Kinetic Theory: NEET and JEE questions with solutions

Answer: (D) $n_A = \dfrac{1}{2}n_B$

Concept. Mean free path: $\lambda = \dfrac{1}{\sqrt2\pi d^2n}$, so $\lambda \propto \dfrac{1}{d^2n}$.

Step 1. Ratio. With $d_A = 2d_B$:
$$\frac{\lambda_A}{\lambda_B} = \frac{d_B^2n_B}{4d_B^2n_A} = \frac{n_B}{4n_A}$$

Step 2. Solve. $\dfrac{n_B}{4n_A} = \dfrac12 \Rightarrow n_A = \dfrac12n_B$.

Answer. $n_A = \dfrac12n_B$ (Option D).

Solution by Sreeraj P, M.Sc Physics

Answer: (A) $4$

Concept. $\gamma = 1 + \dfrac{2}{F}$; each vibrational mode adds 2 degrees of freedom (kinetic and potential).

Step 1. Total degrees of freedom. $\dfrac{2}{F} = \dfrac87 - 1 = \dfrac17 \Rightarrow F = 14$.

Step 2. Vibrational modes. $6 + 2f = 14 \Rightarrow f = 4$.

Answer. $4$ (Option A).

Solution by Sreeraj P, M.Sc Physics

Answer: (C) $0.116$ kg

Concept. Absolute pressure $=$ gauge $+$ atmospheric; $n = \dfrac{PV}{RT}$ gives the moles left.

Step 1. Pressure. $P = 12$ atm $= 12.12 \times 10^5$ Pa.

Step 2. Moles left. $V = 30 \times 10^{-3}\ \text{m}^3$, $T = 300$ K:
$$n_2 = \frac{12.12 \times 10^5 \times 30 \times 10^{-3}}{\frac{100}{12} \times 300} = \frac{36360}{2500} = 14.54\ \text{mol}$$

Step 3. Mass withdrawn. $(18.20 - 14.54) \times 32 \approx 117$ g $\approx 0.116$ kg.

Answer. $0.116$ kg (Option C).

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $1.6$ atm

Concept. The isolated container exchanges no heat and does no work, so total internal energy $U = \dfrac{f}{2}pV$ is conserved (same gas on both sides).

Step 1. Energy conservation.
$$p(V_1 + V_2) = p_1V_1 + p_2V_2$$

Step 2. Solve.
$$p = \frac{1 \times 2 + 2 \times 3}{5} = 1.6\ \text{atm}$$
(The mole numbers are not needed.)

Answer. $1.6$ atm (Option B).

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $P_1 > P_2 > P_3$

Concept. At constant pressure $T = \dfrac{P}{nR}V$: the slope of a $T$–$V$ line is proportional to the pressure.

Step 1. Compare slopes. $P_1$ is steepest, then $P_2$, then $P_3$.

Answer. $P_1 > P_2 > P_3$ (Option B).

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $3295°$ C

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

Step 1. New speed. "Increased by 3 times" means $v + 3v = 4v$, so $T_2 = 16T_1$.

Step 2. Temperature.
$$T_2 = 16 \times 223 = 3568\ \text{K} = 3295°\text{C}$$
(Reading the new speed as $3v$ gives $2007$ K, which is not an option.)

Answer. $3295°$C (Option B).

Solution by Sreeraj P, M.Sc Physics

Answer: (D) (A)-(Q), (B)-(P), (C)-(S), (D)-(R)

Concept. Standard kinetic-theory results.

Step 1. Match.

• (A) $v_{\text{rms}} = \sqrt{\dfrac{3RT}{M}}$ → (Q).
• (B) $P = \dfrac13nm\bar{v}^2$ → (P).
• (C) Average KE of a molecule $\dfrac32k_BT$ → (S).
• (D) One mole of a diatomic gas (5 degrees of freedom): $\dfrac52RT$ → (R).

Answer. (A)-(Q), (B)-(P), (C)-(S), (D)-(R) (Option D).

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $0.2\ \text{kg/m}^3$

Concept. From $PV = \dfrac{m}{M}RT$: $\rho = \dfrac{PM}{RT}$.

Step 1. Substitute. $M = 2 \times 10^{-3}$ kg/mol, $T = 300$ K:
$$\rho = \frac{249 \times 10^3 \times 2 \times 10^{-3}}{8.3 \times 300} = \frac{498}{2490} = 0.2\ \text{kg/m}^3$$

Answer. $0.2\ \text{kg/m}^3$ (Option B).

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $\dfrac{1}{\sqrt{2}\,n\pi d^2}$

Concept. A molecule sweeps a collision cylinder of cross-section $\pi d^2$; the motion of the others adds the factor $\sqrt2$.

Step 1. Mean free path.
$$\lambda = \frac{1}{\sqrt2n\pi d^2}$$

Answer. $\dfrac{1}{\sqrt2n\pi d^2}$ (Option B).

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $\dfrac{3}{2}k_BT$

Concept. Equipartition: $\dfrac12k_BT$ per degree of freedom.

Step 1. Monatomic. 3 translational degrees of freedom: $E = \dfrac32k_BT$.

Answer. $\dfrac32k_BT$ (Option B).

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $11.7$ atm

Concept. At constant volume, $P \propto nT$.

Step 1. Substitute. Half the gas, $T$ from $300$ K to $350$ K:
$$P_2 = 20 \times \frac12 \times \frac{350}{300} \approx 11.7\ \text{atm}$$

Answer. $11.7$ atm (Option B).

Solution by Sreeraj P, M.Sc Physics

Answer: (A) $2P$

Concept. Dalton's law: partial pressures add.

Step 1. Each gas. Alone in $V$ at $T$, each exerts $P$.

Step 2. Mixture. $P + P = 2P$.

Answer. $2P$ (Option A).

Solution by Sreeraj P, M.Sc Physics

Answer: (D) $64.6$ cm

Concept. Gas law on each side of the piston; at equilibrium both sides have the same pressure.

