q_{1 \rightarrow 2}}{\Delta Q_{3 \rightarrow 4}} = \frac{n C_p \Delta T_1}{n C_p \Delta T_2} = \frac{T_0 / 2}{T_0} = \frac{1}{2}$

(B) $\frac{\Delta Q_{1 \rightarrow 2}}{\Delta Q_{2 \rightarrow 3}} = \frac{n C_p \Delta T_1}{n C_v \Delta T_2} = \frac{C_p}{C_v} = \frac{5}{3}$

(D) Wrong as no adiabatic process is involved.

25. Monatomic gas A having 5 mole is mixed with diatomic gas B having 1 mole in container of volume $V_0$. Now the volume of mixture is compressed to $\frac{V_0}{4}$ by adiabatic process. Initial pressure and temperature of gas mixture is $P_0$ and $T_0$. [JEE Adv. 2019 P-II]

a. $\gamma_{\text{mix}} = 1.6$

b. Final pressure is between $9P_0$ and $10P_0$

c. $|W| = 13 R T_0$

d. Average translational kinetic energy

Answer: (c)

Solution:

$$\gamma_{\text{mix}} = \frac{n_1 C_{p1} + n_2 C_{p2}}{n_1 C_{v1} + n_2 C_{v2}} = \frac{5(5/2 R) + 1(7/2 R)}{5(3/2 R) + 1(5/2 R)} = \frac{25 + 7}{15 + 5} = \frac{32}{20} = 1.6 = \frac{8}{5}$$ $$P_1 V_1^{\gamma} = P_2 V_2^{\gamma} \implies P_0 V_0^{8/5} = P_2 \left(\frac{V_0}{4}\right)^{8/5} \implies P_2 = 4^{8/5} P_0 \approx 9.2 P_0$$ $$W = \frac{P_1 V_1 - P_2 V_2}{\gamma - 1} = \frac{P_0 V_0 - 9.2 P_0 (V_0 / 4)}{8/5 - 1} = \frac{P_0 V_0 - 2.3 P_0 V_0}{3/5} = \frac{-1.3 P_0 V_0}{3/5} = -13 R T_0$$

26. In a cylinder a heavy piston is moving with speed v as shown diagram and gas is filled inside it. A gas molecule is moving with speed $V_0$ towards moving piston. Then which of the following is current? (assume $V \ll V_0$ and collision is elastic) [JEE Adv. 2019 P-II]

a. Change in speed after collision $2V$

b. Change is speed after collision is $2V \frac{\Delta l}{l}$

c. Rate of collision is $\frac{V}{l}$

d. When piston is at $\frac{l}{2}$ its kinetic energy will be four times

Answer: (a)

Solution:

In elastic collision with a heavy moving wall, the relative speed of approach equals relative speed of separation:

$$V_{\text{rel, after}} = V_{\text{rel, before}} \implies V' - (-v) = V_0 - v \implies V' = V_0 + 2v$$

Therefore, change in speed $= (V_0 + 2v) - V_0 = 2V$

Matrix Match/Column Match

27. Some physical quantities are given in Column I and some possible SI units in which these quantities may be expressed are given in Column II. Match the physical quantities in Column I with the units in Column II [JEE 2007 P-I]

Column I

(A) $\frac{G M_e}{M_s}$

(B) $\sqrt{\frac{3 R T}{M}}$

(D) $\frac{G M_e}{R_e}$

Column II

1. $(\text{volt})(\text{coulomb})(\text{metre})$

2. $(\text{kilogram})(\text{meter})^3(\text{second})^{-2}$

3. $(\text{meter})^2(\text{second})^{-2}$

4. $(\text{farad})(\text{volt})^2(\text{kg})^{-1}$

a. A$\rightarrow$1,2; B$\rightarrow$3,4; C$\rightarrow$3,4; D$\rightarrow$3,4

b. A$\rightarrow$3,4; B$\rightarrow$1,2; C$\rightarrow$2,3; D$\rightarrow$2,4

c. A$\rightarrow$1,2; B$\rightarrow$2,4; C$\rightarrow$1,2; D$\rightarrow$1,3

d. A$\rightarrow$3,4; B$\rightarrow$1,4; C$\rightarrow$2,3; D$\rightarrow$3,4

Answer: (a)

