JEE Physics Kinetic Theory MCQs Set 10

Answer: (c) 17.4 J/mol-K

Explanation:
1. \( C_V \) of He \( =\tfrac32R \), \( C_V \) of rigid \( \text{H}_2=\tfrac52R \).
2. \( C_{V,mix}=\dfrac{2(1.5R)+3(2.5R)}{5}=2.1R=2.1\times8.3\approx17.4 \) J/mol-K.

Teacher's Note:
1. Use the equipartition result: each degree of freedom carries kT/2 per molecule (RT/2 per mole), so \( U = f \) nRT/2 with \( f = 3, 5, 7 \) for monatomic, diatomic rigid and diatomic with vibration.
2. For rms, mean and most probable speeds use \( \sqrt{3RT/M}, \sqrt{8RT/(\pi M)} \) and \( \sqrt{2RT/M} \); all vary as \( \sqrt{T/M} \).

Answer: (a) \( \sim10^{9} \)

Explanation:
1. \( n=\dfrac{6.02\times10^{23}}{25\times10^{-3}}=2.4\times10^{25} m ^{-3} \); mean free path \( \lambda=\dfrac{1}{\sqrt2\pi d^2n}=\dfrac{1}{1.414(3.14)(9\times10^{-20})(2.4\times10^{25})}\approx1.0\times10^{-7} m \).
2. Collision rate \( =\dfrac v\lambda=\dfrac{200}{1.0\times10^{-7}}\approx1.9\times10^{9} \) per second, of the order of \( 10^9 \).

Teacher's Note:
1. Use the equipartition result: each degree of freedom carries kT/2 per molecule (RT/2 per mole), so \( U = f \) nRT/2 with \( f = 3, 5, 7 \) for monatomic, diatomic rigid and diatomic with vibration.
2. For rms, mean and most probable speeds use \( \sqrt{3RT/M}, \sqrt{8RT/(\pi M)} \) and \( \sqrt{2RT/M} \); all vary as \( \sqrt{T/M} \).

Answer: (c) A has a vibrational mode but B has none

Explanation:
1. \( \gamma=\dfrac{C_p}{C_V} \): for \( A \), \( \dfrac{29}{22}=1.32\approx\dfrac97 \); for \( B \), \( \dfrac{30}{21}=1.43\approx\dfrac75 \).
2. \( \gamma=\dfrac75 \) belongs to a rigid diatomic molecule, and the lower \( \gamma=\dfrac97 \) shows the extra vibrational degree of freedom in \( A \).

Teacher's Note:
1. Use the equipartition result: each degree of freedom carries kT/2 per molecule (RT/2 per mole), so \( U = f \) nRT/2 with \( f = 3, 5, 7 \) for monatomic, diatomic rigid and diatomic with vibration.
2. For rms, mean and most probable speeds use \( \sqrt{3RT/M}, \sqrt{8RT/(\pi M)} \) and \( \sqrt{2RT/M} \); all vary as \( \sqrt{T/M} \).

Answer: (a) \( 100\sqrt5\ \text{m/s} \)

Explanation:
1. \( v_{rms}\propto\sqrt T \) and does not depend on the pressure.
2. \( v=200\sqrt{\dfrac{500}{400}}=200\times\dfrac{\sqrt5}{2}=100\sqrt5 m/s \).

Teacher's Note:
1. Use the equipartition result: each degree of freedom carries kT/2 per molecule (RT/2 per mole), so \( U = f \) nRT/2 with \( f = 3, 5, 7 \) for monatomic, diatomic rigid and diatomic with vibration.
2. For rms, mean and most probable speeds use \( \sqrt{3RT/M}, \sqrt{8RT/(\pi M)} \) and \( \sqrt{2RT/M} \); all vary as \( \sqrt{T/M} \).

Answer: (c) \( 4\times10^{-8}\ \text{s} \)

Explanation:
1. Mean time \( \tau=\dfrac{\lambda}{v_{rms}}\propto\dfrac{1}{n\sqrt T} \) and \( n\propto\dfrac PT \), so \( \tau\propto\dfrac{\sqrt T}{P} \).
2. \( \tau'=6\times10^{-8}\times\sqrt{\dfrac{500}{300}}\times\dfrac12\approx3.9\times10^{-8} s \), close to \( 4\times10^{-8} s \).

