Physics · Thermodynamics

JEE Advanced 2018 — Paper 2 — Question 16

One mole of a monatomic ideal gas undergoes four thermodynamic processes as shown schematically in the PVdiagram below. Among these four processes, one is isobaric, one is isochoric, one is isothermal and one is adiabatic. Match the processes mentioned in List-1 with the corresponding statements in List-II.

List-IList-II
P. In process I1. Work done by the gas is zero
Q. In process II2. Temperature of the gas remains unchanged
R. In process III3. No heat is exchanged between the gas and its surroundings
S. In process IV4. Work done by the gas is 6P0 V06 \mathrm{P}_{0} \mathrm{~V}_{0}
Question figure
  1. Option A:

    P→4;Q→3;R→1; S →2\mathrm{P} \rightarrow 4 ; \mathrm{Q} \rightarrow 3 ; \mathrm{R} \rightarrow \mathbf{1 ; ~ S ~} \rightarrow 2

  2. Option B:

    P→1;Q→3;R→2;S→4\mathrm{P} \rightarrow 1 ; \mathrm{Q} \rightarrow 3 ; \mathrm{R} \rightarrow 2 ; \mathrm{S} \rightarrow 4

  3. Option C:

    P→3;Q→4;R→1;S→2\mathrm{P} \rightarrow 3 ; \mathrm{Q} \rightarrow 4 ; \mathrm{R} \rightarrow 1 ; \mathrm{S} \rightarrow 2

    Correct
  4. Option D:

    P→3;Q→4;R→2;S→1\mathbf{P} \rightarrow \mathbf{3} ; \mathbf{Q} \rightarrow \mathbf{4} ; \mathbf{R} \rightarrow 2 ; \mathrm{S} \rightarrow \mathbf{1}

Answer: C

Step-by-step solution

Process 1 is adiabatic process, hence Q=0\mathrm{Q}=0

Process 2 is isobaric, hence W=P⋅ΔV=6P0 V0\mathrm{W}=\mathrm{P} \cdot \Delta \mathrm{V}=6 \mathrm{P}_{0} \mathrm{~V}_{0}

Process 3 is isochoric, hence W=0\mathrm{W}=0

Process 4 is isothermal

Answer key and solution verified before publishing.

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Exam
JEE Advanced 2018
Paper
Paper 2
Subject
Physics
Chapter
Thermodynamics
Topic
Different Thermodynamic Processes
One mole of a monatomic ideal gas undergoes four thermodynamic… | JEE Advanced 2018 PYQ with Solution · DhiX AI