Step 1. Equations. Each half has length $x$ at pressure $P$:

• heated side: $\dfrac{P'(x + 5)}{373} = \dfrac{Px}{273}$;
• other side (still $273$ K): $P'(x - 5) = Px$.

Step 2. Divide.
$$\frac{x + 5}{x - 5} = \frac{373}{273} \;\Rightarrow\; 100x = 3230 \;\Rightarrow\; x = 32.3\ \text{cm}$$

Step 3. Length. $2x = 64.6$ cm.

Answer. $64.6$ cm (Option D).

Solution by Sreeraj P, M.Sc Physics

Answer: (A) $483$ m s$^{-1}$

Concept. $v_{\text{rms}} = \sqrt{\dfrac{3RT}{M}}$ with $M$ in kg/mol.

Step 1. Substitute.
$$v_{\text{rms}} = \sqrt{\frac{3 \times 8.31 \times 300}{0.032}} \approx 483\ \text{m s}^{-1}$$

Answer. $483$ m s$^{-1}$ (Option A).

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $\sqrt{3} : \sqrt{8/\pi} : \sqrt{2}$

Concept. $v_{\text{rms}} = \sqrt{\dfrac{3RT}{M}}$, $v_{\text{avg}} = \sqrt{\dfrac{8RT}{\pi M}}$, $v_{\text{mp}} = \sqrt{\dfrac{2RT}{M}}$.

Step 1. Ratio. $\sqrt3 : \sqrt{8/\pi} : \sqrt2$ (about $1.73 : 1.60 : 1.41$).

Answer. $\sqrt3 : \sqrt{8/\pi} : \sqrt2$ (Option B).

Solution by Sreeraj P, M.Sc Physics

Answer: 500

Concept. Kinetic theory: $p = \dfrac13\rho v_{\text{rms}}^2$.

Step 1. Solve.
$$v_{\text{rms}} = \sqrt{\frac{3p}{\rho}} = \sqrt{\frac{3 \times 10^5}{1.2}} = \sqrt{2.5 \times 10^5} = 500\ \text{m s}^{-1}$$

Answer. $500$.

Solution by Sreeraj P, M.Sc Physics

Answer: (C) $7.5$ kJ

Concept. Translational KE of $n$ moles: $E = \dfrac32nRT$.

Step 1. Substitute.
$$E = \frac32 \times 2 \times 8.31 \times 300 \approx 7479\ \text{J} \approx 7.5\ \text{kJ}$$

Answer. $7.5$ kJ (Option C).

Solution by Sreeraj P, M.Sc Physics

Answer: (D) $1.5$

Concept. For a mixture, $C_v$ is the mole-weighted average; $C_p = C_v + R$.

Step 1. $C_v$. $\dfrac{\frac32R + \frac52R}{2} = 2R$.

Step 2. $\gamma$. $C_p = 3R$, so $\gamma = \dfrac{3R}{2R} = 1.5$.

Answer. $1.5$ (Option D).

Solution by Sreeraj P, M.Sc Physics

Answer: (A) $\dfrac{9}{7}$

Concept. $\gamma = 1 + \dfrac{2}{f}$; vibration adds 2 degrees of freedom.

Step 1. Degrees of freedom. $3 + 2 + 2 = 7$.

Step 2. $\gamma$. $1 + \dfrac27 = \dfrac97$.

Answer. $\dfrac97$ (Option A).

Solution by Sreeraj P, M.Sc Physics

Answer: (B) is halved

Concept. $\lambda = \dfrac{kT}{\sqrt2\pi d^2p}$, so at constant $T$, $\lambda \propto \dfrac1p$.

Step 1. Double $p$. The molecules are packed twice as closely: $\lambda$ is halved.

Answer. Is halved (Option B).

Solution by Sreeraj P, M.Sc Physics

Answer: (C) $6.2 \times 10^{-21}$ J

Concept. Average translational KE $= \dfrac32k_BT$.

Step 1. Substitute. $1.5 \times 1.38 \times 10^{-23} \times 300 \approx 6.2 \times 10^{-21}$ J.

Answer. $6.2 \times 10^{-21}$ J (Option C).

Solution by Sreeraj P, M.Sc Physics

Answer: (D) $20$ K

Concept. $v_{\text{rms}} \propto \sqrt{\dfrac{T}{M}}$.

Step 1. Equal speeds. $\dfrac{T_H}{2} = \dfrac{320}{32} \Rightarrow T_H = 20$ K.

Answer. $20$ K (Option D).

Solution by Sreeraj P, M.Sc Physics

Answer: (A) $20$ m

Concept. Boyle's law, then $p_{\text{bottom}} = p_0 + \rho gh$.

Step 1. Pressure at the bottom. $p_{\text{bottom}}V = p_0(3V) \Rightarrow p_{\text{bottom}} = 3 \times 10^5$ Pa.

Step 2. Depth. $\rho gh = 2 \times 10^5 \Rightarrow h = \dfrac{2 \times 10^5}{10^4} = 20$ m.

Answer. $20$ m (Option A).

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $\dfrac{5}{2}RT$

Concept. Equipartition: $\dfrac12RT$ per mole per degree of freedom.

Step 1. Rigid diatomic. 3 translational $+$ 2 rotational $= 5$: $U = \dfrac52RT$.

Answer. $\dfrac52RT$ (Option B).

Solution by Sreeraj P, M.Sc Physics

Answer: (C) $2.7 \times 10^{25}$

Concept. $p = nk_BT$.

Step 1. Solve.
$$n = \frac{1.01 \times 10^5}{1.38 \times 10^{-23} \times 273} \approx 2.7 \times 10^{25}\ \text{m}^{-3}$$

Answer. $2.7 \times 10^{25}$ (Option C).

Solution by Sreeraj P, M.Sc Physics

Answer: (D) four times

Concept. $v_{\text{rms}} \propto \sqrt{T}$; at constant pressure $V \propto T$.

Step 1. Temperature. Doubling $v_{\text{rms}}$ makes $T$ four times.

Step 2. Volume. Four times.

Answer. Four times (Option D).

