I know that free energy is defined as the amount of work that a system can do at constant temperature and and its formula is F=U-TS and that by definition Helmholtz free energy is the amount of work a system can do at constant temperature and constant volume but the formula is still A=U-TS. Why is that , why is there no difference in the formula or some kind of correction factor to account for the different conditions (like the PV in Gibbs free energy ) ?

  • $\begingroup$ Please rewrite your question ! You get two different parameters (F and A) for the same definition. What is the correct definition ? Is it F = U-TS or A = U-TS ? $\endgroup$
    – Maurice
    Oct 10, 2020 at 16:25
  • $\begingroup$ Well that is the thing , they have the same formula but one is said to happen at constant T only while the other is at constant T and constant V $\endgroup$ Oct 10, 2020 at 18:16
  • $\begingroup$ That is false interpretation. They do not happen and they do say identical thing. A/F nor U definitions are NOT limited to constant V, same as G or H definitions are not limited to constant p. But their usage at the respective constant p or V is with advantage. $\endgroup$
    – Poutnik
    Oct 11, 2020 at 6:15
  • $\begingroup$ BTW, the pV term is rather in enthalpy than In Gibbs energy, that is derived from enthalpy. $\endgroup$
    – Poutnik
    Oct 11, 2020 at 12:14

1 Answer 1


By "normal free energy", you must mean Gibbs free energy $G$, i.e. free energy at constant pressure ( = the same start and final pressure, as state functions do not depend on the process path.)

Both symbols $A$ and $F$ are eventually used for the Helmholtz free energy ( free energy at constant volume ) for historical reasons, so the particular symbol usage is source dependent. They mean the same thing. The fact each of them was used in the context of your question in different cases does not change that.

Note that $F/A$ nor $U$ state function are NOT limited to be used at constant volume ( see e.g. adiabatic expansion ), same as $G$ nor $H$ are not limited to be used at constant pressure. But their usage at the respective constant state variable is with an advantage, as you can see below.

To refresh the relations of $F(A), G, U, H$:

$$H=U+p \cdot V$$

$$G=H-T \cdot S=U + p \cdot V - T \cdot S=A(F) + p \cdot V$$

$$A(F)=U-T \cdot S = H - p \cdot V - T \cdot S = G - p \cdot V$$

For closed, not isolated systems with zero non volume work:
(Non-volume work is e.g. a system charging a battery. )

For constant volume: $$\mathrm{d}U=\text{đ}Q$$

For variable volume: $$\mathrm{d}U=\text{đ} Q - p \cdot \mathrm{d}V$$

For constant pressure and zero non volume work, $$\mathrm{d}H=\mathrm{d}U + \mathrm{d}(pV) = \mathrm{d}U + p \cdot \mathrm{d}V= \text{đ}Q$$

For variable pressure, it would be $$\mathrm{d}H=\mathrm{d}U + \mathrm{d}(pV) = \mathrm{d}U + p \cdot \mathrm{d}V + V \cdot \mathrm{d}p= \text{đ}Q + V \cdot \mathrm{d}p$$

Now involving free energies:

$$\mathrm{d}F = \mathrm{d}U - \mathrm{d}(T \cdot S ) = \mathrm{d}U - T \cdot \mathrm{d}S - S \cdot \mathrm{d}T$$

$$\mathrm{d}G = \mathrm{d}H - \mathrm{d}(T \cdot S ) = \mathrm{d}H - T \cdot \mathrm{d}S - S \cdot \mathrm{d}T$$

Isothermal cases:

For $\mathrm{d}F$ and $\mathrm{d}G$, there would be just the additional term $-T \cdot \mathrm{d}S$

$$\mathrm{d}F= \text{đ}Q - p \cdot \mathrm{d}V - T \cdot \mathrm{d}S$$

$$\mathrm{d}G=\mathrm{d}F + \mathrm{d}(pV)= \text{đ}Q + V \cdot \mathrm{d}p - T \cdot \mathrm{d}S$$

Non-isothermal cases:

There appears another term: $-S \cdot \mathrm{d}T$

$$\mathrm{d}F= \text{đ}Q - p \cdot \mathrm{d}V - T \cdot \mathrm{d}S - S \cdot \mathrm{d}T $$

$$\mathrm{d}G=\mathrm{d}F + \mathrm{d}(pV)= \text{đ}Q + V \cdot \mathrm{d}p - T \cdot \mathrm{d}S - S \cdot \mathrm{d}T$$

Then for reversible processes, as $dS = \text{đ}Q_\mathrm{rev}/T$:

$$\mathrm{d}F= - p \cdot \mathrm{d}V - S \cdot \mathrm{d}T $$

$$\mathrm{d}G= V \cdot \mathrm{d}p - S \cdot \mathrm{d}T$$

You can see now, that for zero non-volume work and reversible processes, $\mathrm{d}F=0$ for constant $T$,$V$ and $\mathrm{d}G=0$ for constant $T$,$p$.

For irreversible processes, as $\mathrm{d}S_\mathrm{irr} \gt \text{đ}Q/T$, $\mathrm{d}F \lt 0$ for constant $T$,$V$ and $\mathrm{d}G \lt 0$ for constant $T$,$p$.

  • $\begingroup$ @ Poutnik. I repeat the question given by David Sterlinsky ; " Why is there no difference in the formula or some kind of correction factor to account for the different conditions ? ". Well ! Don't you think the answer might be : There is no difference because A is sometimes called F ? $\endgroup$
    – Maurice
    Oct 10, 2020 at 17:03
  • $\begingroup$ @Maurice Sure, it can be. But the question formulation is quite unclear about the question scope, so I took answering in a broader way as well. $\endgroup$
    – Poutnik
    Oct 10, 2020 at 17:04
  • $\begingroup$ I am sorry for the unclear question , what I meant was that if one is said to be happening at constant temperature only (P and V are able to vary) but the other one (Helmholtz) has T and V constant (as in the case of Gibbs energy would be constant P and constant T ) why don't we have some sort of factor to account for the difference in conditions (in one case V is able to change as is P) and in the other (only P is variable) ? $\endgroup$ Oct 10, 2020 at 18:15
  • $\begingroup$ Such question clarifications by the OP should be expressed by modification of the original question, rather than by a comment, where should me rather just mentioning of the question update. It is helpful to read the question after its writing, if it is clear to you, impersonating yourself to an eventual reader. $\endgroup$
    – Poutnik
    Oct 11, 2020 at 5:01
  • $\begingroup$ I have updated the answer to address your clarification $\endgroup$
    – Poutnik
    Oct 11, 2020 at 9:38

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