Three moles of an ideal gas expand isothermally against a constant opposing pressure of 100 kPa from 20 dm3 to 60 dm3 . Compute Q, W, ∆U, and ∆H. [Q = W = 4 kJ ; ∆U = ∆H = 0]
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Since, the process is isothermal,
ΔU=ΔH=0
From the first law of thermodynamics
ΔU=q+w=0
q=−w
w=−2.303nRTlogV1V2=−2.303×1×8.314×400log1020
=−2.303×1×8.314×0.3010=−2305.3J (Work done by system)
q=−w=2305.3J (Heat absorbed by the system
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