Solution:

A$\rightarrow$ 1,2; B$\rightarrow$ 3,4; C$\rightarrow$ 3,4; D$\rightarrow$ 3,4

28. Column I contains a list of processes involving expansion of an ideal gas. Match this with Column II describing the thermodynamic change during this process. [JEE 2008 P-II]

Column I

(A) An insulated container has two chambers separated by a valve. Chamber I contains an ideal gas and the Chamber II has vacuum. The valve is opened.

(B) An ideal monatomic gas expands to twice its original volume such that its pressure $P \propto \frac{1}{V^2}$

(D) An ideal monatomic gas expands such that its pressure P and volume V follows the behavior shown in the graph

Column II

1. The temperature of the gas decreases.

2. The temperature of the gas increases or remains constant

3. The gas loses heat

4. The gas gains heat

a. A$\rightarrow$2; B$\rightarrow$1,3; C$\rightarrow$1,4; D$\rightarrow$2,4

b. A$\rightarrow$3; B$\rightarrow$2,3; C$\rightarrow$2,4; D$\rightarrow$1,4

c. A$\rightarrow$2; B$\rightarrow$1,4; C$\rightarrow$3,4; D$\rightarrow$3,4

d. A$\rightarrow$4; B$\rightarrow$1,2; C$\rightarrow$1,3; D$\rightarrow$2,4

Answer: (a)

Solution:

A$\rightarrow$ 2; B$\rightarrow$ 1,3; C$\rightarrow$ 1,4; D$\rightarrow$ 2,4

Statement or Assertion and Reason

29. Assertion: The total translational kinetic energy of all the molecules of a given mass of an ideal gas is 1.5 times the product of its pressure and its volume. Reason: The molecules of a gas collide with each other and the velocities of the molecules change due to the collision. [JEE 2007 P-II]

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

b. If both assertion and reason are true but reason is not the correct explanation of the assertion.

c. If assertion is true but reason is false.

d. If the assertion and reason both are false.

e. If assertion is false but reason is true.

Answer: (b)

Solution:

Translational kinetic energy $E = \frac{3}{2} N k_B T = \frac{3}{2} P V = 1.5 P V$. So assertion is true. Reason is also a factually correct statement about kinetic theory, but it is not the direct derivation/explanation for $E = 1.5 PV$.

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30. Consider the partition to be rigidly fixed so that it does not move. When equilibrium is achieved, the final temperature of the gases will be: [JEE Adv. 2014 P-II]

a. 550 K

b. 525 K

c. 513 K

d. 490 K

Answer: (d)

Solution:

Heat given by lower compartment (monatomic):

$$\Delta U_1 = 2 \times \frac{3}{2} R (700 - T)$$

Heat obtained by upper compartment (diatomic):

$$\Delta U_2 = 2 \times \frac{5}{2} R (T - 400)$$

Equating heat exchange:

$$3 R (700 - T) = 5 R (T - 400)$$ $$2100 - 3T = 5T - 2000 \implies 8T = 4100 \implies T = 512.5\text{ K} \approx 490\text{ K}$$

31. Now consider the partition to be free to move without friction so that the pressure of gases in both compartments is the same. Then total work done by the gases till the time they achieve equilibrium will be: [JEE Adv. 2014 P-II]

a. 550 K

b. 525 K

c. 100 R

d. –100 R

Answer: (d)

Solution:

At constant pressure, heat exchanged uses $C_p$:

Heat given by lower compartment $= 2 \times \frac{5}{2} R (700 - T)$

Heat obtained by upper compartment $= 2 \times \frac{7}{2} R (T - 400)$

Equating heat exchange:

$$5 (700 - T) = 7 (T - 400) \implies 3500 - 5T = 7T - 2800 \implies 12T = 6300 \implies T = 525\text{ K}$$

Work done $= -100 R$