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• • Contributed by Jeremy Tatum
• Emeritus Professor (Physics & Astronomy) at University of Victoria

We are going to calculate an expression for $$(∂T/∂P)_S$$. The expression will be positive, since T and P increase together. We shall consider the entropy as a function of temperature and pressure, and, with the variables $\left(\frac{\partial S}{\partial T}\right)_{P}\left(\frac{\partial T}{\partial P}\right)_{S}\left(\frac{\partial P}{\partial S}\right)_{T}=-1. \label{15.2.1}$

The middle term is the one we want. Let’s find expressions for the first and third partial derivatives in terms of things that we can measure.

In a reversible process $$dS = dQ/T$$, and, in an isobaric process, $$dQ = C_PdT$$. Therefore

$\left(\frac{\partial S}{\partial T}\right)_{p}=\frac{C_{p}}{T}.$

Also, we have a Maxwell relation (Equation 12.6.16). $$\left(\frac{\partial S}{\partial P}\right)_{T}=-\left(\frac{\partial V}{\partial T}\right)_{P}$$. Thus equation \ref{15.2.1} becomes

$\left(\frac{\partial T}{\partial P}\right)_{S}=\frac{T}{C_{P}}\left(\frac{\partial V}{\partial T}\right)_{P}. \label{15.2.2}$

Check the dimensions of this. Note also that CP can be total, specific or molar, provided that V is correspondingly total, specific or molar. (∂T/∂P)S is, of course, intensive.

If the gas is an ideal gas, the equation of state is $$PV = RT$$, so that

$\left(\frac{\partial V}{\partial T}\right)_{P}=\frac{R}{P}=\frac{V}{T}.$

Equation \ref{15.2.2} therefore becomes

$\left(\frac{\partial T}{\partial P}\right)_{S}=\frac{V}{C_{P}}.$