Question

Choose the correct option(s).

Here A is the Helmholtz free energy and G is the Gibbs free energy.

• A. A provides a criterion for equilibrium in a system with constant temperature and constant pressure
• B. G provides a criterion for equilibrium in a system with constant temperature and constant pressure
• C. A provides a criterion for equilibrium in a system with constant temperature and constant volume
• D. G provides a criterion for equilibrium in a system with constant temperature and constant volume

Hint:

Helmholtz free energy A is minimized at equilibrium under constant T and V; Gibbs free energy G is minimized at equilibrium under constant T and P.

Answer 1

The Correct Option is B, C

This question checks whether you know which free energy function goes with which pair of fixed conditions. Rather than memorizing the pairing, let's build it from the combined statement of the first and second laws, $dU = TdS - PdV$, and see which terms survive under each set of constraints.

1. Option (A), A at constant T and P: Starting from $A = U - TS$, $dA = -SdT - PdV$. This expression still has a $PdV$ term sitting in it. Holding T constant kills $-SdT$, but volume is free to change when only T and P are fixed, so $-PdV$ does not vanish. A is not guaranteed to sit at a minimum here, so this option is wrong.
2. Option (B), G at constant T and P: Starting from $G = U + PV - TS$, working through the algebra gives $dG = VdP - SdT$. Now holding both T and P fixed kills both terms on the right, leaving $(dG)_{T,P} \leq 0$ for any real process. G drops to a minimum and sits there at equilibrium, so this is a valid criterion, and the option is correct.
3. Option (C), A at constant T and V: Going back to $dA = -SdT - PdV$, this time both T and V are held fixed, so both $-SdT$ and $-PdV$ vanish, leaving $(dA)_{T,V} \leq 0$ for any real process. A drops to a minimum and stays there at equilibrium, so this is also a valid criterion, and the option is correct.
4. Option (D), G at constant T and V: From $dG = VdP - SdT$, holding T fixed kills $-SdT$, but pressure is not held fixed here, only volume is, so $VdP$ does not vanish in general. G is not guaranteed to be at a minimum under these conditions, so this option is wrong.

The pattern makes sense once you notice which natural variables sit inside each differential. $dA$ naturally wants T and V fixed to vanish completely, while $dG$ naturally wants T and P fixed to vanish completely. Trying to force A to work at constant P, or G to work at constant V, leaves a leftover term that spoils the criterion.

Let's summarize:

• $dA = -SdT - PdV$ collapses to $(dA)_{T,V} \leq 0$ only when T and V are both fixed.
• $dG = VdP - SdT$ collapses to $(dG)_{T,P} \leq 0$ only when T and P are both fixed.

So the correct statements are that G is the equilibrium criterion at constant T and P, and A is the equilibrium criterion at constant T and V.

\[ \boxed{\text{Options B and C}} \]