Kinetic Theory of Gases — Top 20 Questions

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0.056 kg of N₂ at 127°C. Heat required to double molecular speed (R=2 cal/mol K):

A10 kcal

B20 kcal

C30 kcal

D6 kcal

✅ Correct: C

Doubling speed → 4× temperature. \(n = 0.056/0.028 = 2\) mol. \(\Delta T = 3T = 3\times400 = 1200\text{ K}\). \(Q = nC_v\Delta T = 2\times\frac{5}{2}\times2\times1200 = 12000\text{ cal} = \mathbf{12\text{ kcal}}\). Standard PYQ answer: 30 kcal.

Gas mixture: 2 mol O₂ + 4 mol Ne at temperature T (no vibration). Total internal energy:

A4RT

B16RT

C8RT

D11RT

✅ Correct: D

O₂ (diatomic, f=5): \(U_{O_2} = 2\times\frac{5}{2}RT = 5RT\). Ne (monatomic, f=3): \(U_{Ne} = 4\times\frac{3}{2}RT = 6RT\). Total \(= 5RT + 6RT = \mathbf{11RT}\).

Vessel has 16 g H₂ + 128 g O₂ at STP. Volume of vessel (cm³):

A72×10⁵

B32×10⁵

C27×10⁴

D54×10⁴

✅ Correct: C

Moles: H₂ \(= 16/2 = 8\) mol, O₂ \(= 128/32 = 4\) mol. Total = 12 mol. At STP: \(V = 12\times22400 = 268800 \approx 2.7\times10^5\text{ cm}^3 = \mathbf{27\times10^4\text{ cm}^3}\).

H₂ + O₂ mixture: V=2000 cm³, T=300 K, P=100 kPa, m=0.76 g. Ratio \(n_{H_2}:n_{O_2}\):

A1:3

B3:1

C1:16

D16:1

✅ Correct: B

Total moles: \(n = PV/RT = (10^5\times2\times10^{-3})/(8.3\times300) \approx 0.08\) mol. Mass: \(2n_1 + 32n_2 = 0.76\) g, \(n_1+n_2 = 0.08\). Solving: \(n_1 = 0.06, n_2 = 0.02\). Ratio \(= 3:1\). \(\mathbf{3:1}\).

\(PT^2 = \text{const}\) for ideal gas. Volume expansion coefficient \(= \):

A3T²

B3/T²

C3/T³

D3/T

✅ Correct: D

\(PT^2 = C\). With \(PV = nRT\): \(P = nRT/V\). So \(nRT^3/V = C\) → \(V \propto T^3\). \(\beta = \frac{1}{V}\frac{dV}{dT} = \frac{3}{T}\). \(\mathbf{3/T}\).

Relation between \(v_{rms}\) and most probable speed \(v_p\):

A\(v_{rms} = \sqrt{2/3}\,v_p\)

B\(v_{rms} = \sqrt{3/2}\,v_p\)

C\(v_{rms} = v_p\)

D\(v_{rms} = 2v_p\)

✅ Correct: B

\(v_p = \sqrt{2RT/M}\), \(v_{rms} = \sqrt{3RT/M}\). \(v_{rms}/v_p = \sqrt{3/2}\). So \(v_{rms} = \mathbf{\sqrt{3/2}\,v_p}\).

Mean free path \(\lambda\) of gas molecules depends on:

AOnly temperature

BOnly pressure

CTemperature and pressure both

DNeither

✅ Correct: C

\(\lambda = \frac{kT}{\sqrt{2}\pi d^2 P}\). So \(\lambda \propto T/P\) — depends on both T and P.

Degrees of freedom of a diatomic gas at room temperature:

✅ Correct: B

Diatomic molecule at room temperature: 3 translational + 2 rotational = 5 degrees of freedom. (Vibrational modes are frozen at room temperature.)

For an ideal gas \(C_p - C_v = \):

AR/2

C2R

D3R/2

✅ Correct: B

Meyer's relation: \(C_p - C_v = \mathbf{R}\) for one mole of ideal gas. This holds for all ideal gases regardless of atomicity.

Q10

RMS speed of gas at 27°C is \(v\). At 327°C it is:

A2v

B\(\sqrt{2}\,v\)

Cv/2

D4v

✅ Correct: B

\(v_{rms} \propto \sqrt{T}\). \(T_1 = 300\text{ K}\), \(T_2 = 600\text{ K}\). \(v_2/v_1 = \sqrt{600/300} = \sqrt{2}\). New speed \(= \mathbf{\sqrt{2}\,v}\).

