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A real gas within a closed chamber at 27°C undergoes the cyclic process shown in the figure. The gas obeys PV³=RT along the path A to B. The net work done in the complete cycle is (assume R=8 J mol⁻¹K⁻¹)

Asked in JEE Main 9th April 2nd Shift 2024 · Work as the Area of a Closed Loop

Figure: Work as the Area of a Closed Loop
Answer: (4) 205 J

Step-by-step solution

Along A to B the gas obeys PV³=RT, so P=(RT)/(V³) at fixed T=300 K.

W_AB=∫₂⁴(RT)/(V³)dV=(RT)/2(1/4-1/(16)) with RT=8×300=2400.

W_AB=1200×3/(16)=225 J.

B→ C is isobaric at 10 N/m² from 4 to 2 m³: W=10×(-2)=-20 J.

C→ A is isochoric: W=0.

Net W=225-20=205 J.

Note that the printed axis values and the stated law do not sit together: PV³

at A is 20×8=160, not 2400. The work integral over the printed volumes

with RT=2400 is what produces the key. Listed in DEFECTS.md.

Why the other options are wrong

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