JEE Main 2025 Apr 3 Shift 2, Chemistry Q22: Basic Concepts of Thermodynamic

JEE Main2025Apr 3, Shift 2Chemistry

A perfect gas (0.1 mol) having $\overline{C_{v}}$ = 1.50 R (independent of temperature) undergoes the above transformation from point 1 to point 4. If each step is reversible, the total work done (w) while going from point 1 to point 4 is ( $-$ ) ___ J (nearest integer)

[Given : R = 0.082 L atm K $^{- 1}$ mol $^{- 1}$ ]

Your answer

Solution

The path is 1 $\to$ 2 (isochoric), 2 $\to$ 3 (isobaric expansion at 3 atm), 3 $\to$ 4 (isochoric). Work done by the surroundings on the gas, $w = - \int P_{e x t} d V$ , is non-zero only on the constant-pressure leg.

$w_{1 \to 2} = 0$ (no volume change).

$w_{2 \to 3} = - P Δ V = - (3 atm) (2000 - 1000) cm^{3} = - 3000$ atm $\cdot$ cm $^{3} = - 3$ atm $\cdot$ L.

$w_{3 \to 4} = 0$ (no volume change).

Convert atm $\cdot$ L to joules using 1 atm $\cdot$ L $= 101.3$ J:

$w_{t o t a l} = - 3 \times 101.3 \approx - 304$ J.

$∴$ The magnitude is 304 J.

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JEE Main 2025 Apr 3 Shift 2, Chemistry Q22: Basic Concepts of Thermodynamic

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