I have been studying about the Helmholtz free energy function ($A$) that $$A=U-TS$$ Evidently the change in $A$ is given by (generally for isothermal reversible processes) $$\Delta A = \Delta U - T\Delta S \tag1$$ and we know that $$\Delta S = \frac{Q}{T} \tag2$$

Now this Wikipedia page describes two different sign conventions informally named Clausius's sign convention and Planck's sign convention.

If I follow Clausius's convention then $$\Delta U = Q-W \tag3$$ Using (1),(2) and (3), $$W= -\Delta A \tag4$$ which makes complete sense to me that the decrease in work function ($A$) should be equal to work done ($W$) by the system.

But confusion occurs when I use Planck's convention for the same thing. According to it $$\Delta U= Q+W \tag5$$ Using (1),(2) and (5), $$W=+\Delta A \tag6$$ Now I know that obviously both the sign conventions are correct and paint the same picture. But Clausius' convention seems to be correct and Planck's convention do not both intuitively as well as logically. I am not calling any of these conventions wrong. I am asking why (4) and (6) gives different interpretations of the same thing?

  • 1
    $\begingroup$ Is not it obvious ? If W is work done to system then W= +ΔA. If by the system then W=-ΔA. It is not about intuition but about historical context. Work done by the system was used by engineers/physicists, more interested what the system can do. Chemists/scientists were interested more in what happens to the system. $\endgroup$
    – Poutnik
    Commented Aug 18, 2023 at 5:09
  • $\begingroup$ So can I say (4) and (6) are fundamentally same in meaning? And sign is a matter of preference? $\endgroup$ Commented Aug 18, 2023 at 5:16
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    $\begingroup$ It is not a matter of preference. It is a matter of defining the internal energy of a system. Either add the work done on the system or subtract the work done by the system. The work must result in random change in molecular momentum and change the heat content of the system. $\endgroup$
    – jimchmst
    Commented Aug 22, 2023 at 2:46

1 Answer 1


The Helmholtz energy change does not have a sign convention. Work does.

If there is provided reversible work 10 J to an isothermic, isochoric, closed system, then

  • in the Clausius convention (focusing on work done by thermal engines):
    • $\Delta U = Q - W$
    • $W=\pu{−10 J}$
    • $\Delta A=\Delta U - T\Delta S=\\ T\Delta S - W - T\Delta S =\\ −W =\pu{+10 J}$
  • in the Planck convention (focusing on changes of the system state):
    • $\Delta U = Q + W$
    • $W=\pu{+10 J}$
    • $\Delta A=\Delta U - T\Delta S=\\ T\Delta S + W -T\Delta S =\\ +W =\pu{+10 J}$

Note that while both conventions are correct, if not mutually confused, physical chemistry prefers the Planck convention (work done on the system). It is a good habit to explicitly mention which convention is used.

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    $\begingroup$ Helmholtz energy does not have a sign convention. Work does. This sentence was the key to end the confusion. Thanks. $\endgroup$ Commented Aug 19, 2023 at 4:13

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