Solution by Sreeraj P, M.Sc Physics

Answer: (A) $100$

Concept. The vessels are joined, so the gas can move but its total amount stays fixed, and the final pressure is the same in both. Count moles with $n=\dfrac{PV}{RT}$ in each vessel before and after.

Step 1. Before (both at $400$ K): $n=\dfrac{90V}{R}\left(\dfrac1{400}+\dfrac1{400}\right)$.

Step 2. After (pressure $P$, temperatures $400$ and $500$ K): $n=\dfrac{PV}{R}\left(\dfrac1{400}+\dfrac1{500}\right)$.

Step 3. Equate. $90\times\dfrac{2}{400}=P\times\dfrac{9}{2000}\Rightarrow P=\dfrac{0.45}{0.0045}=100$ kPa.

Answer. Option A.

Solution by Sreeraj P, M.Sc Physics

Answer: (D) $\dfrac{7}{4}T$

Concept. When gases mix in isolation, the total internal energy is conserved. For monatomic gases $U=\tfrac32nRT$, so $\sum n_iT_i=(\sum n_i)T$.

Step 1. Energy balance. $2T+6(2T)=8T_f$.

Step 2. Solve. $T_f=\dfrac{14T}{8}=\dfrac74T$.

Answer. Option D.

Solution by Sreeraj P, M.Sc Physics

Answer: (C) $\dfrac{4}{T}$

Concept. Coefficient of volume expansion $\gamma=\dfrac1V\dfrac{dV}{dT}$. Combine the given relation with $PV=nRT$ to get $V$ as a power of $T$.

Step 1. Eliminate $P$. $P=\dfrac{nRT}{V}$, so $PT^3=\dfrac{nRT^4}{V}=$ constant, i.e. $V\propto T^4$.

Step 2. Coefficient. $\dfrac1V\dfrac{dV}{dT}=\dfrac4T$.

Answer. Option C. In general $V\propto T^k$ gives $\gamma=\dfrac kT$.

Solution by Sreeraj P, M.Sc Physics

Answer: (C) A is true but $\mathbf{R}$ is false

Concept. Mean translational KE $\tfrac32k_BT$ depends only on $T$, not on the gas or the container. The rms speed $\sqrt{3RT/M}$ depends on the molar mass.

Step 1. Assertion. True. Reason. False (H$_2$ is four times faster than O$_2$).

Answer. Option C.

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $\dfrac{T_1T_2PV}{T_2P_1V_1 + T_1P_2V_2}$

Concept. Mixing conserves the number of moles: $\dfrac{P_1V_1}{T_1}+\dfrac{P_2V_2}{T_2}=\dfrac{PV}{T}$.

Step 1. Solve. $T=\dfrac{PV}{P_1V_1/T_1+P_2V_2/T_2}=\dfrac{T_1T_2PV}{T_2P_1V_1+T_1P_2V_2}$.

Answer. Option B.

Solution by Sreeraj P, M.Sc Physics

Answer: (C) $0.21$ and $0.12$

Concept. The gas law gives the total number of moles; the mass gives a second equation. Solve the pair.

Step 1. Total moles. $n=\dfrac{PV}{RT}=\dfrac{10^5\times8.31\times10^{-3}}{8.31\times300}=\dfrac13$ mol.

Step 2. Mass equation. $44a+32b=13.2$ with $a+b=\dfrac13$.

Step 3. Solve. $44a+32\left(\dfrac13-a\right)=13.2\Rightarrow12a=2.53\Rightarrow a=0.21$, $b=0.12$.

Answer. Option C.

Solution by Sreeraj P, M.Sc Physics

Answer: (C) $2\sqrt2\times10^{-8}$

Concept. Mean free path $\lambda=\dfrac{1}{\sqrt2\pi d^2n}=\dfrac{k_BT}{\sqrt2\pi d^2P}$.

Step 1. Values. $T=314$ K, $d^2=25\times10^{-20}$ m$^2$.

Step 2. Compute.
$$\lambda=\frac{1.38\times10^{-23}\times314}{1.414\times3.14\times25\times10^{-20}\times1.38\times10^5}=\frac{314}{1.11\times10^{10}}=2.83\times10^{-8}\ \text{m}$$

Answer. $2\sqrt2\times10^{-8}$ m (Option C).

Solution by Sreeraj P, M.Sc Physics

Answer: 6460

Concept. In a closed cylinder (fixed volume), $P \propto T$ (absolute).

Step 1. Double the absolute temperature. $323$ K → $646$ K:
$$P' = 2 \times 3.23\ \text{kPa} = 6460\ \text{Pa}$$

(Reading "double" as $50°$C → $100°$C gives $3230 \times \dfrac{373}{323} = 3730$ Pa; the official key accepted both 3730 and 6460.)

Answer. $6460$.

Solution by Sreeraj P, M.Sc Physics

Answer: (C) $\gamma_2 < \gamma_1$

Concept. $\gamma=1+\dfrac2f$: extra vibrational modes raise $f$ and so lower $\gamma$.

Answer. $\gamma_2\lt\gamma_1$ (Option C).

Solution by Sreeraj P, M.Sc Physics

Answer: (B) see figure

Concept. $\overline{v^2}=\dfrac{3k_BT}{m}$: proportional to $T$.

Step 1. Choose. A straight line through the origin.

Answer. Option B.

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $4102.8\ \text{J}$

Concept. Every gas has translational kinetic energy $\tfrac32nRT$, whatever its atomicity.

Step 1. Moles. $\dfrac{50}{44}=1.136$ mol; $T=290$ K.

Step 2. Energy. $\tfrac32\times1.136\times8.3\times290=4103$ J.

Answer. Option B. Rotational energy is not counted here.

Solution by Sreeraj P, M.Sc Physics

Answer: (C) $\dfrac{5}{2}$

Concept. $v_{\text{rms}}=\sqrt{\dfrac{3RT}{M}}$. Vapour density is proportional to molar mass, so at the same temperature $v_{\text{rms}}\propto\dfrac{1}{\sqrt{\rho}}$.

Step 1. Ratio. $\dfrac{v_1}{v_2}=\sqrt{\dfrac{\rho_2}{\rho_1}}=\sqrt{\dfrac{25}{4}}=\dfrac52$.

