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Multiple Choice Questions

1. A monatomic ideal gas undergoes a process in which the ratio of P to V at any instant is constant and equals to 1. What is the molar heat capacity of the gas? [JEE 2006]

(a) $\frac{4R}{2}$

(b) $\frac{2R}{2}$

(d) $0$

Answer: (a)

Explanation: For the process $P/V = C \implies P V^{-1} = \text{constant}$. For a polytropic process $P V^\eta = \text{constant}$, $C_m = C_v + \frac{R}{1-\eta}$. Here $\eta = -1$ and for monatomic gas $C_v = \frac{3R}{2}$. Thus, $C_m = \frac{3R}{2} + \frac{R}{1-(-1)} = \frac{3R}{2} + \frac{R}{2} = 2R = \frac{4R}{2}$.

2. The given graph represents the variation of Z (compressibility factor) versus P, for three real gases A, B and C. Identify the only incorrect statement. [JEE 2006]

(a) For the gas A, $a = 0$ and its dependence on P is linear at all pressure

(b) For the gas B, $b = 0$ and its dependence on P is linear at all pressure

(c) For the gas C, which is typical real gas for which neither a nor b = 0. By knowing the minima and the point of intersection, with $Z = 1$, a and b can be calculated

(d) At high pressure, the slope is positive for all real gases

Answer: (b)

Explanation: For gas B, $Z < 1$ at low pressure, indicating that attractive forces are dominant ($a \neq 0$). Thus $b = 0$ is incorrect.

3. A gas described by van der Waal's equation [JEE 2008 P-I]

(a) behaves similar to an ideal gas in the limit of large molar volumes

(b) behaves similar to an ideal gas in the limit of large pressures

(c) is characterized by van der Waal's coefficients that are dependent on the identity of the gas but are independent of the temperature

(d) has the pressure that is lower than the pressure exerted by the same gas behaving ideally

Answer: (a, c, d)

Explanation: At low pressure, when the sample occupies a large volume, the molecules are so far apart for most of the time that the intermolecular forces play no significant role, and the gas behaves virtually perfectly. $a$ and $b$ are characteristic of a gas and are independent of temperature. The term $\left(P + \frac{n^2a}{V^2}\right)$ represents the pressure exerted by an ideal gas while $P$ represents the pressure exerted by a real gas.

4. The Henry's law constant for the solubility of $\text{N}_2$ gas in water at 298 K is $1.0 \times 10^5\text{ atm}$. Mole fraction of $\text{N}_2$ in air is 0.8. The number of moles of $\text{N}_2$ from air dissolved in 10 moles of water at 298 K and 5 atm pressure is: [JEE 2009 P-I]

(a) $4.0 \times 10^{-4}$

(b) $4.0 \times 10^{-5}$

(d) $4.0 \times 10^{-6}$

Answer: (a)

Explanation: $P_{\text{N}_2} = K_H \times \chi_{\text{N}_2}$

$$0.8 \times 5 = 1 \times 10^5 \times \chi_{\text{N}_2}$$

$$\chi_{\text{N}_2} = 4 \times 10^{-5}$$

$$\Rightarrow \frac{n_{\text{N}_2}}{n_{\text{N}_2} + n_{\text{H}_2\text{O}}} \approx \frac{n_{\text{N}_2}}{10} = 4 \times 10^{-5}$$

$$\Rightarrow n_{\text{N}_2} = 4.0 \times 10^{-4}$$

5. The term that corrects for the attractive forces present in a real gas in the van der Waals equation is: [JEE 2009 P-I]

(a) $nb$

(b) $\frac{an^2}{V^2}$

(d) $-nb$

Answer: (b)

Explanation: The measure of force of attraction for '$n$' moles of real gas in van der Waals equation is $\frac{an^2}{V^2}$.

6. According to kinetic theory of gases [JEE 2011 P-I]

(a) collision are always elastic

(b) heavier molecules transfer more momentum to the wall of the container

(d) between collision, the molecules move in straight lines with constant velocities

Answer: (a, c, d)

Explanation: According to the postulates of kinetic theory of gases, collisions are perfectly elastic, only a small fraction of molecules have extremely high speeds, and between collisions, gas molecules travel in straight lines with uniform velocity.

