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This is for an isochoric system held at a temperature $T$ by its surroundings, and that heat $dQ$ is added to the system by its surroundings.

According to chapter 16.5 in Concepts in Thermal Physics by Blundell and Blundell, for a system held at a constant temperature and volume, the inequality $$ dw \geq dF $$ shows that Helmholtz free energy of the system increases if work is added to the system. But I don't see why the RHS of the inequality is more than 0 if $dw$ is positive.

Note that it is likely that, due to my misunderstandings, I am possibly not representing what the book said correctly.

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  • $\begingroup$ To learn how to format, click the question mark on the edit bar of the post, then click on "Advanced Help" that opens underneath. To get math in a line put them between two dollar signs, to make a centered equation enclose them between two dollar signs. $\endgroup$
    – Themis
    May 1, 2023 at 11:13

2 Answers 2

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The definition of F tells us that, at constant temperature, $$\Delta F=\Delta U-T\Delta S=Q+W-T\Delta S$$ where W is the work done by the surroundings on the system. For a system in contact with a reservoir at constant temperature T, the 2nd law tells us that $$\Delta S=\frac{Q}{T}+\sigma$$where $\sigma$ is the (positive) generated entropy. So if we combine these equations, we obtain: $$\Delta F=W-T\sigma$$So,if the process is reversible, F will increase if non-PV work is done on the system, but, if irreversibilites are present to a large enough extent, F can decrease.

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I think your confusion arises because you assume that $dF$ is always negative. It is negative for a spontaneous process, in which case the system has the ability to produce work, i.e. $dw<0$.

If we must add work to the system, the process is not spontaneous and its $dF$ is positive, as is $dw$.

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  • $\begingroup$ But I do not understand why dF has to be positive, surely it can still be less than 0 regardless of dw. $\endgroup$
    – Tarnish3d
    May 1, 2023 at 15:08
  • $\begingroup$ @Tarnish3d "surely it can still be less than 0 regardless of dw". What is your reason for saying this? As a state function, if $dF$ is negative for a process, it is positive for the reverse process. $\endgroup$
    – Themis
    May 1, 2023 at 15:23
  • $\begingroup$ Oh, I see now. I didn't think of that. I was just focused on the inequality. Thank you very much. $\endgroup$
    – Tarnish3d
    May 1, 2023 at 16:29

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