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Physics · Thermodynamics

JEE Advanced 2019 — Paper 1 — Question 10

One mole of a monatomic ideal gas goes through a thermodynamic cycle, as shown in the volume versus temperature (V−T)(\mathrm{V}-\mathrm{T}) diagram. The correct statement(s) is/are:

[ R is the gas constant]

Question figure
  1. Option A:

    Work done in this thermodynamic cycle (1→2→3→4→1)(1 \rightarrow 2 \rightarrow 3 \rightarrow 4 \rightarrow 1) is ∣W∣=12RT0|\mathrm{W}|=\frac{1}{2} R T_{0}

    Correct
  2. Option B:

    The ratio of heat transfer during processes 1→21 \rightarrow 2 and 2→32 \rightarrow 3 is ∣Q1→2Q2→3∣=53\left|\frac{Q_{1 \rightarrow 2}}{Q_{2 \rightarrow 3}}\right|=\frac{5}{3}

    Correct
  3. Option C:

    The above thermodynamic cycle exhibits only isochoric and adiabatic processes.

  4. Option D:

    The ratio of heat transfer during processes 1→21 \rightarrow 2 and 3→43 \rightarrow 4 is ∣Q1→2Q3→4∣=12\left|\frac{Q_{1 \rightarrow 2}}{Q_{3 \rightarrow 4}}\right|=\frac{1}{2}

Answer: A, B

Step-by-step solution

Wcycle =P0 V0=RT02\mathrm{W}_{\text {cycle }}=\mathrm{P}_{0} \mathrm{~V}_{0}=\frac{\mathrm{RT}_{0}}{2} ∣Q1→2Q2→3∣=∣nCp( T2−T1)nCV( T3−T2)∣=∣−53∣=53\left|\frac{\mathrm{Q}_{1 \rightarrow 2}}{\mathrm{Q}_{2 \rightarrow 3}}\right|=\left|\frac{\mathrm{nC}_{\mathrm{p}}\left(\mathrm{~T}_{2}-\mathrm{T}_{1}\right)}{\mathrm{nC}_{\mathrm{V}}\left(\mathrm{~T}_{3}-\mathrm{T}_{2}\right)}\right|=\left|-\frac{5}{3}\right|=\frac{5}{3} ∣Q1→2Q2→3∣=∣nCp( T2−T1)nCV( T4−T3)∣=2\left|\frac{\mathrm{Q}_{1 \rightarrow 2}}{\mathrm{Q}_{2 \rightarrow 3}}\right|=\left|\frac{\mathrm{nC}_{\mathrm{p}}\left(\mathrm{~T}_{2}-\mathrm{T}_{1}\right)}{\mathrm{nC}_{\mathrm{V}}\left(\mathrm{~T}_{4}-\mathrm{T}_{3}\right)}\right|=2
Solution figure

Answer key and solution verified before publishing.

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Exam
JEE Advanced 2019
Paper
Paper 1
Subject
Physics
Chapter
Thermodynamics
Topic
Calculation of Work, Heat and Internal Energy

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