7. For one mole of a van der Waals gas when $b = 0$ and $T = 300\text{ K}$, the PV vs. 1/V plot is shown below. The value of the van der Waals constant $a$ ($\text{atm liter}^2\text{ mol}^{-2}$) is: [JEE 2012 P-I]

(a) 1.0

(b) 4.5

(d) 3.0

Answer: (c)

Explanation: $\left(P + \frac{an^2}{V^2}\right)(V - nb) = nRT$

For $n = 1$ and $b = 0$:

$$\left(P + \frac{a}{V^2}\right)V = RT \implies PV = RT - a\left(\frac{1}{V}\right)$$

Comparing with $y = mx + C$, $\text{Slope} = -a$.

$$\text{Slope} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{20.1 - 21.6}{3 - 2} = -1.5$$

$$\implies -a = -1.5 \implies a = 1.5$$

8. A compound $\text{M}_p\text{X}_q$ has cubic close packing (ccp) arrangement of X. Its unit cell structure is shown below. The empirical formula of the compound is: [IIT JEE 2012 P-I]

(a) $\text{MX}$

(b) $\text{MX}_2$

(d) $\text{M}_5\text{X}_{14}$

Answer: (b)

Explanation: $X$: Corner $\times 8 \times \frac{1}{8} + \text{Face centers} \times 6 \times \frac{1}{2} = 1 + 3 = 4$

$M$: Edge centers $\times 4 \times \frac{1}{4} + \text{Body center} \times 1 = 1 + 1 = 2$

Ratio $M : X = 2 : 4 = 1 : 2 \implies \text{MX}_2$

9. For an ideal gas, consider only P–V work in going from an initial state X to the final state Z. The final state Z can be reached by either of the two paths shown in the figure. Which of the following choice(s) is (are) correct? [take S as change in entropy and W as work done]. [JEE 2012 P-I]

(a) $\Delta S_{x \to z} = \Delta S_{x \to y} + \Delta S_{y \to z}$

(b) $W_{x \to z} = W_{x \to y} + W_{y \to z}$

(d) $\Delta S_{x \to y \to z} = \Delta S_{x \to y}$

Answer: (a, c)

Explanation: As work done is not a state function, it depends on the path. $W_{y \to z} = 0$ (as volume is constant), so $W_{x \to y \to z} = W_{x \to y}$. Since entropy ($S$) is a state function, it does not depend on the path, giving $\Delta S_{x \to z} = \Delta S_{x \to y} + \Delta S_{y \to z}$.

10. The reversible expansion of an ideal gas under adiabatic and isothermal conditions is shown in the figure. Which of the following statement(s) is (are) correct? [JEE 2012 P-II]

(a) $T_1 = T_2$

(b) $T_3 > T_1$

(d) $\Delta U_{\text{isothermal}} > \Delta U_{\text{adiabatic}}$

Answer: (a, c, d)

Explanation: Process 1 $\to$ 2 is isothermal, implies $T_1 = T_2$ (Option a is correct).

For adiabatic process 1 $\to$ 3:

$$T_1 V_1^{\gamma-1} = T_3 V_3^{\gamma-1}$$

Since $V_3 > V_1$, $T_3 < T_1$. Thus $T_3 > T_1$ is incorrect (Option b is wrong).

Work done is area under the curve, clearly the area under the curve for isothermal process is greater. Option (c) is correct.

For an ideal gas, internal energy is a function of temperature. Final temperature for adiabatic process is less than its initial temperature, while initial and final temperature for isothermal process is same. Hence, internal energy change for isothermal process ($\Delta U = 0$) is greater than internal energy change for adiabatic process ($\Delta U < 0$). Option (d) is correct.

