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8.4: Volume Dependence of Helmholtz Energy

  • Page ID
    199221
  • 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)