Answer. Option C.

Solution by Sreeraj P, M.Sc Physics

Answer: 327

Concept. Fixed amount at fixed volume: $P\propto T$ in kelvin.

Step 1. Double the pressure. $T_2=2\times300=600$ K $=327^\circ$C.

Answer. $327$.

Solution by Sreeraj P, M.Sc Physics

Answer: (A) $\dfrac{9}{4}R$

Concept. For a mixture, $C_V=\dfrac{\sum n_iC_{V,i}}{\sum n_i}$ (energies add).

Step 1. Apply. $\dfrac{2\times\frac32R+6\times\frac52R}{8}=\dfrac{3R+15R}{8}=\dfrac94R$.

Answer. Option A.

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $0.5$

Concept. Same temperature: $v_{rms}\propto\dfrac1{\sqrt M}$.

Step 1. Scale. $2\sqrt{\dfrac{2}{32}}=0.5$ km s$^{-1}$.

Answer. Option B.

Solution by Sreeraj P, M.Sc Physics

Answer: (A) $\rho_1 > \rho_2$

Concept. $P=\dfrac{\rho}{M}RT$: on a $P$–$T$ graph the slope is proportional to the density.

Step 1. Compare. $\rho_1$ has the steepest line: $\rho_1>\rho_2>\rho_3$.

Answer. Option A.

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $f_t = 3$ and $f_r = 3$

Concept. A non-linear molecule can move along 3 axes and rotate about 3 axes. CH$_4$ is tetrahedral (non-linear).

Answer. $f_t=3$, $f_r=3$ (Option B).

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $\dfrac{2}{\sqrt3}Z$

Concept. Collision frequency $=\dfrac{v}{\lambda}$. In a closed chamber $n$ is fixed, so $\lambda$ is fixed, and $v\propto\sqrt T$.

Step 1. Ratio. $\sqrt{\dfrac{400}{300}}=\dfrac{2}{\sqrt3}$.

Answer. $\dfrac{2}{\sqrt3}Z$ (Option B).

Solution by Sreeraj P, M.Sc Physics

Answer: (C) $\dfrac{1}{\sqrt2\,n\pi d^2}$

Concept. Mean free path $\lambda=\dfrac{1}{\sqrt2\,\pi d^2n}$.

Answer. Option C.

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $\dfrac{2\sqrt2}{1}$

Concept. In a mixture both gases share the same temperature, so $v_{\text{rms}} \propto \dfrac{1}{\sqrt M}$.

Step 1. Ratio.
$$\frac{v_{He}}{v_{O_2}} = \sqrt{\frac{32}{4}} = 2\sqrt2$$

Answer. $\dfrac{2\sqrt2}{1}$ (Option B).

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $35\,k_BT$

Concept. Equipartition: $\tfrac12k_BT$ per degree of freedom. A non-rigid diatomic has $3$ translational $+2$ rotational $+2$ vibrational (kinetic and potential) $=7$.

Step 1. Per molecule. $\tfrac72k_BT$.

Step 2. Ten molecules. $35k_BT$.

Answer. Option B.

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $3200\ \text{K}$

Concept. Monatomic molecule: average KE $= \dfrac32k_BT$.

Step 1. Solve for $T$.
$$T = \frac{2E}{3k_B} = \frac{2 \times 0.414 \times 1.6 \times 10^{-19}}{3 \times 1.38 \times 10^{-23}} = \frac{1.325 \times 10^{-19}}{4.14 \times 10^{-23}} \approx 3200\ \text{K}$$

Answer. $3200$ K (Option B).

Solution by Sreeraj P, M.Sc Physics

Answer: (C) $6232.5\ \text{J}$

Concept. Total KE of 1 mol of a diatomic gas (rigid, 5 degrees of freedom): $\tfrac52RT$.

Step 1. Substitute. $2.5\times8.31\times300=6232.5$ J.

Answer. Option C.

Solution by Sreeraj P, M.Sc Physics

Answer: (A) 16

Concept. Same $V$ and $T$: $P \propto n$.

Step 1. Ratio.
$$\frac{P_A}{P_B} = \frac{1/2}{1/32} = 16$$

Answer. $16$ (Option A).

Solution by Sreeraj P, M.Sc Physics

Answer: (A) $500\ \text{K}$

Concept. $P=nk_BT$ with $n$ the number density.

Step 1. Solve. $T=\dfrac{P}{nk_B}=\dfrac{1.38\times10^5}{2\times10^{25}\times1.38\times10^{-23}}=500$ K (with $1$ atm $\approx10^5$ Pa).

Answer. Option A.

Solution by Sreeraj P, M.Sc Physics

Answer: (C) 4

Concept. A mixture's $C_V$ is the mole-weighted mean: $C_V=\dfrac{\sum n_iC_{V,i}}{\sum n_i}$. "Behaves as diatomic" means $C_V=\tfrac52R$.

Step 1. Equation. $\dfrac{3N+2\times\tfrac32}{N+2}=\dfrac52\Rightarrow3N+3=2.5N+5$.

Step 2. Solve. $N=4$.

Answer. Option C.

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $\dfrac35$

Concept. $C_v = \dfrac{f}{2}R$.

Step 1. Values. Monatomic $\dfrac32R$, rigid diatomic $\dfrac52R$.

Step 2. Ratio. $\dfrac35$.

Answer. $\dfrac35$ (Option B).

Solution by Sreeraj P, M.Sc Physics

Answer: (C) Both Statement I and Statement II are true

Concept. $\lambda\propto\dfrac1{d^2}$; mean KE $=\tfrac32k_BT$.

Answer. Both true (Option C).

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $117\,{}^\circ\text{C}$

Concept. Mean translational KE $\propto T$ in kelvin.

Step 1. Double it. $T=2\times195=390$ K $=117^\circ$C.

Answer. Option B.

Solution by Sreeraj P, M.Sc Physics

Answer: (D) $20\ \text{K}$

Concept. $v_{\text{rms}}=\sqrt{\dfrac{3RT}{M}}$; equal speeds need equal $\dfrac TM$.

