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CodeVID-P11-12-CH-01
Kinetic Theory — Full Chapter Test
Name: ____________________
Roll No.: __________
Date: ____________
General Instructions
- This is a full-length test covering the whole chapter — every topic is included.
- 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.
Section A — Multiple Choice Questions
6 × 1 = 6 marks
1.
The ideal gas equation is:
- A.$PV=nRT$
- B.$PV=\frac{nRT}{2}$
- C.$\frac{P}{V}=nRT$
- D.$PVT=nR$
2.
The kinetic theory expression for pressure is:
- A.$P=\frac{1}{3}\rho\,\overline{v^2}$
- B.$P=\rho\,\overline{v^2}$
- C.$P=\frac{1}{2}\rho\,\overline{v^2}$
- D.$P=\frac{2}{3}\rho\,\overline{v^2}$
3.
The average translational KE of a molecule is:
- A.$\frac{1}{2}k_BT$
- B.$k_BT$
- C.$\frac{3}{2}k_BT$
- D.$3k_BT$
4.
rms speed varies with temperature as:
- A.$v_{rms}\propto T$
- B.$v_{rms}\propto\sqrt{T}$
- C.$v_{rms}\propto T^2$
- D.$v_{rms}\propto\frac{1}{T}$
5.
The degrees of freedom of a diatomic gas are:
- A.3
- B.5
- C.6
- D.7
6.
The value of $\gamma$ for a monatomic gas is:
- A.$\frac{4}{3}$
- B.$\frac{7}{5}$
- C.$\frac{5}{3}$
- D.$1$
Section B — Short Answer (2 marks)
4 × 2 = 8 marks
7.
State the ideal gas equation and define each symbol.
8.
Write the relation between average translational KE and temperature.
9.
State the law of equipartition of energy.
10.
Define mean free path.
Section C — Short Answer (3 marks)
2 × 3 = 6 marks
11.
Find the rms speed of nitrogen molecules ($M=0.028\ \text{kg/mol}$) at 300 K.
12.
Find $C_v$, $C_p$ and $\gamma$ for a diatomic gas.
Section D — Long Answer (5 marks)
2 × 5 = 10 marks
13.
Derive the kinetic theory expression for pressure and use it to show that the average translational KE of a molecule is $\frac{3}{2}k_BT$.
14.
Using the law of equipartition, obtain $C_v$, $C_p$ and $\gamma$ for monatomic, diatomic and polyatomic gases.
Answer Key
Section A — Multiple Choice Questions
- (A) $PV=nRT$
- (A) $P=\frac{1}{3}\rho\,\overline{v^2}$
- (C) $\frac{3}{2}k_BT$
- (B) $v_{rms}\propto\sqrt{T}$
- (B) 5
- (C) $\frac{5}{3}$
Section B — Short Answer (2 marks)
- $PV=nRT$: $P$ pressure, $V$ volume, $n$ moles, $R$ gas constant, $T$ absolute temperature.
- $\frac{1}{2}m\,\overline{v^2}=\frac{3}{2}k_BT$, so temperature measures average molecular KE.
- Energy is shared equally among all degrees of freedom, each carrying $\frac{1}{2}k_BT$ per molecule.
- The average distance a molecule travels between two successive collisions, $\lambda=\frac{1}{\sqrt{2}\,\pi d^2 n}$.
Section C — Short Answer (3 marks)
- $v_{rms}=\sqrt{\frac{3RT}{M}}=\sqrt{\frac{3\times8.314\times300}{0.028}}\approx517\ \text{m/s}$.
- $C_v=\frac{5}{2}R$, $C_p=\frac{7}{2}R$, $\gamma=\frac{7}{5}=1.40$.
Section D — Long Answer (5 marks)
- From $N$ molecules in a cube, the pressure is $P=\frac{1}{3}\frac{Nm\,\overline{v^2}}{V}$. Then $PV=\frac{2}{3}N\left(\frac{1}{2}m\overline{v^2}\right)$; comparing with $PV=Nk_BT$ gives $\frac{1}{2}m\overline{v^2}=\frac{3}{2}k_BT$.
- $U=\frac{f}{2}RT$ gives $C_v=\frac{f}{2}R$, $C_p=\frac{f+2}{2}R$, $\gamma=1+\frac{2}{f}$. Mono ($f=3$): $\frac{3}{2}R,\frac{5}{2}R,\frac{5}{3}$. Di ($f=5$): $\frac{5}{2}R,\frac{7}{2}R,\frac{7}{5}$. Poly ($f=6$): $3R,4R,\frac{4}{3}$.
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