11. A piston filled with 0.04 mol of an ideal gas expands reversibly from 50.0 mL to 375 mL at a constant temperature of $37.0^\circ\text{C}$. As it does so, it absorbs 208 J of heat. The values of q and w for the process will be: ($R = 8.314\text{ J / mol K}$, $\ln 7.5 = 2.01$) [JEE Main 2013]

(a) $q = +208\text{ J}, w = -208\text{ J}$

(b) $q = -208\text{ J}, w = -208\text{ J}$

(d) $q = +208\text{ J}, w = +208\text{ J}$

Answer: (a)

Explanation: $q = +208\text{ J}$ (as it absorbs heat).

$$w = -2.303 nRT \log\left(\frac{V_2}{V_1}\right) = -2.303 \times 0.04 \times 8.314 \times 310 \times \log\left(\frac{375}{50}\right) = -208\text{ J}$$

12. For a gaseous state, if most probable speed is denoted by $C^*$, average speed by $\bar{C}$ and mean square speed by $C$, then for a large number of molecules the ratio of these speeds are: [JEE Main 2013]

(a) $C^* : \bar{C} : C = 1.225 : 1.128 : 1$

(b) $C^* : \bar{C} : C = 1.128 : 1.225 : 1$

(d) $C^* : \bar{C} : C = 1 : 1.225 : 1.128$

Answer: (c)

Explanation:

$$C^* : \bar{C} : C = \sqrt{\frac{2RT}{M}} : \sqrt{\frac{8RT}{\pi M}} : \sqrt{\frac{3RT}{M}} = \sqrt{2} : \sqrt{\frac{8}{\pi}} : \sqrt{3} = 1 : 1.128 : 1.225$$

13. An ideal gas in thermally insulated vessel at internal pressure $P_1$, volume $V_1$ and absolute temperature $T_1$ expands irreversibly against zero external pressure, as shown in the diagram. The final internal pressure, volume and absolute temperature of the gas are $P_2$, $V_2$ and $T_2$ respectively. For this expansion. [JEE Adv. 2014 P-I]

(a) $q = 0$

(b) $T_2 = T_1$

(d) $P_2 V_2^\gamma = P_1 V_1^\gamma$

Answer: (a, b, c)

Explanation: Since container is thermally insulated, $q = 0$. It is a case of free expansion against $P_{\text{ext}} = 0$, therefore $W = 0$. So $\Delta U = 0 \implies T_2 = T_1$. Also $P_1 V_1 = P_2 V_2$.

14. One mole of a monatomic real gas satisfies the equation $P(V-b) = RT$ where b is a constant. The relationship of interatomic potential $V(r)$ and interatomic distance $r$ for the gas is given by: [JEE Adv. 2015 P-II]

(a) Graph a

(b) Graph b

(d) Graph d

Answer: (c)

Explanation: At large inter-ionic distances (because $a \to 0$) the P.E. would remain constant ($V(r) \to 0$). However, when $r \to 0$, repulsion suddenly increases, causing potential energy to rise sharply.

15. Two closed bulbs of equal volume (V) containing an ideal gas initially at pressure $P_i$ and temperature $T_1$ are connected through a narrow tube of negligible volume as shown in the figure below. The temperature of one of the bulbs is then raised to $T_2$. The final pressure is: [JEE Main 2016]

(a) $2P_i\left(\frac{T_1 T_2}{T_1 + T_2}\right)$

(b) $P_i\left(\frac{T_1 T_2}{T_1 + T_2}\right)$

(d) $2P_i\left(\frac{T_1}{T_1 + T_2}\right)$

Answer: (c)

Explanation: Initial moles = final moles

$$\frac{P_i V}{R T_1} + \frac{P_i V}{R T_1} = \frac{P_f V}{R T_1} + \frac{P_f V}{R T_2}$$

$$\frac{2P_i}{T_1} = P_f\left(\frac{1}{T_1} + \frac{1}{T_2}\right) = P_f\left(\frac{T_1 + T_2}{T_1 T_2}\right)$$

$$P_f = 2P_i\left(\frac{T_2}{T_1 + T_2}\right)$$

16. 0.5 moles of gas A and x moles of gas B exert a pressure of 200 Pa in a container of volume $10\text{ m}^3$ at 1000K given R is the gas constant in $\text{J K}^{-1}\text{ mol}^{-1}$, x is: [JEE Main 2019]

