24 Jul 2025

T-dS Equations, Heat Capacities, and Joule-Kelvin Effect

Maxwell-relation response identities and throttling coefficients for ideal and van der Waals gases.

bsc semester-iv mj-6 heat-and-thermodynamics joule-kelvin-effect heat-capacities

For a fixed-composition simple compressible system, the Maxwell relations turn entropy differentials into measurable response functions.

The two $T\,dS$ equations

Using $S=S(T,V)$,

\[dS=\left(\frac{\partial S}{\partial T}\right)_VdT +\left(\frac{\partial S}{\partial V}\right)_TdV.\]

Since $(\partial S/\partial T)_V=C_V/T$ and $(\partial S/\partial V)_T=(\partial p/\partial T)_V$,

\[\boxed{T\,dS=C_V\,dT +T\left(\frac{\partial p}{\partial T}\right)_VdV}.\]

With $(\partial p/\partial T)_V=\alpha/\kappa_T$,

\[\boxed{T\,dS=C_V\,dT+\frac{T\alpha}{\kappa_T}\,dV}.\]

Using $S=S(T,p)$ and $(\partial S/\partial p)_T=-(\partial V/\partial T)_p=-V\alpha$ gives

\[\boxed{T\,dS=C_P\,dT-TV\alpha\,dp}.\]

General value of $C_P-C_V$

Evaluate the first $T\,dS$ equation at constant pressure. Then $dV=V\alpha\,dT$ and $T\,dS=C_P\,dT$, so

\[C_P=C_V+T\left(\frac{\partial p}{\partial T}\right)_VV\alpha.\]

Therefore

\[\boxed{C_P-C_V=\frac{TV\alpha^2}{\kappa_T}}.\]

For the van der Waals molar equation

\[p=\frac{RT}{V_m-b}-\frac{a}{V_m^2},\]

direct differentiation gives

\[\left(\frac{\partial p}{\partial T}\right)_{V_m} =\frac{R}{V_m-b},\] \[\left(\frac{\partial p}{\partial V_m}\right)_T =-\frac{RT}{(V_m-b)^2}+\frac{2a}{V_m^3}.\]

Using $C_{P,m}-C_{V,m}=-T(p_T)^2/p_V$,

\[\boxed{C_{P,m}-C_{V,m} =\frac{R}{1-\dfrac{2a(V_m-b)^2}{RTV_m^3}}}.\]

The ideal-gas limit $a,b\to0$ is $C_{P,m}-C_{V,m}=R$.

Joule-Kelvin coefficient

In steady throttling through a porous plug or valve, assume adiabatic walls, no shaft work, and negligible changes of bulk kinetic and gravitational potential energy. The steady-flow energy equation then gives

\[H_1=H_2.\]

The Joule-Kelvin coefficient is

\[\mu_{\mathrm{JT}}=\left(\frac{\partial T}{\partial p}\right)_H.\]

From

\[dH=C_P\,dT+ \left[V-T\left(\frac{\partial V}{\partial T}\right)_p\right]dp,\]

set $dH=0$ to obtain

\[\boxed{\mu_{\mathrm{JT}} =\frac{T(\partial V/\partial T)_p-V}{C_P} =\frac{V}{C_P}(\alpha T-1)}.\]

Its SI unit is $\mathrm{K\,Pa^{-1}}$. A pressure drop has $dp<0$: the gas cools when $\mu_{\mathrm{JT}}>0$ and warms when $\mu_{\mathrm{JT}}<0$.

For an ideal gas, $V=nRT/p$ and $\alpha=1/T$, so

\[\boxed{\mu_{\mathrm{JT}}=0}.\]

For one mole of van der Waals gas, implicit differentiation at constant pressure gives

\[\left(\frac{\partial V_m}{\partial T}\right)_p =\frac{R/(V_m-b)} {RT/(V_m-b)^2-2a/V_m^3}.\]

Hence the exact coefficient is

\[\boxed{ \mu_{\mathrm{JT}}= \frac1{C_{P,m}} \left[ \frac{TR/(V_m-b)}{RT/(V_m-b)^2-2a/V_m^3}-V_m \right]}.\]

In the dilute, low-pressure limit $V_m\gg b$,

\[\boxed{\mu_{\mathrm{JT}}\simeq \frac1{C_{P,m}}\left(\frac{2a}{RT}-b\right)}.\]

This approximation predicts an inversion temperature $T_i\simeq2a/(Rb)=2T_B$: below it attractions dominate and throttling cools; above it excluded volume dominates and throttling warms.

The linked Unit III Maxima worksheet verifies the van der Waals $C_P-C_V$ and exact Joule-Kelvin expressions. Every displayed symbolic residual is zero.