What is correct sign convention for Helmholtz free energy change?

Question

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?

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.