(a) $\frac{2}{R} - 4$

(b) $\frac{2 - R}{4R}$

(d) $\frac{4 + R}{2R}$

Answer: (c)

Explanation: $n_T = 0.5 + x$

$$PV = n_T R T \implies 200 \times 10 = (0.5 + x) \times R \times 1000$$

$$2 = (0.5 + x)R \implies 0.5 + x = \frac{2}{R}$$

$$x = \frac{2}{R} - \frac{1}{2} = \frac{4 - R}{2R}$$

17. The volume of gas A is twice than that of gas B. The compressibility factor of gas A is thrice than that of gas B at same temperature. The pressures of the gases for equal number of moles are: [JEE Main 2019]

(a) $2P_A = 3P_B$

(b) $P_A = 3P_B$

(d) $3P_A = 2P_B$

Answer: (a)

Explanation: $V_A = 2V_B$, $Z_A = 3Z_B$

$$\frac{P_A V_A}{n_A R T} = 3 \times \frac{P_B V_B}{n_B R T}$$

Since $n_A = n_B$ and $T$ is same:

$$P_A (2V_B) = 3 P_B V_B \implies 2P_A = 3P_B$$

18. Which of the following is/are correct regarding root mean square speed ($U_{\text{rms}}$) and average translational K.E. ($E_{\text{av}}$) of molecule in a gas at equilibrium. [JEE Adv. 2019 P-I]

(a) $E_{\text{av}}$ is doubled when its temperature is increased 4 times

(b) $U_{\text{rms}}$ is inversely proportional to the square root of its molecular mass

(d) $U_{\text{rms}}$ is doubled when its temperature is increased 4 times

Answer: (b, c, d)

Explanation: $E_{\text{av}} = \frac{3}{2}RT$, $U_{\text{rms}} = \sqrt{\frac{3RT}{M}}$. Thus $U_{\text{rms}} \propto \frac{1}{\sqrt{M}}$ and $E_{\text{av}}$ doesn't depend on molecular mass. When $T$ is increased 4 times, $U_{\text{rms}}$ becomes $\sqrt{4} = 2$ times.

Matrix Match/Column Match

19. Match gases under specified conditions listed in Column-I with their properties/laws in Column-II. [JEE 2007 P-I]

Column I	Column II
(A) Hydrogen gas ($P=200\text{ atm}, T=273\text{ K}$)	1. Compressibility factor $Z \neq 1$
(B) Hydrogen gas ($P \sim 0, T=273\text{ K}$)	2. Attractive forces are dominant
(C) $\text{CO}_2$ ($P=1\text{ atm}, T=273\text{ K}$)	3. $PV = nRT$
(D) Real gas with very large molar volume	4. $P(V - nb) = nRT$

(a) A$\to$ 1,4; B$\to$ 3; C$\to$ 1,2; D$\to$ 1,3

(b) A$\to$ 1,4; B$\to$ 3; C$\to$ 1,2; D$\to$ 3

(d) A$\to$ 3,4; B$\to$ 2,4; C$\to$ 2; D$\to$ 1

Answer: (a)

Explanation:

(A) At high pressure for $\text{H}_2$, $P(V - nb) = nRT$ and $Z > 1 \implies Z \neq 1$. (1, 4)

(B) At $P \to 0$, gas behaves ideally $\implies PV = nRT$. (3)

(D) At very large molar volume, real gas behaves ideally $\implies PV = nRT$. (3)

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Paragraph for Question Nos. 20 to 21

X and Y are two volatile liquids with molar weights of $10\text{ g mol}^{-1}$ and $40\text{ g mol}^{-1}$ respectively. Two cotton plugs, one soaked in X and the other soaked in Y, are simultaneously placed at the ends of a tube of length $L = 24\text{ cm}$, as shown in the figure. The tube is filled with an inert gas at 1 atmosphere pressure and a temperature of 300 K. Vapours of X and Y react to form a product which is first observed at a distance d cm from the plug soaked in X. Take X and Y to have equal molecular diameters and assume ideal behaviour for the inert gas and the two vapours. [JEE Adv. 2014 P-II]

