Q50 / 54 · JEE Advanced 2022 (Paper 2) · Physics

In the given P-V diagram, a monoatomic gas γ=53 is first compressed adiabatically from state A state B. Then it expands isothermally from state B to state C. [Given : 130.6=0.5,ln2≈0.7].

Figure

Which of the following statement(s) is(are) correct?

  • A
    The magnitude of the total work done in the process A→B→C is 144 kJ
  • Correct answer
    B
    The magnitude of the work done in the process B→C is 84kJ.
  • Correct answer
    C
    The magnitude of the work done in the process A→B is 60kJ.
  • Correct answer
    D
    The magnitude of the work done in the process C→A is zero.

Solution

For adiabatic process PVγ=C

⇒1000.853=300VB53

⇒VB=0.833/5=0.4 m3

Work done in an adiabatic process is given by,

W=P1V1-P2V2γ-1

⇒WAB=PAVA-PBVB53-1=80-300×0.42/3 kJ

⇒WAB=-60kJ

Therefore, option C is correct.

C→A is isochoric ⇒D is correct.

For process BC

WBC=nRTlnV2V1=PVlnV2V1

⇒WBC=300×0.4ln0.80.4

⇒WBC=120×ln2 kJ⇒WBC=84 kJ

Therefore, option B is correct.

Now,

WAB+WBC=-60+84=24 kJ

Therefore, option A is incorrect.

Hence, the correct answer is options: 'B', 'C', 'D'.

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