Teacher's Note:
1. Use the equipartition result: each degree of freedom carries kT/2 per molecule (RT/2 per mole), so \( U = f \) nRT/2 with \( f = 3, 5, 7 \) for monatomic, diatomic rigid and diatomic with vibration.
2. For rms, mean and most probable speeds use \( \sqrt{3RT/M}, \sqrt{8RT/(\pi M)} \) and \( \sqrt{2RT/M} \); all vary as \( \sqrt{T/M} \).

Answer: (d) \( \dfrac{nRT}{g}\left[\dfrac{l_1-l_2}{l_1l_2}\right] \)

Explanation:
1. With cross-section \( A \): upper pressure \( p_1=\dfrac{nRT}{Al_1} \) and lower pressure \( p_2=\dfrac{nRT}{Al_2} \).
2. Equilibrium: \( mg=(p_2-p_1)A=nRT\left(\dfrac1{l_2}-\dfrac1{l_1}\right)\Rightarrow m=\dfrac{nRT}{g}\dfrac{l_1-l_2}{l_1l_2} \).

Teacher's Note:
1. Use the equipartition result: each degree of freedom carries kT/2 per molecule (RT/2 per mole), so \( U = f \) nRT/2 with \( f = 3, 5, 7 \) for monatomic, diatomic rigid and diatomic with vibration.
2. For rms, mean and most probable speeds use \( \sqrt{3RT/M}, \sqrt{8RT/(\pi M)} \) and \( \sqrt{2RT/M} \); all vary as \( \sqrt{T/M} \).

Answer: (b) 15 RT

Explanation:
1. Oxygen (diatomic): \( 3\times\tfrac52RT=7.5RT \).
2. Argon (monoatomic): \( 5\times\tfrac32RT=7.5RT \). Total \( U=15RT \).

Teacher's Note:
1. Use the equipartition result: each degree of freedom carries kT/2 per molecule (RT/2 per mole), so \( U = f \) nRT/2 with \( f = 3, 5, 7 \) for monatomic, diatomic rigid and diatomic with vibration.
2. For rms, mean and most probable speeds use \( \sqrt{3RT/M}, \sqrt{8RT/(\pi M)} \) and \( \sqrt{2RT/M} \); all vary as \( \sqrt{T/M} \).

Answer: (d) 10 kJ

Explanation:
1. \( v_{rms}\propto\sqrt T \): doubling it needs \( T'=4T=1200 K \), so \( \Delta T=900 K \). \( n=\dfrac{15}{28}=0.536 \).
2. \( Q=n\cdot\tfrac52R\Delta T=0.536\times2.5\times8.3\times900\approx1.0\times10^4 J = 10 \) kJ.

Teacher's Note:
1. Use the equipartition result: each degree of freedom carries kT/2 per molecule (RT/2 per mole), so \( U = f \) nRT/2 with \( f = 3, 5, 7 \) for monatomic, diatomic rigid and diatomic with vibration.
2. For rms, mean and most probable speeds use \( \sqrt{3RT/M}, \sqrt{8RT/(\pi M)} \) and \( \sqrt{2RT/M} \); all vary as \( \sqrt{T/M} \).

Answer: (b) \( 2.35\times10^3\ \text{N/m}^2 \)

Explanation:
1. Change of momentum per molecule (normal component reverses) \( =2mv\cos45^\circ=2(3.32\times10^{-27})(10^3)(0.707)=4.7\times10^{-24} \) kg \( m/s \).
2. Force \( =10^{23}\times4.7\times10^{-24}=0.47 N \); pressure \( =\dfrac{0.47}{2\times10^{-4}}\approx2.35\times10^3 N/m ^2 \).

