Vidaara.orgClass 11 · Physics
CodeVID-P11-12-DOF-01
Degrees of Freedom & Specific Heat — Assignment
Name: ____________________
Roll No.: __________
Date: ____________
General Instructions
- All questions are compulsory.
- Section A carries 1 mark each, Section B 2 marks, Section C 3 marks and Section D 5 marks.
- Show all working for Sections B, C and D. Only final answers are given at the end — for full solutions, raise your doubts with your teacher.
Section A — Multiple Choice Questions
5 × 1 = 5 marks
1.
The degrees of freedom of a diatomic gas molecule at ordinary temperature are:
- A.3
- B.5
- C.6
- D.7
2.
The molar specific heat at constant volume of a monatomic gas is:
- A.$\frac{3}{2}R$
- B.$\frac{5}{2}R$
- C.$3R$
- D.$R$
3.
Mayer's relation between the molar specific heats is:
- A.$C_p-C_v=R$
- B.$C_p+C_v=R$
- C.$C_pC_v=R$
- D.$\frac{C_p}{C_v}=R$
4.
The adiabatic constant $\gamma$ for a polyatomic gas ($f=6$) is:
- A.$\frac{5}{3}$
- B.$\frac{7}{5}$
- C.$\frac{4}{3}$
- D.$2$
5.
The mean free path is largest when the gas pressure is:
- A.very high
- B.very low
- C.exactly 1 atm
- D.independent of pressure
Section B — Short Answer (2 marks)
3 × 2 = 6 marks
6.
State the law of equipartition of energy.
7.
Write the relation between $\gamma$ and the degrees of freedom $f$.
8.
Find $C_p$ for a diatomic gas in terms of $R$.
Section C — Short Answer (3 marks)
2 × 3 = 6 marks
9.
Find $C_v$, $C_p$ and $\gamma$ for a polyatomic gas ($f=6$).
10.
Define mean free path and write its formula, stating how it depends on number density.
Section D — Long Answer (5 marks)
1 × 5 = 5 marks
11.
Using the law of equipartition, derive the molar specific heats $C_v$ and $C_p$ and the ratio $\gamma$ for monatomic and diatomic gases.
Answer Key
Section A — Multiple Choice Questions
- (B) 5
- (A) $\frac{3}{2}R$
- (A) $C_p-C_v=R$
- (C) $\frac{4}{3}$
- (B) very low
Section B — Short Answer (2 marks)
- In thermal equilibrium, energy is shared equally among all degrees of freedom, each carrying an average of $\frac{1}{2}k_BT$ per molecule.
- $\gamma=\frac{C_p}{C_v}=1+\frac{2}{f}$.
- $C_v=\frac{5}{2}R$, so $C_p=C_v+R=\frac{7}{2}R$.
Section C — Short Answer (3 marks)
- $C_v=\frac{6}{2}R=3R$, $C_p=3R+R=4R$, $\gamma=\frac{4R}{3R}=\frac{4}{3}\approx1.33$.
- Average distance between successive collisions; $\lambda=\frac{1}{\sqrt{2}\,\pi d^2 n}$, inversely proportional to the number density $n$.
Section D — Long Answer (5 marks)
- Internal energy of one mole is $U=\frac{f}{2}RT$, so $C_v=\frac{f}{2}R$ and $C_p=C_v+R=\frac{f+2}{2}R$, giving $\gamma=1+\frac{2}{f}$. Monatomic ($f=3$): $C_v=\frac{3}{2}R$, $C_p=\frac{5}{2}R$, $\gamma=\frac{5}{3}$. Diatomic ($f=5$): $C_v=\frac{5}{2}R$, $C_p=\frac{7}{2}R$, $\gamma=\frac{7}{5}$.
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