JEE Challenger
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Monatomic Ideal Gas Thermodynamic Cycle in TV Diagram

One mole of a monatomic ideal gas undergoes a cyclic thermodynamic process as represented in the accompanying TT-VV diagram, where VV denotes the volume and TT denotes the temperature. Which of the following statements is (are) correct?

Question Diagram 1

Options

A

Process I corresponds to an isochoric process

B

In process II, heat is absorbed by the gas

Correct
C

In process IV, heat is released by the gas

Correct
D

Processes I and III are not isobaric

Correct

Step-by-Step Solution

To determine the correct statements for the given thermodynamic cycle of an ideal gas in the TT-VV plane:

  1. Process I: The volume VV changes as temperature TT varies, meaning the process cannot be isochoric. Therefore, statement (A) is incorrect.

  2. Processes I and III: For an isobaric process, Charles's law requires T∝VT \propto V, which represents a straight line passing through the origin with a positive slope. Since neither Process I nor Process III follows this relation, neither process is isobaric. Therefore, statement (D) is correct.

  3. Process II: This is an isothermal expansion at constant temperature (T=constantT = \text{constant}) with increasing volume (dV>0dV > 0). Since ΔU=nCvΔT=0\Delta U = nC_v \Delta T = 0 and work done W=∫P dV>0W = \int P\,dV > 0, the first law of thermodynamics gives Q=ΔU+W>0Q = \Delta U + W > 0. Thus, heat is absorbed by the gas, making statement (B) correct.

  4. Process IV: This is an isothermal compression at constant temperature (T=constantT = \text{constant}) with decreasing volume (dV<0dV < 0). With ΔU=0\Delta U = 0 and work done W<0W < 0, the first law yields Q=W<0Q = W < 0. Thus, heat is released by the gas, making statement (C) correct.

Hence, the correct statements are (B), (C), and (D).