Question

One mole of an ideal monoatomic gas initially at STP expand isothermally and reversible to 224 lit. Calculate ∆S.

Answer 1

Here at STP,
Temperature, T = 273K
Intial volume, Vi= 22.4l
Final volume, Vf= 224l

Work done in an isothermal reversible process is given by:

W = -nRTlnVf/Vi
=-1 mol ×
8.314 Jmol^−1K^−1×273K
×ln(224/22.4)
=-5227.17J
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Now, From first law of thermodynamics,
∆U = Qrev+W
In an isothermal process

Delta U= 0

=> Qrev= - W
= 5227.17J

There fore
Entropy change
Delta S = Qrev/T
=5227. 17/273
=19.147J K^-1

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