Teacher's Note:
1. Use the equipartition result: each degree of freedom carries kT/2 per molecule (RT/2 per mole), so \( U = f \) nRT/2 with \( f = 3, 5, 7 \) for monatomic, diatomic rigid and diatomic with vibration.
2. For rms, mean and most probable speeds use \( \sqrt{3RT/M}, \sqrt{8RT/(\pi M)} \) and \( \sqrt{2RT/M} \); all vary as \( \sqrt{T/M} \).

Answer: (d) \( n=\dfrac{C-C_p}{C-C_V} \)

Explanation:
1. For \( pV^n= \) const: \( C=C_V+\dfrac{R}{1-n} \). Using \( R=C_p-C_V \): \( C-C_V=\dfrac{C_p-C_V}{1-n} \).
2. \( 1-n=\dfrac{C_p-C_V}{C-C_V}\Rightarrow n=\dfrac{C-C_p}{C-C_V} \). Check: \( C=C_V \) gives \( n\to\infty \) (isochoric) and \( C=C_p \) gives \( n=0 \) (isobaric).

Teacher's Note:
1. Use the equipartition result: each degree of freedom carries kT/2 per molecule (RT/2 per mole), so \( U = f \) nRT/2 with \( f = 3, 5, 7 \) for monatomic, diatomic rigid and diatomic with vibration.
2. For rms, mean and most probable speeds use \( \sqrt{3RT/M}, \sqrt{8RT/(\pi M)} \) and \( \sqrt{2RT/M} \); all vary as \( \sqrt{T/M} \).

Answer: (b) pressure is low and temperature is high

Explanation:
1. Ideal behaviour needs negligible intermolecular forces and negligible molecular volume.
2. This holds when the gas is dilute (low pressure) and the molecules are fast (high temperature).

Teacher's Note:
1. Use the equipartition result: each degree of freedom carries kT/2 per molecule (RT/2 per mole), so \( U = f \) nRT/2 with \( f = 3, 5, 7 \) for monatomic, diatomic rigid and diatomic with vibration.
2. For rms, mean and most probable speeds use \( \sqrt{3RT/M}, \sqrt{8RT/(\pi M)} \) and \( \sqrt{2RT/M} \); all vary as \( \sqrt{T/M} \).

Answer: (c) 11 RT

Explanation:
1. Oxygen: \( 2\times\tfrac52RT=5RT \). Argon: \( 4\times\tfrac32RT=6RT \).
2. Total \( U=11RT \).

Teacher's Note:
1. Use the equipartition result: each degree of freedom carries kT/2 per molecule (RT/2 per mole), so \( U = f \) nRT/2 with \( f = 3, 5, 7 \) for monatomic, diatomic rigid and diatomic with vibration.
2. For rms, mean and most probable speeds use \( \sqrt{3RT/M}, \sqrt{8RT/(\pi M)} \) and \( \sqrt{2RT/M} \); all vary as \( \sqrt{T/M} \).

Answer: (a) 1 : 1

Explanation:
1. Both are diatomic with two rotational degrees of freedom.
2. By equipartition each molecule has average rotational energy \( 2\times\tfrac12kT=kT \), independent of moment of inertia, so the ratio is 1 : 1.

Teacher's Note:
1. Use the equipartition result: each degree of freedom carries kT/2 per molecule (RT/2 per mole), so \( U = f \) nRT/2 with \( f = 3, 5, 7 \) for monatomic, diatomic rigid and diatomic with vibration.
2. For rms, mean and most probable speeds use \( \sqrt{3RT/M}, \sqrt{8RT/(\pi M)} \) and \( \sqrt{2RT/M} \); all vary as \( \sqrt{T/M} \).

Answer: (c) 0.048

Explanation:
1. Average translational kinetic energy per molecule \( =\tfrac32kT \).
2. It depends only on temperature, not on the kind of gas, so \( N _2 \) also has 0.048 eV.

Teacher's Note:
1. Use the equipartition result: each degree of freedom carries kT/2 per molecule (RT/2 per mole), so \( U = f \) nRT/2 with \( f = 3, 5, 7 \) for monatomic, diatomic rigid and diatomic with vibration.
2. For rms, mean and most probable speeds use \( \sqrt{3RT/M}, \sqrt{8RT/(\pi M)} \) and \( \sqrt{2RT/M} \); all vary as \( \sqrt{T/M} \).

