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CodeVID-P11-12-KPT-01
Kinetic Interpretation of Pressure & Temperature — 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.
Pressure of a gas in kinetic theory is proportional to:
- A.$\overline{v^2}$
- B.$\frac{1}{\overline{v^2}}$
- C.$v_{rms}$
- D.$T^2$
2.
The average translational KE of a molecule depends on:
- A.the type of gas only
- B.the temperature only
- C.the pressure only
- D.the volume only
3.
rms speed varies with temperature as:
- A.$v_{rms}\propto T$
- B.$v_{rms}\propto\sqrt{T}$
- C.$v_{rms}\propto T^2$
- D.$v_{rms}\propto\frac{1}{T}$
4.
At a given temperature, a lighter gas molecule has a rms speed that is:
- A.smaller
- B.larger
- C.equal
- D.zero
5.
The factor $\frac{1}{3}$ in the pressure formula arises because:
- A.there are three gas laws
- B.motion is shared equally among three directions
- C.pressure acts on three walls
- D.temperature has three scales
Section B — Short Answer (2 marks)
3 × 2 = 6 marks
6.
Write the kinetic theory expression for pressure in terms of density.
7.
What is the average translational KE of a molecule at temperature $T$?
8.
Find the rms speed of a gas where $P=10^{5}\ \text{Pa}$ and $\rho=1.0\ \text{kg/m}^3$.
Section C — Short Answer (3 marks)
2 × 3 = 6 marks
9.
Show that the average translational KE of a molecule is $\frac{3}{2}k_BT$.
10.
Find the temperature at which oxygen molecules ($M=0.032\ \text{kg/mol}$) have a rms speed of 500 m/s.
Section D — Long Answer (5 marks)
1 × 5 = 5 marks
11.
Derive the kinetic theory expression for the pressure of an ideal gas, $P=\frac{1}{3}\frac{Nm\,\overline{v^2}}{V}$, and hence explain the kinetic interpretation of temperature.
Answer Key
Section A — Multiple Choice Questions
- (A) $\overline{v^2}$
- (B) the temperature only
- (B) $v_{rms}\propto\sqrt{T}$
- (B) larger
- (B) motion is shared equally among three directions
Section B — Short Answer (2 marks)
- $P=\frac{1}{3}\rho\,\overline{v^2}$, where $\rho$ is the density and $\overline{v^2}$ the mean-square speed.
- $\frac{1}{2}m\,\overline{v^2}=\frac{3}{2}k_BT$, the same for all gases at a given $T$.
- $v_{rms}=\sqrt{\frac{3P}{\rho}}=\sqrt{3\times10^{5}}\approx548\ \text{m/s}$.
Section C — Short Answer (3 marks)
- From $P=\frac{1}{3}\frac{Nm\,\overline{v^2}}{V}$, $PV=\frac{2}{3}N\left(\frac{1}{2}m\overline{v^2}\right)$; comparing with $PV=Nk_BT$ gives $\frac{1}{2}m\overline{v^2}=\frac{3}{2}k_BT$.
- $T=\frac{Mv_{rms}^2}{3R}=\frac{0.032\times500^2}{3\times8.314}\approx321\ \text{K}$.
Section D — Long Answer (5 marks)
- Consider $N$ molecules of mass $m$ in a cube of side $L$; the momentum change per wall collision is $2mv_x$ and the force from all molecules gives $P=\frac{1}{3}\frac{Nm\,\overline{v^2}}{V}$ (the $\frac{1}{3}$ from $\overline{v_x^2}=\frac{1}{3}\overline{v^2}$). Comparing $PV=\frac{2}{3}N\left(\frac{1}{2}m\overline{v^2}\right)$ with $PV=Nk_BT$ gives $\frac{1}{2}m\overline{v^2}=\frac{3}{2}k_BT$, so temperature measures the average translational kinetic energy of the molecules.
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