# 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

$dG = VdP - SdT \nonumber$

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

$dV=\alpha VdT-\beta VdP \nonumber$

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