Answer: (a) 2v

Explanation:
1. \( v_{rms}\propto\sqrt T \).
2. \( \sqrt{480/120}=2 \), so the speed becomes \( 2v \).

Teacher's Note:
1. Use the equipartition result: each degree of freedom carries kT/2 per molecule (RT/2 per mole), so \( U = f \) nRT/2 with \( f = 3, 5, 7 \) for monatomic, diatomic rigid and diatomic with vibration.
2. For rms, mean and most probable speeds use \( \sqrt{3RT/M}, \sqrt{8RT/(\pi M)} \) and \( \sqrt{2RT/M} \); all vary as \( \sqrt{T/M} \).

Answer: (c) 1.50

Explanation:
1. \( C_V=\tfrac32R+\tfrac52R=4R \) and \( C_p=\tfrac52R+\tfrac72R=6R \) for the two moles together.
2. \( \gamma=\dfrac{6R}{4R}=1.5 \).

Teacher's Note:
1. Use the equipartition result: each degree of freedom carries kT/2 per molecule (RT/2 per mole), so \( U = f \) nRT/2 with \( f = 3, 5, 7 \) for monatomic, diatomic rigid and diatomic with vibration.
2. For rms, mean and most probable speeds use \( \sqrt{3RT/M}, \sqrt{8RT/(\pi M)} \) and \( \sqrt{2RT/M} \); all vary as \( \sqrt{T/M} \).

Answer: (a) Both statements are true, but Statement II is not the correct explanation of Statement I

Explanation:
1. Total translational energy \( =\tfrac32nRT=\tfrac32PV \): Statement I is true. Collisions do change velocities: Statement II is true.
2. The factor 1.5 comes from three translational degrees of freedom, not from collisions, so II does not explain I.

Teacher's Note:
1. Use the equipartition result: each degree of freedom carries kT/2 per molecule (RT/2 per mole), so \( U = f \) nRT/2 with \( f = 3, 5, 7 \) for monatomic, diatomic rigid and diatomic with vibration.
2. For rms, mean and most probable speeds use \( \sqrt{3RT/M}, \sqrt{8RT/(\pi M)} \) and \( \sqrt{2RT/M} \); all vary as \( \sqrt{T/M} \).

Answer: (b) (a), (b) and (d) only

Explanation:
1. \( (a) U=\tfrac52RT+\tfrac32RT=4RT \) for 2 moles, i.e. 2RT per mole: true. \( (b) \) Mixture: \( C_V=2R \), \( C_p=3R \), \( \gamma=1.5 \), \( M=3 g \); helium: \( \gamma=\tfrac53 \), \( M=4 \). \( \dfrac{v_{mix}}{v_{He}}=\sqrt{\dfrac{1.5/3}{(5/3)/4}}=\sqrt{\dfrac65} \): true.
2. \( (c), (d) \dfrac{v_{He}}{v_{H_2}}=\sqrt{\dfrac{2}{4}}=\dfrac1{\sqrt2} \): \( (d) \) true, \( (c) \) false.

Teacher's Note:
1. Use the equipartition result: each degree of freedom carries kT/2 per molecule (RT/2 per mole), so \( U = f \) nRT/2 with \( f = 3, 5, 7 \) for monatomic, diatomic rigid and diatomic with vibration.
2. For rms, mean and most probable speeds use \( \sqrt{3RT/M}, \sqrt{8RT/(\pi M)} \) and \( \sqrt{2RT/M} \); all vary as \( \sqrt{T/M} \).

Answer: (d) (a) and (c) only

Explanation:
1. \( (a) \gamma_V=\dfrac1T \) for every ideal gas: true. \( (c) \lambda=\dfrac{kT}{\sqrt2\pi d^2p} \) grows as \( p \) falls: true.
2. \( (b) \) The average translational energy is \( \tfrac32kT \), not 3kT: false. \( (d) \) At one temperature every component has the same average translational energy: false.