Step 1. Oxygen. $\dfrac{T}{M}=\dfrac{320}{32}=10$.

Step 2. Hydrogen. $T=10\times2=20$ K.

Answer. Option D.

Solution by Sreeraj P, M.Sc Physics

Answer: (C) $1.52$

Concept. Mix $C_V$ by moles, then $C_P=C_V+R$ and $\gamma=\dfrac{C_P}{C_V}$. (Do not average the $\gamma$ values.)

Step 1. $C_V$. $\dfrac{3\times1.5R+2\times2.5R}{5}=1.9R$.

Step 2. $\gamma$. $\dfrac{2.9}{1.9}=1.53$.

Answer. Option C.

Solution by Sreeraj P, M.Sc Physics

Answer: (D) $P_1 > P_2$

Concept. $V=\dfrac{nR}{P}T$: on a $V$–$T$ graph the slope is inversely proportional to the pressure.

Step 1. Compare. $P_2$'s line is steeper, so $P_2\lt P_1$.

Answer. Option D.

Solution by Sreeraj P, M.Sc Physics

Answer: (A) kinetic energy

Concept. At a given temperature every gas has the same mean translational KE, $\tfrac32k_BT$; masses, speeds and momenta differ.

Answer. Option A.

Solution by Sreeraj P, M.Sc Physics

Answer: (C) $27RT$

Concept. Total internal energy $=\sum n_i\dfrac{f_i}{2}RT$. Argon is monatomic ($f=3$), oxygen rigid diatomic ($f=5$).

Step 1. Add. $8\times\tfrac32RT+6\times\tfrac52RT=12RT+15RT=27RT$.

Answer. Option C.

Solution by Sreeraj P, M.Sc Physics

Answer: (C) $278\ \text{kPa}$

Concept. A tyre's volume is nearly fixed: $P\propto T$ (kelvin).

Step 1. Pressure. $270\times\dfrac{309}{300}\approx278$ kPa.

Answer. Option C.

Solution by Sreeraj P, M.Sc Physics

Answer: (B) 28

Concept. $\dfrac{v_{rms}}{\bar v}=\sqrt{\dfrac{3RT/M}{8RT/\pi M}}=\sqrt{\dfrac{3\pi}{8}}$, at any temperature.

Step 1. With $\pi=\tfrac{22}{7}$. $\dfrac{3\pi}{8}=\dfrac{66}{56}=\dfrac{33}{28}=\dfrac{28+5}{28}$.

Answer. $\alpha=28$ (Option B).

Solution by Sreeraj P, M.Sc Physics

Answer: (D) $\dfrac{3}{T}$

Concept. Volume expansion coefficient $\gamma=\dfrac1V\dfrac{dV}{dT}$. Combine the process with $PV=nRT$ to get $V(T)$.

Step 1. $V(T)$. $P=\dfrac{c}{T^2}$, so $V=\dfrac{nRT}{P}=\dfrac{nR}{c}T^3$.

Step 2. Coefficient. $\dfrac{3}{T}$.

Answer. Option D.

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $1:1$

Concept. Average translational KE per molecule $\dfrac32kT$ depends only on temperature.

Step 1. Same temperature. Ratio $1 : 1$ (the mass ratio is irrelevant).

Answer. $1 : 1$ (Option B).

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $\gamma\propto T^0$

Concept. In the ideal-gas model $C_P$ and $C_V$ are fixed numbers ($\gamma=1+2/f$), independent of temperature.

Answer. $\gamma\propto T^0$ (Option B).

Solution by Sreeraj P, M.Sc Physics

Answer: (B) Statement I is true but Statement II is false

Concept. $v_{\text{rms}}\propto\sqrt T$ ($T$ in kelvin). Kinetic theory gives $PV=\tfrac23E_{\text{trans}}$, not $PV=E_{\text{trans}}$.

Step 1. Statement I. $-73^\circ$C $=200$ K, $527^\circ$C $=800$ K; $T$ quadruples, so $v_{\text{rms}}$ doubles. True.

Step 2. Statement II. $PV=\tfrac23E$, so it is false.

Answer. Option B.

Solution by Sreeraj P, M.Sc Physics

Answer: (C) $\dfrac{25}{21}$

Concept. $\gamma = 1 + \dfrac2f$: monatomic $f = 3$, rigid diatomic $f = 5$.

Step 1. Values. $\gamma_1 = \dfrac53$, $\gamma_2 = \dfrac75$.

Step 2. Ratio. $\dfrac53 \times \dfrac57 = \dfrac{25}{21}$.

Answer. $\dfrac{25}{21}$ (Option C).

Solution by Sreeraj P, M.Sc Physics

Answer: (C) Proportional to square root of temperature $\sqrt T$

Concept. $v_{rms}=\sqrt{\dfrac{3RT}{M}}\propto\sqrt T$.

Answer. Option C.

Solution by Sreeraj P, M.Sc Physics

Answer: (D) $\dfrac72R$

Concept. Equipartition: each translational and rotational degree gives $\tfrac12R$ to $C_V$; a vibrational mode gives $R$ (kinetic and potential).

Step 1. Add. $\tfrac32R+R+R=\tfrac72R$.

Answer. Option D.

Solution by Sreeraj P, M.Sc Physics

Answer: (C) $2049d$

Concept. $\lambda=\dfrac{k_BT}{\sqrt2\pi d^2P}$: at constant pressure $\lambda\propto T$.

Step 1. Apply. $1500d\times\dfrac{373}{273}=2049d$.

Answer. Option C.

Solution by Sreeraj P, M.Sc Physics

Answer: (B) 1

Concept. Average translational KE per molecule is $\dfrac32kT$, independent of the gas.

Step 1. Same temperature. Ratio $= 1$.

Answer. $1$ (Option B).

Solution by Sreeraj P, M.Sc Physics

Answer: (A) $11RT$

Concept. $U=\sum n_i\dfrac{f_i}{2}RT$: O$_2$ $f=5$, Ne $f=3$.

