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CodeVID-P11-11-TDP-01
Thermodynamic Processes — 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 isothermal curve on a $P$–$V$ diagram is a:
- A.straight line
- B.rectangular hyperbola
- C.parabola
- D.circle
2.
For a monatomic ideal gas, $\gamma$ is approximately:
- A.1.40
- B.1.67
- C.1.00
- D.2.00
3.
In an adiabatic expansion the temperature of the gas:
- A.rises
- B.falls
- C.stays the same
- D.first rises then falls
4.
Work done in an isobaric process is:
- A.$nR\,\Delta T$
- B.zero
- C.$nC_v\Delta T$
- D.$nRT\ln(V_2/V_1)$
5.
Heat supplied at constant volume goes entirely into:
- A.work
- B.internal energy
- C.expansion
- D.radiation
Section B — Short Answer (2 marks)
3 × 2 = 6 marks
6.
Why is $C_p$ greater than $C_v$ for a gas?
7.
One mole of a gas with $C_v=12.5\ \text{J}\,\text{mol}^{-1}\text{K}^{-1}$ — find $C_p$.
8.
Why does an adiabatic curve fall more steeply than an isothermal curve?
Section C — Short Answer (3 marks)
2 × 3 = 6 marks
9.
Derive the work done in an isothermal expansion of an ideal gas.
10.
A diatomic gas ($\gamma=1.4$) at 300 K is expanded adiabatically to twice its volume. Find the final temperature.
Section D — Long Answer (5 marks)
1 × 5 = 5 marks
11.
Describe the four basic thermodynamic processes, write the work done in each, and derive Mayer's relation $C_p-C_v=R$.
Answer Key
Section A — Multiple Choice Questions
- (B) rectangular hyperbola
- (B) 1.67
- (B) falls
- (A) $nR\,\Delta T$
- (B) internal energy
Section B — Short Answer (2 marks)
- At constant pressure extra heat is needed to do work in expansion, in addition to raising the internal energy, so $C_p>C_v$ and $C_p-C_v=R$.
- $C_p=C_v+R=12.5+8.314\approx20.8\ \text{J}\,\text{mol}^{-1}\text{K}^{-1}$.
- Adiabatic slope $\propto\gamma P/V$ while isothermal slope $\propto P/V$; since $\gamma>1$, the adiabatic curve is steeper.
Section C — Short Answer (3 marks)
- $W=\int_{V_1}^{V_2}P\,dV=\int_{V_1}^{V_2}\frac{nRT}{V}\,dV=nRT\ln\!\left(\frac{V_2}{V_1}\right)$.
- $T_2=T_1\left(\frac{V_1}{V_2}\right)^{\gamma-1}=300\times(1/2)^{0.4}=300\times0.758\approx227\ \text{K}$.
Section D — Long Answer (5 marks)
- Isothermal: $W=nRT\ln(V_2/V_1)$, $\Delta U=0$. Adiabatic: $W=\frac{P_1V_1-P_2V_2}{\gamma-1}$, $\Delta Q=0$. Isobaric: $W=P\Delta V=nR\Delta T$. Isochoric: $W=0$. For one mole at constant pressure: $C_p\,dT=C_v\,dT+P\,dV$ and $P\,dV=R\,dT$, so $C_p=C_v+R$, i.e. $C_p-C_v=R$.
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