Teacher's Note:
1. Use the equipartition result: each degree of freedom carries kT/2 per molecule (RT/2 per mole), so \( U = f \) nRT/2 with \( f = 3, 5, 7 \) for monatomic, diatomic rigid and diatomic with vibration.
2. For rms, mean and most probable speeds use \( \sqrt{3RT/M}, \sqrt{8RT/(\pi M)} \) and \( \sqrt{2RT/M} \); all vary as \( \sqrt{T/M} \).

Answer: (b) \( \sqrt2\,T \)

Explanation:
1. \( p^2V=c\Rightarrow p\propto V^{-1/2} \).
2. \( T\propto pV\propto V^{1/2} \), so \( T'=T\sqrt2 \).

Teacher's Note:
1. Use the equipartition result: each degree of freedom carries kT/2 per molecule (RT/2 per mole), so \( U = f \) nRT/2 with \( f = 3, 5, 7 \) for monatomic, diatomic rigid and diatomic with vibration.
2. For rms, mean and most probable speeds use \( \sqrt{3RT/M}, \sqrt{8RT/(\pi M)} \) and \( \sqrt{2RT/M} \); all vary as \( \sqrt{T/M} \).

Answer: (d) 2v

Explanation:
1. \( v_{rms}=\sqrt{\dfrac{3RT}{M}} \).
2. Temperature doubles (factor \( \sqrt2 \)) and the molar mass halves (another factor \( \sqrt2 \)), so the speed is \( 2v \).

Teacher's Note:
1. Use the equipartition result: each degree of freedom carries kT/2 per molecule (RT/2 per mole), so \( U = f \) nRT/2 with \( f = 3, 5, 7 \) for monatomic, diatomic rigid and diatomic with vibration.
2. For rms, mean and most probable speeds use \( \sqrt{3RT/M}, \sqrt{8RT/(\pi M)} \) and \( \sqrt{2RT/M} \); all vary as \( \sqrt{T/M} \).

Answer: (b) \( p_1\lt p_2 \)

Explanation:
1. \( V=\dfrac{nR}{p}T \), so the slope of the \( V-T \) line is \( \dfrac{nR}p \).
2. A steeper line means a smaller pressure: \( p_1\lt p_2 \).

Teacher's Note:
1. Use the equipartition result: each degree of freedom carries kT/2 per molecule (RT/2 per mole), so \( U = f \) nRT/2 with \( f = 3, 5, 7 \) for monatomic, diatomic rigid and diatomic with vibration.
2. For rms, mean and most probable speeds use \( \sqrt{3RT/M}, \sqrt{8RT/(\pi M)} \) and \( \sqrt{2RT/M} \); all vary as \( \sqrt{T/M} \).

Answer: (d) \( 3.3\times10^{-21}\ \text{J} \)

Explanation:
1. Average translational energy \( =\tfrac32kT=1.5\times1.38\times10^{-23}\times160 \).
2. \( \approx3.3\times10^{-21} J \) (equal to \( \tfrac12mv^2=\tfrac12\times6.6\times10^{-27}\times10^6 \)).

Teacher's Note:
1. Use the equipartition result: each degree of freedom carries kT/2 per molecule (RT/2 per mole), so \( U = f \) nRT/2 with \( f = 3, 5, 7 \) for monatomic, diatomic rigid and diatomic with vibration.
2. For rms, mean and most probable speeds use \( \sqrt{3RT/M}, \sqrt{8RT/(\pi M)} \) and \( \sqrt{2RT/M} \); all vary as \( \sqrt{T/M} \).

Answer: (a) 0.3 g

Explanation:
1. \( n=\dfrac{pV}{RT}=\dfrac{100\times1}{(25/3)(160)}=0.075 \) mol.
2. Mass \( =0.075\times4=0.3 g \).

Teacher's Note:
1. Use the equipartition result: each degree of freedom carries kT/2 per molecule (RT/2 per mole), so \( U = f \) nRT/2 with \( f = 3, 5, 7 \) for monatomic, diatomic rigid and diatomic with vibration.
2. For rms, mean and most probable speeds use \( \sqrt{3RT/M}, \sqrt{8RT/(\pi M)} \) and \( \sqrt{2RT/M} \); all vary as \( \sqrt{T/M} \).