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10.5: Expressing Thermodynamic Functions with Independent Variables P and T

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    We can follow a parallel development to express these thermodynamic functions with \(P\) and \(T\) as the independent variables. We have the differential relationship \(dH=TdS+VdP\). We expand \(\ dH\) with \(P\) and \(T\) as the independent variables. Equating these, we obtain

    \[\begin{align} dH &= \left(\frac{\partial H}{\partial P}\right)_TdP + \left(\frac{\partial H}{\partial T}\right)_PdT \\[4pt] &= TdS + VdP \end{align} \nonumber \]

    so that we have

    \[dS = \frac{\mathrm{1}}{T}{\left(\frac{\partial H}{\partial T}\right)}_PdT + \frac{\mathrm{1}}{T}\left[{\left(\frac{\partial H}{\partial P}\right)}_T - V\right]dP \nonumber \]

    From the coefficient of \(dT\) and the definition \({\left({\partial H}/{\partial T}\right)}_P=C_P\), we have

    \[\left(\frac{\partial S}{\partial T}\right)_P = \frac{\mathrm{1}}{T}{\left(\frac{\partial H}{\partial T}\right)}_P = \frac{C_P}{T} \nonumber \]

    (When we are describing a reversible system that is not a pure substance, \(C_P\) is just an abbreviation for \({\left({\partial H}/{\partial T}\right)}_P\).) From the coefficient of \(dP\) and the relationship \({\left({\partial S}/{\partial P}\right)}_T=-{\left({\partial V}/{\partial T}\right)}_P\) that we find in Section 10.1, we have

    \[{\left(\frac{\partial S}{\partial P}\right)}_T = \frac{\mathrm{1}}{T}\left[{\left(\frac{\partial H}{\partial P}\right)}_T - V\right] = -{\left(\frac{\partial V}{\partial T}\right)}_P \nonumber \]

    Substituting into the expression for \(dS\), we find

    \[dS = \frac{C_P}{T}dT - {\left(\frac{\partial V}{\partial T}\right)}_PdP \nonumber \]

    Using the same approach as in the previous section, we can now obtain

    \[\begin{align} dE &= \left[C_P - P{\left(\frac{\partial V}{\partial T}\right)}_P\right]dT - \left[{P\left(\frac{\partial V}{\partial P}\right)}_T + T{\left(\frac{\partial V}{\partial T}\right)}_P\right]dP \\[4pt] dH &= C_PdT + \left[V - T{\left(\frac{\partial V}{\partial T}\right)}_P\right]dP \\[4pt] dA &= -\left[S + P{\left(\frac{\partial V}{\partial T}\right)}_P\right]dT - {P\left(\frac{\partial V}{\partial P}\right)}_TdP \end{align} \nonumber \]

    and, we already have

    \[dG = VdP - SdT \nonumber \]

    Finally, we can write \(V=V\left(P,T\right)\) to find

    \[{dV = \left(\frac{\partial V}{\partial P}\right)}_TdP + {\left(\frac{\partial V}{\partial T}\right)}_PdT \nonumber \]

    so that we have total differentials for all of the principal thermodynamic functions when they are expressed as functions of \(P\) and \(T\). If an equation of state is not known but the coefficients of thermal expansion and isothermal compressibility are available, we have \({\left({\partial V}/{\partial T}\right)}_P=\alpha V\) and \({\left({\partial V}/{\partial P}\right)}_T=-\beta V\). Then we can estimate a volume change, for example, as a line integral of

    \[dV=\alpha VdT-\beta VdP \nonumber \]

    This page titled 10.5: Expressing Thermodynamic Functions with Independent Variables P and T is shared under a CC BY-SA 4.0 license and was authored, remixed, and/or curated by Paul Ellgen via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.