How to Derive a Thermodynamic Relation from Helmholtz Free Energy

The Helmholtz free energy is a useful thermodynamic potential when temperature and volume are the natural variables. Its differential gives direct formulas for entropy and pressure. From those formulas, you can derive a frequently used identity connecting the volume dependence of internal energy to the temperature dependence of pressure:

$$\left(\frac{\partial U}{\partial V}\right)_T = T\left(\frac{\partial P}{\partial T}\right)_V-P.$$

This relation applies to an equilibrium, simple compressible system with fixed composition. The derivation also shows why it gives zero for an ideal gas and a nonzero result for a van der Waals gas.

Start with the Helmholtz free energy

Define the Helmholtz free energy as

$$F=U-TS,$$

where $U$ is internal energy, $T$ is absolute temperature, and $S$ is entropy. For a fixed amount of a simple compressible substance, the first law in differential form is

$$dU=T\,dS-P\,dV.$$

Differentiate $F=U-TS$:

$$dF=dU-T\,dS-S\,dT.$$

Substituting the first-law expression for $dU$ cancels the $T\,dS$ terms, leaving

$$dF=-S\,dT-P\,dV.$$

Thus, when $F$ is regarded as a function of its natural variables $T$ and $V$,

$$S=-\left(\frac{\partial F}{\partial T}\right)_V, \qquad P=-\left(\frac{\partial F}{\partial V}\right)_T.$$

These are the entropy and pressure identities for the Helmholtz free energy.

Derive the internal-energy relation

Rearrange the definition of $F$ to write the internal energy as

$$U=F+TS.$$

At fixed temperature, differentiate with respect to volume:

$$\left(\frac{\partial U}{\partial V}\right)_T = \left(\frac{\partial F}{\partial V}\right)_T +T\left(\frac{\partial S}{\partial V}\right)_T.$$

The first term is $-P$. To rewrite the second term, use the fact that the mixed second derivatives of a well-behaved state function are equal. From the entropy and pressure identities,

$$\left(\frac{\partial S}{\partial V}\right)_T =-\frac{\partial^2 F}{\partial V\,\partial T} =\left(\frac{\partial P}{\partial T}\right)_V.$$

This is a Maxwell relation. Substituting it gives

$$\boxed{\left(\frac{\partial U}{\partial V}\right)_T =T\left(\frac{\partial P}{\partial T}\right)_V-P.}$$

The right-hand side uses the equation of state: how pressure varies with temperature when volume is held fixed. The left-hand side is the change in internal energy per change in volume at fixed temperature.

Worked example: a van der Waals gas

For $n$ moles, the van der Waals equation of state is

$$P=\frac{nRT}{V-nb}-\frac{a n^2}{V^2},$$

where $a$ and $b$ are constants describing intermolecular attraction and excluded volume. Differentiate with respect to temperature at fixed volume:

$$\left(\frac{\partial P}{\partial T}\right)_V =\frac{nR}{V-nb}.$$

Now substitute this and the equation of state into the derived identity:

$$\begin{aligned} \left(\frac{\partial U}{\partial V}\right)_T &=T\frac{nR}{V-nb}-\left(\frac{nRT}{V-nb}-\frac{a n^2}{V^2}\right)\\ &=\frac{a n^2}{V^2}. \end{aligned}$$

So at fixed temperature, increasing the volume increases the internal energy in this model. Integrating with respect to $V$ gives

$$U(T,V)=f(T)-\frac{a n^2}{V},$$

where $f(T)$ is an undetermined function of temperature. The equation of state alone determines the volume-dependent part, not the full temperature-dependent contribution.

Check: the ideal gas

For an ideal gas, $P=nRT/V$, so

$$T\left(\frac{\partial P}{\partial T}\right)_V-P =T\frac{nR}{V}-\frac{nRT}{V}=0.$$

Therefore $(\partial U/\partial V)_T=0$: the internal energy of an ideal gas depends only on temperature, not volume. This is a result of the ideal-gas equation of state, not a general rule for all gases.

Common mistakes

• Dropping the minus sign in the pressure identity: Since $dF=-S\,dT-P\,dV$, pressure is $-(\partial F/\partial V)_T$.
• Holding the wrong variable fixed: The pressure derivative in the result is at constant volume; the internal-energy derivative is at constant temperature.
• Assuming the ideal-gas result applies to every substance: Real-gas interactions can make internal energy depend on volume even at fixed temperature.
• Confusing Helmholtz and Gibbs free energy: Helmholtz free energy has natural variables $T,V$; Gibbs free energy has natural variables $T,P$ for a simple compressible system.