# 8.4: Volume Dependence of Helmholtz Energy

If one needs to know how the Helmholtz function changes with changing volume at constant temperature, the following expression can be used:

$\Delta A = \int_{V_1}^{V_2} \left( \dfrac{\partial A}{\partial V} \right)_T dV \label{eq1}$

But how does one derive an expression for the partial derivative in Equation \ref{eq1}? This is a fairly straight forward process that begins with the definition of $$A$$:

$A = U - TS$

Differentiating (and using the chain rule to evaluate $$d(TS)$$ yields

$dA = dU - TdS - SdT \label{eq4}$

Now, it is convenient to use the combined first and second laws

$dU = TdS - pdV \label{eq5}$

which assumes:

1. a reversible change and
2. only $$pV$$ work is being done.

Substituting Equation \ref{eq5} into Equation \ref{eq4} yields

$dA = \cancel{TdS} - pdV - \cancel{TdS} - SdT \label{eq6}$

Canceling the $$TdS$$ terms gives the important result

$dA = - pdV - SdT \label{eq6.5}$

The natural variables of $$A$$ are therefore $$V$$ and $$T$$! So the total differential of $$A$$ is conveniently expressed as

$dA = \left( \dfrac{\partial A}{\partial V} \right)_T dV + \left( \dfrac{\partial A}{\partial T} \right)_V dT \label{Total2}$

and by simple comparison of Equations \ref{eq6.5} and \ref{Total2}, it is clear that

$\left( \dfrac{\partial A}{\partial V} \right)_T = - p$

$\left( \dfrac{\partial A}{\partial T} \right)_V = - S$

And so, one can evaluate Equation \ref{eq1} as

$\Delta A = - \int_{V_1}^{V_2} p\, dV$

If the pressure is independent of the temperature, it can be pulled out of the integral.

$\Delta A = - p \int_{V_1}^{V_2} dV = -p (V_2-V_1)$

Otherwise, the temperature dependence of the pressure must be included.

$\Delta A = - \int_{V_1}^{V_2} p(V)\, dV$

Fortunately, this is easy if the substance is an ideal gas (or if some other equation of state can be used, such as the van der Waals equation.)

Example $$\PageIndex{1}$$: Ideal Gas Expansion

Calculate $$\Delta A$$ for the isothermal expansion of 1.00 mol of an ideal gas from 10.0 L to 25.0 L at 298 K.

Solution:

For an ideal gas,

$p =\dfrac{nRT}{V}$

So

$\left( \dfrac{\partial A}{\partial V} \right)_T = -p$

becomes

$\left( \dfrac{\partial A}{\partial V} \right)_T = -\dfrac{nRT}{V}$

And so (Equation \ref{eq1})

$\Delta A = \int_{V_1}^{V_2} \left( \dfrac{\partial A}{\partial V} \right)_T dV$

becomes

$\Delta A = -nRT \int_{V_1}^{V_2} \dfrac{dV}{V} dT$

or

$\Delta A = -nRT \ln \left( \dfrac{V_2}{V_1} \right)$

Substituting the values from the problem

$\Delta A = -(1.00\,mol)(8.314 \, J/(mol\,K))(298\,K) \ln \left( \dfrac{25.0\,L}{10.0\,L} \right)$

But further, it is easy to show that the Maxwell relation that arises from the simplified expression for the total differential of $$A$$ is

$\left( \dfrac{\partial p}{\partial T} \right)_V = \left( \dfrac{\partial S}{\partial V} \right)_T$

This particular Maxwell relation is exceedingly useful since one of the terms depends only on $$p$$, $$V$$, and $$T$$. As such it can be expressed in terms of our old friends, $$\alpha$$ and $$\kappa_T$$!

$\left( \dfrac{\partial p}{\partial T} \right)_V = \dfrac{\alpha}{\kappa_T}$

## Contributors

• Patrick E. Fleming (Department of Chemistry and Biochemistry; California State University, East Bay)