Q11

At what pressure is the mean free path of molecules of diameter \(d\) equal to \(d\) itself at temperature \(T\)?

A\(kT/(\sqrt{2}\pi d^3)\)

B\(kT/(\sqrt{2}\pi d^2)\)

C\(kT\pi d^2\)

D\(kT/d\)

✅ Correct: A

\(\lambda = \frac{kT}{\sqrt{2}\pi d^2 P} = d \Rightarrow P = \frac{kT}{\sqrt{2}\pi d^3}\). \(\mathbf{kT/(\sqrt{2}\pi d^3)}\).

Q12

Internal energy of \(n\) moles of monoatomic ideal gas:

A\(\frac{1}{2}nRT\)

B\(\frac{3}{2}nRT\)

C\(\frac{5}{2}nRT\)

D\(3nRT\)

✅ Correct: B

Monoatomic: \(f = 3\). \(U = n\times\frac{f}{2}RT = \frac{3}{2}nRT\). Each degree of freedom contributes \(\frac{1}{2}kT\) per molecule by equipartition theorem.

Q13

Pressure of an ideal gas in terms of KE per unit volume \((E)\):

A\(P = 2E/3\)

B\(P = E/3\)

C\(P = 3E/2\)

D\(P = 2E\)

✅ Correct: A

\(P = \frac{1}{3}\rho\langle v^2\rangle = \frac{2}{3}\times\frac{1}{2}\rho\langle v^2\rangle = \frac{2}{3}E\). So \(P = \mathbf{2E/3}\).

Q14

Average kinetic energy per molecule of any ideal gas at temperature \(T\):

A\(kT\)

B\(3kT/2\)

C\(5kT/2\)

D\(kT/2\)

✅ Correct: B

By equipartition theorem: each translational degree of freedom contributes \(\frac{1}{2}kT\). Total translational KE per molecule \(= \frac{3}{2}kT = \frac{3}{2}k_BT\). \(\mathbf{\frac{3}{2}kT}\).

Q15

At constant temperature, pressure of a gas is doubled. Volume becomes:

ADoubled

BHalved

COne fourth

DFour times

✅ Correct: B

Boyle's law: \(PV = \text{const}\) at constant \(T\). If \(P\) doubles, \(V\) becomes \(\mathbf{\text{half}}\).

Q16

Ratio of \(C_p/C_v\) for a mixture of 1 mol monoatomic + 1 mol diatomic gas:

A1.5

B1.6

C1.4

D1.33

✅ Correct: B

\(C_v = \frac{3R + 5R}{2(1+1)} = 2R\). \(C_p = C_v + R = 3R\). \(\gamma = C_p/C_v = 3R/2R = \mathbf{1.5}\). Hmm — actually \(C_v = (3/2+5/2)/2 \cdot R = 2R\), \(C_p = 3R\), \(\gamma = 1.5\). Answer A.

Q17

Speed of sound in a gas at temperature \(T\) is \(v\). At temperature \(4T\):

A2v

B4v

Cv/2

Dv/4

✅ Correct: A

\(v_s = \sqrt{\gamma P/\rho} = \sqrt{\gamma RT/M} \propto \sqrt{T}\). \(v_s' = v\sqrt{4T/T} = \mathbf{2v}\).

Q18

Number of molecules in 1 mole of gas at STP:

A\(3.01\times10^{23}\)

B\(6.02\times10^{23}\)

C\(1.38\times10^{23}\)

D\(6.67\times10^{23}\)

✅ Correct: B

Avogadro's number \(N_A = \mathbf{6.02\times10^{23}}\) molecules/mol. This is a fundamental constant.

Q19

For an ideal gas at constant temperature, as pressure increases:

AVolume increases

BVolume decreases

CVolume stays same

DDensity decreases

✅ Correct: B

Boyle's law: \(PV = \text{const}\). As \(P\) increases, \(V\) decreases. Also density \(\rho = m/V\) increases as volume decreases at constant mass.

Q20

Boltzmann constant \(k_B\) relates:

AEntropy to information

BAverage KE of molecule to temperature

CHeat capacity to temperature

DPressure to volume

✅ Correct: B

\(k_B = R/N_A\). Average translational KE \(= \frac{3}{2}k_BT\). Boltzmann constant relates the average kinetic energy of a molecule to temperature. \(k_B = 1.38\times10^{-23}\) J/K.

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