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JEE Advanced Physics Thermodynamics 2022 JEE Advanced 2022 (Paper 2)

JEE Advanced Physics Question (2022) — Solution

Question

In the given  P - V diagram, a monoatomic gas γ = 5 3  is first compressed adiabatically from state A state B . Then it expands isothermally from state B to state C . [Given : 1 3 0 . 6 = 0 . 5 , ln 2 ≈ 0 . 7 ]. Which of the following statement(s) is(are) correct?

Options

  1. A. The magnitude of the total work done in the process A → B → C  is  144   kJ
  2. B. The magnitude of the work done in the process  B → C  is  84 kJ .
  3. C. The magnitude of the work done in the process  A → B  is  60 kJ .
  4. D. The magnitude of the work done in the process  C → A  is zero.

Answer

D. The magnitude of the work done in the process  C → A  is zero.

Step-by-step solution

For adiabatic process  P V γ = C ⇒ 100 0 . 8 5 3 = 300 V B 5 3 ⇒ V B = 0 . 8 3 3 / 5 = 0 . 4   m 3 Work done in an adiabatic process is given by, W = P 1 V 1 - P 2 V 2 γ - 1 ⇒ W A B = P A V A - P B V B 5 3 - 1 = 80 - 300 × 0 . 4 2 / 3   kJ ⇒ W A B = - 60 kJ Therefore, option  C  is correct. C → A  is isochoric  ⇒ D  is correct. For process  B C W B C = n R T ln V 2 V 1 = P V ln V 2 V 1 ⇒ W B C = 300 × 0 . 4 ln 0 . 8 0 . 4 ⇒ W B C = 120 × ln 2   kJ ⇒ W B C = 84   kJ Therefore, option  B  is correct. Now, W A B + W B C = - 60 + 84 = 24   kJ Therefore, option  A  is incorrect.

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