JEE Main 2025 Apr 2 Shift 2, Physics Q10: Thermodynamic Processes

JEE Main2025Apr 2, Shift 2Physics

Identify the characteristics of an adiabatic process in a monoatomic gas.

(A) Internal energy is constant.

(B) Work done in the process is equal to the charge in internal energy.

(D) The product of pressure and volume is a constant.

(E) The work done to change the temperature from $T_{1}$ to $T_{2}$ is proportional to $(T_{2} - T_{1})$

Choose the correct answer from the options given below :

A
(A), (C), (D) only
B
(A), (C), (E) only
C
(B), (E) only
D
(B), (D) only

Solution

In an adiabatic process $Q = 0$ , so the first law gives $W = - Δ U$ . Therefore the magnitude of the work done equals the change in internal energy. Statement (B) is correct.

For an ideal gas $Δ U = n C_{v} Δ T$ , so $∣ W ∣ = n C_{v} (T_{2} - T_{1})$ , which is directly proportional to $(T_{2} - T_{1})$ . Statement (E) is correct.

Statement (A) is wrong because the internal energy does change in an adiabatic process (it absorbs/releases the work).

For an adiabatic process the correct invariants are $T V^{γ - 1} =$ const and $P V^{γ} =$ const, not simply $T V$ or $P V$ . Statements (C) and (D) are therefore wrong.

$∴$ Only (B) and (E) hold.

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JEE Main 2025 Apr 2 Shift 2, Physics Q10: Thermodynamic Processes

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