Step 1. Add. $2\times\tfrac52RT+4\times\tfrac32RT=11RT$.

Answer. Option A.

Solution by Sreeraj P, M.Sc Physics

Answer: (A) $v_{rms}(\text{mono})>v_{rms}(\text{dia})>v_{rms}(\text{poly})$

Concept. At the same temperature, $v_{\text{rms}}\propto\dfrac{1}{\sqrt M}$: lighter molecules move faster. Atomicity does not matter here.

Step 1. Molar masses. Ne $20$, Cl$_2$ $71$, UF$_6$ $352$.

Answer. Neon fastest, then chlorine, then UF$_6$ (Option A).

Solution by Sreeraj P, M.Sc Physics

Answer: (D) $523\ \text{m s}^{-1}$

Concept. $v_{rms}=\sqrt{\dfrac{3k_BT}{m}}$.

Step 1. Substitute. $\sqrt{\dfrac{3\times1.4\times10^{-23}\times300}{4.6\times10^{-26}}}=\sqrt{2.74\times10^5}\approx523$ m s$^{-1}$.

Answer. Option D.

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $651.7\ \text{m s}^{-1}$

Concept. Same temperature: $v_{rms}\propto\dfrac1{\sqrt M}$.

Step 1. Scale. $490\sqrt{\dfrac{70.9}{39.9}}=490\times1.333\approx653$ m s$^{-1}$, closest to $651.7$.

Answer. Option B.

Solution by Sreeraj P, M.Sc Physics

Answer: (C) 28

Concept. $\dfrac{v_{\text{rms}}}{v_{\text{avg}}} = \sqrt{\dfrac{3\pi}{8}}$.

Step 1. Substitute $\pi = \dfrac{22}{7}$.
$$\sqrt{\frac{33}{28}} = \sqrt{1 + \frac{5}{28}}$$

Step 2. Compare. $x = 28$.

Answer. $28$ (Option C).

Solution by Sreeraj P, M.Sc Physics

Answer: (D) 0.25

Concept. Collision frequency $z=\sqrt2\pi d^2n\bar v$: proportional to the number density at fixed $T$.

Step 1. Ratio. $\dfrac{3}{12}=0.25$.

Answer. Option D.

Solution by Sreeraj P, M.Sc Physics

Answer: (D) $2v_0$

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

Step 1. Ratio. $T$ becomes 4 times, so $v = 2v_0$.

Answer. $2v_0$ (Option D).

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $327^\circ\text{C}$

Concept. Kinetic energy of the molecules $\propto T$ (absolute).

Step 1. Double. $T = 2 \times 300 = 600$ K $= 327°$C.

Answer. $327°$C (Option B).

Solution by Sreeraj P, M.Sc Physics

Answer: (A) A-I, B-III, C-IV, D-II

Concept. Monatomic: 3 translational. Rigid diatomic: 3 + 2. Non-rigid diatomic: 3 + 2 + vibration. Polyatomic: 3 + 3 + vibrations.

Answer. A-I, B-III, C-IV, D-II (Option A).

Solution by Sreeraj P, M.Sc Physics

Answer: (A) proportional to absolute temperature

Concept. Average translational KE per molecule $= \dfrac32kT$.

Step 1. Dependence. Proportional to absolute temperature only.

Answer. Proportional to absolute temperature (Option A).

Solution by Sreeraj P, M.Sc Physics

Answer: (A) $-273^\circ\text{C}$

Concept. At constant volume $P\propto T$ (kelvin): every gas's $P$–$t$ line meets the temperature axis at absolute zero.

Answer. $-273^\circ$C (Option A).

Solution by Sreeraj P, M.Sc Physics

Answer: 12

Concept. Speed $\propto\sqrt T$, so doubling the speed quadruples the absolute temperature. In a closed vessel the heat goes at constant volume: $Q=nC_V\Delta T$.

Step 1. Moles and temperatures. $n=\dfrac{56}{28}=2$; $T$ from $400$ K to $1600$ K, $\Delta T=1200$ K.

Step 2. Heat ($C_V=\tfrac52R=5$ cal mol$^{-1}$ K$^{-1}$): $Q=2\times5\times1200=12\,000$ cal.

Answer. $12$ kcal.

Solution by Sreeraj P, M.Sc Physics

Answer: (C) $v_{\text{rms}} = \sqrt{\dfrac32}\,v_p$

Concept. $v_{\text{rms}}=\sqrt{\dfrac{3RT}{M}}$ and $v_p=\sqrt{\dfrac{2RT}{M}}$.

Step 1. Ratio. $\dfrac{v_{\text{rms}}}{v_p}=\sqrt{\dfrac32}$.

Answer. Option C.

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $1 + \dfrac2f$

Concept. Equipartition: $C_V = \dfrac{f}{2}R$, and $C_P = C_V + R$.

Step 1. Ratio.
$$\gamma = \frac{\left(\frac f2 + 1\right)R}{\frac f2R} = 1 + \frac2f$$

Answer. $1 + \dfrac2f$ (Option B).

Solution by Sreeraj P, M.Sc Physics

Answer: 250

Concept. At fixed volume $\dfrac{\Delta P}{P}=\dfrac{\Delta T}{T}$.

Step 1. Solve. $0.004=\dfrac{1}{T}\Rightarrow T=250$ K.

Answer. $250$.

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $\dfrac{Mv^2}{5R}$

Concept. When the vessel stops, the bulk kinetic energy of the gas becomes internal energy. Per mole: $\tfrac12Mv^2=C_V\Delta T$, with $C_V=\dfrac{R}{\gamma-1}=\dfrac52R$.

Step 1. Solve. $\Delta T=\dfrac{Mv^2}{2\times2.5R}=\dfrac{Mv^2}{5R}$.

Answer. Option B.

Solution by Sreeraj P, M.Sc Physics

Answer: (A) (1) and (4) only

Concept. Mean KE depends on $T$ alone; $PV=nRT$ relates the rest.

Step 1. Check. (1) true; (2) false (fixed $T$, fixed KE); (3) false; (4) true; (5) false (at constant $P$, $V$ rises with $T$).

