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15.2: Adiabatic Decompression

Last updated

Sep 10, 2020
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- 15.1: Introduction
- 15.3: Adiabatic Demagnetization

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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

Screen Shot 2019-07-08 at 12.06.25 PM.png

we shall start with the cyclic relation

$\left(\frac{\partial S}{\partial T}\right)_{P}\left(\frac{\partial T}{\partial P}\right)_{S}\left(\frac{\partial P}{\partial S}\right)_{T}=-1. \label{15.2.1}$

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 C_P 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

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

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