A real gas within a closed chamber at 27 undergoes the cyclic process as shown in figure. The gas obeys PV³ = RT equation for the path A to B. The net work done in the complete cycle is (assuming R = 8 J/mol K): (a) -20 J (b) 205J (c) 225J (d) 20J

Answer: (b). Path A → B: P = RT/V³ with the chamber temperature T = 300 K, so WAB = ∫₂⁴ RT/V³ dV = RT/2(1/2² - 1/4²) WAB = (8 × 300)/(2) × 3/16 = 225 J Path B → C (isobaric, P = 10 N/m ²): WBC = 10 × (2 - 4) = -20 J Path C → A is isochoric: WCA = 0 Net work = 225 - 20 + 0 = 205 J. So the answer is (b) 205 J.

A real gas within a closed chamber at 27 undergoes the cyclic process… | JEE Main 2024 (9 Apr, Shift 2)

Q39JEE Main 2024 · 9 APR · SHIFT 2Difficult

P-V Diagrams & Cyclic Processes

A real gas within a closed chamber at $27^\circ\text{C}$ undergoes the cyclic process as shown in figure. The gas obeys $PV^3 = RT$ equation for the path $A$ to $B$ . The net work done in the complete cycle is (assuming $R = 8\text{ J/mol K}$ ):

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Answer: (b)

Path $A \to B$ : $P = \frac{RT}{V^3}$ with the chamber temperature $T = 300$ K, so $W_{AB} = \int_2^4 \frac{RT}{V^3}\,dV = \frac{RT}{2}\left(\frac{1}{2^2} - \frac{1}{4^2}\right)$
$W_{AB} = \frac{8 \times 300}{2} \times \frac{3}{16} = 225$ J
Path $B \to C$ (isobaric, $P = 10$ N/m $^2$ ): $W_{BC} = 10 \times (2 - 4) = -20$ J
Path $C \to A$ is isochoric: $W_{CA} = 0$
Net work $= 225 - 20 + 0 = 205$ J. So the answer is (b) 205 J.

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