Answer. Option A.

Solution by Sreeraj P, M.Sc Physics

Answer: (B) 2

Concept. In a mixture, both speeds use the average molar mass: $v_{\text{rms}}=\sqrt{\dfrac{3RT}{M}}$, $v_{\text{sound}}=\sqrt{\dfrac{\gamma RT}{M}}$. Their ratio is $\sqrt{\dfrac{3}{\gamma}}$.

Step 1. Find $\gamma$. $\sqrt{\dfrac3\gamma}=\sqrt2\Rightarrow\gamma=\dfrac32$, so $C_V=2R$.

Step 2. Mixture. $\dfrac{2\times\frac32R+n\times\frac52R}{2+n}=2R\Rightarrow n=2$.

Answer. Option B.

Solution by Sreeraj P, M.Sc Physics

Answer: (A) $\dfrac{n}{n+2}$

Concept. $C_V = \dfrac n2R$, $C_P = \left(\dfrac n2 + 1\right)R$.

Step 1. Ratio. $\dfrac{C_V}{C_P} = \dfrac{n}{n + 2}$.

Answer. $\dfrac{n}{n + 2}$ (Option A).

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $\dfrac{3}{1}$

Concept. Total moles from $PV=nRT$, and the mass condition: two equations for $n_{H_2}$ and $n_{O_2}$.

Step 1. Total moles. $\dfrac{10^5\times2\times10^{-3}}{8.3\times300}=0.0803$.

Step 2. Mass. $2n_1+32n_2=0.76$ with $n_1+n_2=0.0803$: $30n_2=0.599\Rightarrow n_2=0.020$, $n_1=0.060$.

Step 3. Ratio. $3:1$.

Answer. Option B.

Solution by Sreeraj P, M.Sc Physics

Answer: (A) $P_1 > P_2$

Concept. $V = \dfrac{nR}{P}T$: the slope of a $V$–$T$ line is $\dfrac{nR}{P}$.

Step 1. Compare. The $P_2$ line is steeper, so $P_2$ is smaller.

Answer. $P_1 > P_2$ (Option A).

Solution by Sreeraj P, M.Sc Physics

Answer: (B) B and C only

Concept. $\lambda=\dfrac{k_BT}{\sqrt2\pi d^2P}=\dfrac{1}{\sqrt2\pi d^2n}$; equipartition gives $\tfrac12k_BT$ per degree of freedom.

Step 1. Check. A false (motion stops only at absolute zero, and not even then classically); B true; C true; D false ($\tfrac12k_BT$).

Answer. B and C only (Option B).

Solution by Sreeraj P, M.Sc Physics

Answer: (C) A and B only

Concept. Same gas, same $T$: same $v_{rms}$. Same $V$ and $T$: $P\propto N$.

Step 1. Check. A true, B true; C and D false.

Answer. Option C.

Solution by Sreeraj P, M.Sc Physics

Answer: (B) B and D only

Concept. Each degree of freedom carries $\tfrac12k_BT$ per molecule ($\tfrac12RT$ per mole).

Step 1. Check. A false (there are $n$ ways, not $n^2$); B true; C false (a monatomic molecule has no rotational degree); D true (CH$_4$, non-linear rigid: $3+3$).

Answer. Option B.

Solution by Sreeraj P, M.Sc Physics

Answer: (D) Statement I is false but Statement II is true

Concept. The average momentum (a vector) of molecules moving at random is zero at any temperature. $v_{rms}\propto\sqrt{T/M}$.

Step 1. Statement I. False.

Step 2. Statement II. $T\times2$ and $M\div2$ (O$_2\to$ O): $v\times\sqrt4=2v$. True.

Answer. Option D.

Solution by Sreeraj P, M.Sc Physics

Answer: (C) $1.5\times10^{11}$

Concept. Kinetic theory: $PV = \dfrac23N\bar E$.

Step 1. Solve. $1$ atm $\approx 10^5$ Pa:
$$N = \frac{3PV}{2\bar E} = \frac{3 \times 2 \times 10^5 \times 10^{-3}}{2 \times 2 \times 10^{-9}} = 1.5 \times 10^{11}$$

Answer. $1.5 \times 10^{11}$ (Option C).

Solution by Sreeraj P, M.Sc Physics

Answer: (B) work done on the gas is close to $582$ J

Concept. In an adiabatic process the work on the gas equals the rise in internal energy: $W_{\text{on}}=nC_V\Delta T$. Count $f$: non-linear polyatomic $3+3$, plus $2$ for each of $4$ vibrational modes: $f=14$, $C_V=7R$.

Step 1. Work on the gas. $7\times8.314\times10=582$ J.

Answer. Option B (temperature rises, so work is done on the gas).

Solution by Sreeraj P, M.Sc Physics

Answer: (D) $\dfrac{P_0}{aeR}$

Concept. Solve the differential equation for $P(V)$, then $T=\dfrac{PV}{R}$ (one mole) and maximise.

Step 1. Pressure. $\dfrac{dP}{P}=-a\,dV\Rightarrow P=P_0e^{-aV}$.

Step 2. Temperature. $T=\dfrac{P_0Ve^{-aV}}{R}$.

Step 3. Maximise. $\dfrac{dT}{dV}=\dfrac{P_0}{R}e^{-aV}(1-aV)=0\Rightarrow V=\dfrac1a$.

Step 4. Value. $T_{\max}=\dfrac{P_0}{aeR}$.

Answer. Option D.

Solution by Sreeraj P, M.Sc Physics

Answer: (A) $16 : 3$

Concept. Total moles from $PV = nRT$, then split using the total mass.

Step 1. Total moles.
$$n = \frac{4 \times 10^5 \times 5 \times 10^{-4}}{8.314 \times 300} \approx 0.08\ \text{mol}$$

Step 2. Split. $n_H + n_O = 0.08$, $2n_H + 32n_O = 0.76$: $30n_O = 0.60 \Rightarrow n_O = 0.02$, $n_H = 0.06$.

