Vidaara.orgClass 11 · Physics
CodeVID-P11-12-KGL-01
Kinetic Theory & Gas Laws — 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 SI unit of the universal gas constant $R$ is:
- A.$\text{J}\,\text{mol}^{-1}\,\text{K}^{-1}$
- B.$\text{J}\,\text{K}^{-1}$
- C.$\text{J}\,\text{mol}^{-1}$
- D.$\text{Pa}\,\text{m}^{-3}$
2.
Charles' law (constant $P$) states that:
- A.$V\propto\frac{1}{T}$
- B.$V\propto T$
- C.$V\propto T^2$
- D.$V=\text{constant}$
3.
The Boltzmann constant equals:
- A.$R\,N_A$
- B.$\frac{R}{N_A}$
- C.$\frac{N_A}{R}$
- D.$RN_A^2$
4.
Temperatures in the gas laws must be expressed in:
- A.degrees Celsius
- B.degrees Fahrenheit
- C.kelvin
- D.any scale
5.
If the pressure of a fixed mass of gas is doubled at constant temperature, its volume becomes:
- A.double
- B.half
- C.four times
- D.unchanged
Section B — Short Answer (2 marks)
3 × 2 = 6 marks
6.
State Boyle's law and write its mathematical form.
7.
A gas at 1 atm occupies 6 L. Find its volume at 3 atm (constant $T$).
8.
Write the ideal gas equation in terms of the number of molecules $N$.
Section C — Short Answer (3 marks)
2 × 3 = 6 marks
9.
List any three assumptions of the kinetic theory of gases.
10.
A gas at $27^\circ\text{C}$ and 2 atm is heated to $327^\circ\text{C}$ at constant volume. Find the new pressure.
Section D — Long Answer (5 marks)
1 × 5 = 5 marks
11.
Starting from Boyle's, Charles' and Avogadro's laws, explain how the ideal gas equation $PV=nRT$ is obtained, and define each symbol with its SI unit.
Answer Key
Section A — Multiple Choice Questions
- (A) $\text{J}\,\text{mol}^{-1}\,\text{K}^{-1}$
- (B) $V\propto T$
- (B) $\frac{R}{N_A}$
- (C) kelvin
- (B) half
Section B — Short Answer (2 marks)
- At constant temperature, $PV=\text{constant}$ for a fixed mass of gas, so $P\propto\frac{1}{V}$.
- $P_1V_1=P_2V_2\Rightarrow V_2=\frac{1\times6}{3}=2\ \text{L}$.
- $PV=Nk_BT$, where $k_B=\frac{R}{N_A}=1.38\times10^{-23}\ \text{J/K}$.
Section C — Short Answer (3 marks)
- Molecules are in constant random motion; molecular volume is negligible compared with the container; collisions are perfectly elastic with no inter-molecular force except during collisions.
- $T_1=300$ K, $T_2=600$ K; $\frac{P_1}{T_1}=\frac{P_2}{T_2}\Rightarrow P_2=2\times\frac{600}{300}=4\ \text{atm}$.
Section D — Long Answer (5 marks)
- Combining $P\propto\frac{1}{V}$ (Boyle), $V\propto T$ (Charles) and $V\propto n$ (Avogadro) gives $PV\propto nT$, i.e. $PV=nRT$. Here $P$ is pressure (Pa), $V$ volume (m^3), $n$ number of moles (mol), $T$ absolute temperature (K) and $R=8.314\ \text{J}\,\text{mol}^{-1}\,\text{K}^{-1}$ is the universal gas constant.
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