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