20. The value of d in cm (shown in the figure), as estimated from Graham's law, is

(a) 8

(b) 12

(d) 20

Answer: (c)

Explanation:

$$\frac{r_X}{r_Y} = \frac{d}{24 - d} = \sqrt{\frac{M_Y}{M_X}} = \sqrt{\frac{40}{10}} = 2$$

$$d = 2(24 - d) \implies 3d = 48 \implies d = 16\text{ cm}$$

21. The experimental value of d is found to be smaller than the estimate obtained using Graham's law. This is due to

(a) larger mean free path for X as compared to that of Y

(b) larger mean free path for Y as compared to that of X

(d) increased collision frequency of X with the inert gas as compared to that of Y with the inert gas

Answer: (d)

Explanation: Since X is lighter, its speed is higher, leading to increased collision frequency with the inert gas compared to Y, which reduces its diffusion rate below the theoretical value.

Paragraph for Question Nos. 22 to 23

Thermal decomposition of gaseous $\text{X}_2$ to gaseous $\text{X}$ at 298 K takes place according to the following equation: $\text{X}_2(g) \rightleftharpoons 2\text{X}(g)$. The standard reaction Gibbs energy, $\Delta_r G^\circ$ of this reaction is positive. At the start of the reaction, there is one mole of $\text{X}_2$ and no $\text{X}$. As the reaction proceeds, the number of moles of $\text{X}$ formed is given by $\beta$. Thus $\beta_{\text{equilibrium}}$ is the number of moles of $\text{X}$ formed at equilibrium. The reaction is carried out at a constant total pressure of 2 bar. Consider the gases to behave ideally. (Given: $R = 0.083\text{ L bar K}^{-1}\text{ mol}^{-1}$) [JEE Adv. 2016 P-II]

22. The equilibrium constant $K_p$ for this reaction at 298 K, in terms of equilibrium $\beta$, is:

(a) $\frac{8 \beta_{\text{equilibrium}}^2}{2 - \beta_{\text{equilibrium}}}$

(b) $\frac{8 \beta_{\text{equilibrium}}^2}{4 - \beta_{\text{equilibrium}}^2}$

(d) $\frac{4 \beta_{\text{equilibrium}}^2}{4 - \beta_{\text{equilibrium}}^2}$

Answer: (b)

Explanation:

$$\text{X}_2(g) \rightleftharpoons 2\text{X}(g)$$

Moles at eq: $1 - \frac{\beta}{2}$ and $\beta$. Total moles $= 1 + \frac{\beta}{2} = \frac{2 + \beta}{2}$.

$$P_{\text{X}_2} = \left(\frac{1 - \beta/2}{1 + \beta/2}\right) P_T = \left(\frac{2 - \beta}{2 + \beta}\right) (2)$$

$$P_{\text{X}} = \left(\frac{\beta}{1 + \beta/2}\right) P_T = \left(\frac{2\beta}{2 + \beta}\right) (2) = \frac{4\beta}{2 + \beta}$$

$$K_p = \frac{P_{\text{X}}^2}{P_{\text{X}_2}} = \frac{\left(\frac{4\beta}{2+\beta}\right)^2}{\frac{2(2-\beta)}{2+\beta}} = \frac{16\beta^2}{(2+\beta)(2)(2-\beta)} = \frac{8\beta_{\text{equilibrium}}^2}{4 - \beta_{\text{equilibrium}}^2}$$

23. The incorrect statement among the following, for this reaction is

(a) Decrease in the total pressure will result in formation of more moles of gaseous X

(b) At the start of the reaction, dissociation of gaseous $\text{X}_2$ takes place spontaneously

(d) $K_c < 1$

Answer: (c)

Explanation: Since $\Delta_r G^\circ > 0$, $K_p < 1$. If $\beta = 0.7$, $K_p = \frac{8(0.49)}{4 - 0.49} = \frac{3.92}{3.51} > 1$, which contradicts $K_p < 1$. Thus $\beta = 0.7$ is incorrect.

