Vidaara.orgClass 11 · Physics
CodeVID-P11-11-TEF-01
Thermal Equilibrium & the First Law — 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 first law of thermodynamics is written as:
- A.$\Delta Q=\Delta U-\Delta W$
- B.$\Delta Q=\Delta U+\Delta W$
- C.$\Delta W=\Delta U+\Delta Q$
- D.$\Delta U=\Delta Q+\Delta W$
2.
For an ideal gas, internal energy depends only on:
- A.pressure
- B.volume
- C.temperature
- D.density
3.
In a cyclic process the change in internal energy is:
- A.maximum
- B.zero
- C.negative
- D.equal to the heat absorbed
4.
Work done by a gas is positive when the gas:
- A.is compressed
- B.expands
- C.is cooled
- D.is heated at fixed volume
5.
The quantity that is NOT a state function is:
- A.temperature
- B.internal energy
- C.pressure
- D.heat
Section B — Short Answer (2 marks)
3 × 2 = 6 marks
6.
State the zeroth law of thermodynamics.
7.
A gas absorbs 300 J of heat and its internal energy rises by 120 J. Find the work done by the gas.
8.
Why are heat and work not state functions?
Section C — Short Answer (3 marks)
2 × 3 = 6 marks
9.
Derive the expression for the work done by a gas during a small volume change and write the general formula.
10.
A gas is taken around a cycle and absorbs a net 250 J of heat. Find the net work done and explain.
Section D — Long Answer (5 marks)
1 × 5 = 5 marks
11.
State the first law of thermodynamics, explain the sign convention for $\Delta Q$, $\Delta U$ and $\Delta W$, and apply it to (i) an isothermal change and (ii) a cyclic process.
Answer Key
Section A — Multiple Choice Questions
- (B) $\Delta Q=\Delta U+\Delta W$
- (C) temperature
- (B) zero
- (B) expands
- (D) heat
Section B — Short Answer (2 marks)
- If A is in thermal equilibrium with C and B is in equilibrium with C, then A and B are in equilibrium with each other; this defines temperature.
- $\Delta W=\Delta Q-\Delta U=300-120=180\ \text{J}$.
- Their values depend on the path taken between two states, not only on the initial and final states.
Section C — Short Answer (3 marks)
- For a piston of area $A$ moving $dx$, force $=PA$, so $dW=PA\,dx=P\,dV$; for a finite change $W=\int_{V_1}^{V_2}P\,dV$, the area under the $P$–$V$ curve.
- For a cycle $\Delta U=0$, so $\Delta W=\Delta Q=250\ \text{J}$; the net work equals the area enclosed by the loop on a $P$–$V$ diagram.
Section D — Long Answer (5 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. (i) Isothermal: $\Delta U=0$, so $\Delta Q=\Delta W$. (ii) Cyclic: $\Delta U=0$, so $\Delta Q=\Delta W$ = area enclosed by the loop.
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