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CodeVID-P11-11-CH-01
Thermodynamics — 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 first law of thermodynamics states:
- A.$\Delta Q=\Delta U-\Delta W$
- B.$\Delta Q=\Delta U+\Delta W$
- C.$\Delta U=\Delta Q+\Delta W$
- D.$\Delta W=\Delta U+\Delta Q$
2.
Internal energy of an ideal gas depends only on:
- A.pressure
- B.volume
- C.temperature
- D.density
3.
An adiabatic process obeys:
- A.$PV=\text{const}$
- B.$PV^\gamma=\text{const}$
- C.$P/T=\text{const}$
- D.$V/T=\text{const}$
4.
Mayer's relation is:
- A.$C_p+C_v=R$
- B.$C_p-C_v=R$
- C.$C_p C_v=R$
- D.$C_v-C_p=R$
5.
The efficiency of a Carnot engine is:
- A.$\frac{T_2}{T_1}$
- B.$1-\frac{T_2}{T_1}$
- C.$1-\frac{T_1}{T_2}$
- D.$\frac{T_1}{T_2}$
6.
The coefficient of performance of a refrigerator is:
- A.$\frac{Q_2}{W}$
- B.$\frac{W}{Q_2}$
- C.$1-\frac{Q_2}{Q_1}$
- D.$\frac{Q_1}{Q_2}$
Section B — Short Answer (2 marks)
4 × 2 = 8 marks
7.
State the first law of thermodynamics and its sign convention.
8.
A gas absorbs 250 J of heat and does 100 J of work. Find $\Delta U$.
9.
Why is no work done in an isochoric process?
10.
A Carnot engine works between 500 K and 250 K. Find its efficiency.
Section C — Short Answer (3 marks)
2 × 3 = 6 marks
11.
Derive the work done in an isothermal expansion of an ideal gas.
12.
A heat engine absorbs 1200 J and rejects 900 J per cycle. Find the work done and efficiency.
Section D — Long Answer (5 marks)
2 × 5 = 10 marks
13.
Describe the four basic thermodynamic processes (isothermal, adiabatic, isobaric, isochoric) and write the work done in each.
14.
Explain the working of a heat engine and derive the efficiency of a Carnot engine $\eta=1-\frac{T_2}{T_1}$.
Answer Key
Section A — Multiple Choice Questions
- (B) $\Delta Q=\Delta U+\Delta W$
- (C) temperature
- (B) $PV^\gamma=\text{const}$
- (B) $C_p-C_v=R$
- (B) $1-\frac{T_2}{T_1}$
- (A) $\frac{Q_2}{W}$
Section B — Short Answer (2 marks)
- $\Delta Q=\Delta U+\Delta W$; $\Delta Q>0$ if heat is added, $\Delta W>0$ if the gas does work, $\Delta U>0$ if temperature rises.
- $\Delta U=\Delta Q-\Delta W=250-100=150\ \text{J}$.
- Volume is constant, so $dV=0$ and $W=\int P\,dV=0$.
- $\eta=1-\frac{250}{500}=0.5=50\%$.
Section C — Short Answer (3 marks)
- $W=\int_{V_1}^{V_2}P\,dV=\int_{V_1}^{V_2}\frac{nRT}{V}\,dV=nRT\ln\!\left(\frac{V_2}{V_1}\right)$.
- $W=1200-900=300\ \text{J}$; $\eta=1-\frac{900}{1200}=0.25=25\%$.
Section D — Long Answer (5 marks)
- Isothermal ($T$ const): $\Delta U=0$, $W=nRT\ln(V_2/V_1)$. Adiabatic ($\Delta Q=0$): $W=\frac{P_1V_1-P_2V_2}{\gamma-1}$. Isobaric ($P$ const): $W=P\Delta V=nR\Delta T$. Isochoric ($V$ const): $W=0$, $\Delta Q=\Delta U$.
- An engine absorbs $Q_1$ at $T_1$, does work $W=Q_1-Q_2$, rejects $Q_2$ at $T_2$; $\eta=1-\frac{Q_2}{Q_1}$. For a reversible Carnot cycle $\frac{Q_2}{Q_1}=\frac{T_2}{T_1}$, hence $\eta=1-\frac{T_2}{T_1}$, always less than 1.
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