For an irreversible process involving only pressure-volume work
A. (dF)T, P < 0
B. (dF)T, P = 0
C. (dF)T, P > 0
D. (dA)T, V > 0
Answer: Option A
Solution (By Examveda Team)
Since for an spontaneous process $$Tds > dU + PdV$$For, constant temperature and constant pressure process the above equation can be written as
$$\eqalign{ & d\left( {TS - U} \right) > PdV \cr & \Rightarrow d\left( { - A} \right) > PdV \cr & \Rightarrow {\left( {dF} \right)_{T,\,P}} < 0. \cr} $$
For spontaneous process.
Here the F is Gibbs free energy and A is Helmholtz free energy.
This Question Belongs to Chemical Engineering >> Chemical Engineering Thermodynamics
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