Step 3. Masses. Oxygen $0.64$ g, hydrogen $0.12$ g: $16 : 3$.

Answer. $16 : 3$ (Option A).

Solution by Sreeraj P, M.Sc Physics

Answer: (B) suffer change in momentum when impinge on the walls of container.

Concept. Molecules bounce off the walls; each impact changes their momentum, so the wall exerts (and feels) a force.

Answer. Option B.

Solution by Sreeraj P, M.Sc Physics

Answer: 400

Concept. $v_{\text{rms}}$ depends only on $T$ (for a given gas); pressure does not enter.

Step 1. Ratio. $\dfrac{v_2}{v_1}=\sqrt{\dfrac{400}{300}}=\dfrac{2}{\sqrt3}$.

Step 2. New speed. $200\times\dfrac{2}{\sqrt3}=\dfrac{400}{\sqrt3}$.

Answer. $x=400$.

Solution by Sreeraj P, M.Sc Physics

Answer: (C) $\sqrt{\frac{3\pi}{8}}$

Concept. The three molecular speeds: $v_{\text{rms}}=\sqrt{\dfrac{3RT}{M}}$, $v_{\text{avg}}=\sqrt{\dfrac{8RT}{\pi M}}$, $v_p=\sqrt{\dfrac{2RT}{M}}$. Their ratios do not depend on the gas or the temperature.

Step 1. Ratio. $\dfrac{v_{\text{rms}}}{v_{\text{avg}}}=\sqrt{\dfrac{3}{8/\pi}}=\sqrt{\dfrac{3\pi}{8}}$.

Answer. Option C. (Numerical data in the question are not needed.)

Solution by Sreeraj P, M.Sc Physics

Answer: (D) $\frac{1}{v_A} < \frac{1}{v_B} < \frac{1}{v_C}$

Concept. $v_{\text{rms}} = \sqrt{\dfrac{3kT}{m}}$: at the same temperature the lightest molecule is fastest.

Step 1. Order. $v_A > v_B > v_C$, so $\dfrac{1}{v_A} < \dfrac{1}{v_B} < \dfrac{1}{v_C}$.

Answer. $\dfrac{1}{v_A} < \dfrac{1}{v_B} < \dfrac{1}{v_C}$ (Option D).

Solution by Sreeraj P, M.Sc Physics

Answer: (B) see figure

Concept. For an ideal gas $PV=nRT$: $PV$ is proportional to $T$.

Step 1. Choose. A straight line through the origin.

Answer. Option B.

Solution by Sreeraj P, M.Sc Physics

Answer: (A) $f = \frac{2}{\gamma - 1}$

Concept. $\gamma = 1 + \dfrac2f$.

Step 1. Rearrange. $f = \dfrac{2}{\gamma - 1}$.

Answer. $f = \dfrac{2}{\gamma - 1}$ (Option A).

Solution by Sreeraj P, M.Sc Physics

Answer: (C) $\frac32k_BT$

Concept. Equipartition: $\dfrac12k_BT$ per degree of freedom.

Step 1. Monoatomic. 3 translational degrees: $\dfrac32k_BT$.

Answer. $\dfrac32k_BT$ (Option C).

Solution by Sreeraj P, M.Sc Physics

Answer: 26

Concept. The gases are at different temperatures, so do not assume a common $T$. For an ideal gas $U=\dfrac{f}{2}PV$; in an insulated container the total $U$ is conserved, so (same type of gas) $\sum P_iV_i=PV$.

Step 1. Before. $2.0\times4.5+3.0\times5.5=9+16.5=25.5$ atm L.

Step 2. After. $P\times10=25.5\Rightarrow P=2.55$ atm.

Answer. $x\approx26$.

Solution by Sreeraj P, M.Sc Physics

Answer: (B) polyatomic only

Concept. For an ideal gas $U=nC_VT=\dfrac{C_V}{R}PV$ (plus any constant). Compare with $U=3PV+4$.

Step 1. Match. $\dfrac{C_V}{R}=3\Rightarrow C_V=3R$, i.e. $f=6$.

Answer. Polyatomic (Option B).

Solution by Sreeraj P, M.Sc Physics

Answer: 25

Concept. $\lambda\propto\dfrac{1}{nd^2}$; same number density, so $\lambda\propto\dfrac1{d^2}$.

Step 1. Ratio. $\dfrac{\lambda_A}{\lambda_B}=\left(\dfrac{5}{10}\right)^2=0.25$.

Answer. $25\times10^{-2}$.

Solution by Sreeraj P, M.Sc Physics

Answer: (C) $v_H > v_O > v_C$

Concept. At the same temperature $v_{\text{rms}} \propto \dfrac{1}{\sqrt M}$.

Step 1. Molar masses. H$_2$ (2) $<$ O$_2$ (32) $<$ CO$_2$ (44).

Step 2. Order. $v_H > v_O > v_C$.

Answer. $v_H > v_O > v_C$ (Option C).

Solution by Sreeraj P, M.Sc Physics

Answer: (C) $24.9\times10^5$ Pa

Concept. Dalton: the total pressure of a mixture is $\dfrac{(n_1+n_2)RT}{V}$.

Step 1. Substitute. $P=\dfrac{3\times8.3\times400}{4\times10^{-3}}=2.49\times10^6$ Pa.

Answer. Option C.

Solution by Sreeraj P, M.Sc Physics

Answer: (D) $\frac{4}{T}$

Concept. As above: $P=\dfrac{c}{T^3}\Rightarrow V\propto T^4$.

Step 1. Coefficient. $\dfrac1V\dfrac{dV}{dT}=\dfrac4T$.

Answer. Option D.

Solution by Sreeraj P, M.Sc Physics

Answer: (B) $123.54$ kg

Concept. At fixed volume, the load a balloon can lift is proportional to the density of the outside air, $\rho\propto\dfrac PT$.

Step 1. Ratio of densities. $\dfrac{45}{76}\times\dfrac{300}{266}=0.592\times1.128=0.668$.

Step 2. New load. $185\times0.668=123.5$ kg.

Answer. Option B.

Solution by Sreeraj P, M.Sc Physics