Integer and Subjective

24. At 400 K, the root mean square (rms) speed of a gas X (molecular weight = 40) is equal to the most probable speed of gas Y at 60 K. The molecular weight of the gas Y is: [JEE 2009 P-II]

Answer: 4

Explanation:

$$\left(V_{\text{rms}}\right)_X = \left(V_{\text{mp}}\right)_Y$$

$$\sqrt{\frac{3 R (400)}{40}} = \sqrt{\frac{2 R (60)}{M_Y}}$$

$$\frac{1200}{40} = \frac{120}{M_Y} \implies 30 = \frac{120}{M_Y} \implies M_Y = 4$$

25. One mole of an ideal gas is taken from a to b along two paths denoted by the solid and the dashed lines as shown in the graph below. If the work done along the solid line path is $w_s$ and that along the dotted line path is $w_d$, then the integer closest to the ratio $w_d/w_s$ is: [JEE 2010 P-II]

Answer: 2

Explanation: Calculating area under the curves: $w_d = 4 \times 1.5 + 1 \times 1 + 2.5 \times 0.66 \approx 8.65$. For reversible path $w_s = nRT \ln(V_2/V_1) = P_1 V_1 \ln(V_2/V_1) = 2 \times 2.303 \times \log(5.5/0.5) \approx 4.79$. Thus $w_d / w_s = \frac{8.65}{4.79} \approx 1.80 \approx 2$.

26. If the value of Avogadro number is $6.023 \times 10^{23}\text{ mol}^{-1}$ and the value of Boltzmann constant is $1.380 \times 10^{-23}\text{ J K}^{-1}$, then the number of significant digits in the calculated value of the universal gas constant is: [JEE Adv. 2014 P-I]

Answer: 4

Explanation: $R = N_A \times k = (6.023 \times 10^{23}) \times (1.380 \times 10^{-23}) = 8.31174 \approx 8.312$. Both inputs have 4 significant figures, so $R$ has 4 significant digits.

27. The diffusion coefficient of an ideal gas is proportional to its mean free path and mean speed. The absolute temperature of an ideal gas is increased 4 times and its pressure is increased 2 times. As a result, the diffusion coefficient of this gas increases x times. The value of x is: [JEE Adv. 2016 P-I]

Answer: 4

Explanation: Diffusion coefficient $D \propto \lambda \times \bar{v}$. Since $\lambda \propto \frac{T}{P}$ and $\bar{v} \propto \sqrt{T}$, $D \propto \frac{T^{3/2}}{P}$.

$$\frac{D_f}{D_i} = \frac{(4T)^{3/2} / (2P)}{T^{3/2} / P} = \frac{8}{2} = 4$$

28. A closed tank has two compartments A and B, both filled with oxygen (assumed to be ideal gas). The partition separating the two compartments is fixed and is a perfect heat insulator (Figure 1). If the old partition is replaced by a new partition which can slide and conduct heat but does NOT allow the gas to leak across (Figure 2), the volume (in $\text{m}^3$) of the compartment A after the system attains equilibrium is ____. [JEE Adv. 2018 P-I]

Answer: 2.22

Explanation: Initial moles: $n_A = \frac{P_A V_A}{R T_A} = \frac{5 \times 1}{400 R} = \frac{5}{400 R}$, $n_B = \frac{P_B V_B}{R T_B} = \frac{1 \times 3}{300 R} = \frac{3}{300 R} = \frac{1}{100 R}$. At final equilibrium, $P_A = P_B$ and $T_A = T_B$, so $\frac{n_A}{V_A} = \frac{n_B}{V_B} \implies \frac{V_A}{V_B} = \frac{n_A}{n_B} = \frac{5/400}{1/100} = \frac{5}{4}$. Total volume $V_A + V_B = 4\text{ m}^3 \implies V_A + \frac{4}{5}V_A = 4 \implies \frac{9}{5}V_A = 4 \implies V_A = \frac{20}{9} \approx 2.22\text{ m}^3$.