Is $C_p=T\left(\frac{\partial S}{\partial T}\right)_{P}$ true for both reversible and irreversible process?
I know why is true for the reversible case but can't see if it might be true for the irreversible one.
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$\begingroup$ Think about it. Is pressure constant in an irreversible process? Are temperature and entropy even well defined? $\endgroup$– joigusCommented Mar 21, 2021 at 7:43
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1 Answer
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I know why is true for the reversible case but can't see if it might be true for the irreversible one.
It's not true for the irreversible constant pressure (isobaric) process. That's because the $p$ in $C_p$ refers to the constant equilibrium pressure of the gas during the process, i.e., the gas pressure for a reversible process. For an irreversible process, the gas in not in equilibrium, so its pressure is undefined.
Another reason is while the change in entropy between two equilibrium states is the same for a reversible and irreversible process, you can't connect the same two equilibrium states with an irreversible isobaric process as with a reversible isobaric process.
Fig 1 below shows a reversible isobaric process from state 0 to 1a. The irreversible isobaric process of Fig 2 involves a sudden drop in external pressure which is held at $P_1$ until the gas re-equilibriates at state 1, which is different that state 1a.
Hope this helps.
