Physics · Kinetic theory of gases · chapter 12 of 14

Class 11 Physics Chapter 12: Kinetic Theory case study question with answers

One case-based question of the kind CBSE-pattern school exams ask: a passage and a table, then 3 parts worth 4 marks in all, with the answer and the marking scheme for each. Written from the NCERT chapter.

Case study · 4 marks

In a school laboratory project, Meera looks up the measured molar specific heats of three unlabelled gases X, Y and Z at room temperature, given in the table in J mol⁻¹ K⁻¹. Her teacher asks her to decide the atomicity of each gas from the data alone, using the law of equipartition of energy and the relations C_v = (s/2)R and γ = 1 + 2/s, where s is the number of energy modes of a molecule. She takes R = 8.31 J mol⁻¹ K⁻¹ and treats all three as ideal gases.

Gas	C_v	C_p
X	12.5	20.8
Y	20.8	29.1
Z	24.9	33.2

Part 1 · 1 mark

Which gas is most likely monatomic?

(a) Y(b) Z(c) X(d) none of them, since all three have the same C_p − C_v

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Answer: (c) X

For X, γ = 20.8/12.5 = 1.66 and C_v = 12.5 ≈ (3/2)R, the monatomic value with s = 3.

Part 2 · 1 mark

The nearly equal value of C_p − C_v (about 8.3 J mol⁻¹ K⁻¹) for all three gases shows that

(a) C_p − C_v = R for any ideal gas, whatever its atomicity(b) all three gases have the same number of degrees of freedom(c) the three gases have the same molar mass(d) C_p − C_v equals (3/2)R for every gas

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Answer: (a) C_p − C_v = R for any ideal gas, whatever its atomicity

C_p − C_v = R = 8.31 J mol⁻¹ K⁻¹ for mono, di and polyatomic ideal gases alike; the individual C_v values differ, so the degrees of freedom differ.

Part 3 · 2 marks

Find s for gas Z and suggest what kind of molecule it has.

Answer and marking scheme

s = 2C_v/R = 2 × 24.9/8.31 = 5.99 ≈ 6. Six modes fit a rigid polyatomic (non-linear) molecule with 3 translational and 3 rotational degrees of freedom and no vibration; then γ = 1 + 2/6 = 4/3 = 1.33, matching 33.2/24.9 = 1.33.

s = 2C_v/R ≈ 6 (1)
rigid non-linear polyatomic: 3 translational + 3 rotational, γ = 4/3 checks (1)

Find s for gas Y and suggest what kind of molecule it has.

Answer and marking scheme

s = 2C_v/R = 2 × 20.8/8.31 = 5.01 ≈ 5. Five modes fit a rigid diatomic molecule: 3 translational and 2 rotational degrees of freedom, no vibration; then γ = 1 + 2/5 = 7/5 = 1.40, matching 29.1/20.8 = 1.40.

s = 2C_v/R ≈ 5 (1)
rigid diatomic: 3 translational + 2 rotational, γ = 7/5 checks (1)

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