The gas performs isothermal irreversible work ( W ). where, ∆U=0 (change in internal energy) From, 1st law of thermodynamics, ⇒‌‌∆U‌=∆Q+W ⇒‌‌0‌=∆Q+W ⇒‌‌∆Q‌=−W ‌ Now, ‌‌‌W‌=−p‌ext ‌(V2−V1) ‌=−p‌ext ‌(‌
nRT
p2
−‌
nRT
p1
)=−p‌ext ‌×nRT(‌
1
p2
−‌
1
p1
) Given, p‌ext ‌=4.3MPa,p1=2.1MPa,p2=1.3MPa, n=5mol,T‌‌=293K‌ and ‌R=8.314Jmol−1K−1 ‌‌=−4.3×5×8.314×293(‌