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For an irreversible process involving only pressure-volume work
Answer & Solution
Correct Answer:
Option
A
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.
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.
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