The Joule-Thomson Process | Yucheng (Luca) Jin

This question describes the Joule-Thomson process (also known as the Joule-Kelvin process). The figure shows a thermally insulated pipe which has a porous barrier separating two halves of the pipe. A gas of volume $V_1$, initially on the left-hand side of the pipe, is forced by a piston to go through the porous barrier using a constant pressure $P_1$. Assume the process can be treated quasistatically. As a result the gas flows to the right-hand side, resisted by another piston which applies a constant pressure $P_2$ ($P_2 < P_1$). Eventually all of the gas occupies a volume $V_2$ on the right-hand side.

Figure 1. Joule-Thomson process: gas is pushed through a porous barrier from pressure $P_1$ to $P_2 < P_1$.

Show that enthalpy, $H = E + PV$, is conserved.
Find the Joule-Thomson coefficient $\mu_{\mathrm{JT}} \equiv \left(\dfrac{\partial T}{\partial P}\right)_{\!H}$ in terms of $T$, $V$, the heat capacity at constant pressure $C_P$, and the volume coefficient of expansion $\alpha \equiv \dfrac{1}{V}\!\left(\dfrac{\partial V}{\partial T}\right)_{\!P}$. (Hint: You will need to use a Maxwell relation.)
What is $\mu_{\mathrm{JT}}$ for an ideal gas?
If we wish to use the Joule-Thomson process to cool a real (non-ideal) gas, what must the sign of $\mu_{\mathrm{JT}}$ be?
Derive $\mu_{\mathrm{JT}}$ for a gas obeying the van der Waals equation of state to leading order in the density $N/V$. For what values of temperature $T$ can the gas be cooled?

Label all quantities on the left by 1 and quantities on the right by 2. We have \[dH = dH_1+ dH_2 = dE_1 + P_1dV_1 + V_1dP_1 + dE_2 + P_2dV_2 + V_2dP_2.\tag{1.1}\] Since the process is adiabatic, we have $dE_i = -\delta W_i = -P_idV_i$. Meanwhile, the pressure is constant; thus, all quantities on the left vanish. Hence, we have $dH=0$.

Now, consider the temperature $T$ as a function of $H$ and $P$. The total derivative of $T$ is given by \[dT = \left(\frac{\partial T}{\partial H}\right)_P dH + \left(\frac{\partial T}{\partial P}\right)_H dP = \left(\frac{\partial T}{\partial P}\right)_H dP \equiv \eta_{\text{JT}}dP.\tag{1.2}\] Using Maxwell relations, we have \[\eta_{\text{JT}} = -\left(\frac{\partial T}{\partial H}\right)_P\left(\frac{\partial H}{\partial P}\right)_T = -\frac{1}{C_P}\left[T\left(\frac{\partial S}{\partial P}\right)_T + V\right] = \frac{1}{C_P}\left[T\left(\frac{\partial V}{\partial T}\right)_P - V\right].\tag{1.3}\]

For ideal gas, \[\left(\frac{\partial V}{\partial T}\right)_P = V \implies \eta_{\text{JT}}=0.\tag{1.4}\]

Since $dP<0$, for a cooling process, $dT<0$, we require $\eta_{\text{JT}}>0$.

The van der Waals equation of state is \[\left(P + \frac{aN^2}{V^2}\right)(V-Nb) = NkT.\tag{1.5}\] Expanding and dropping the $ab$ term as second order in the density $N/V$, \[PV - PNb + \frac{aN^2}{V} = NkT.\tag{1.6}\] Differentiating implicitly at constant $P$, \[P\left(\frac{\partial V}{\partial T}\right)_P - \frac{aN^2}{V^2}\left(\frac{\partial V}{\partial T}\right)_P = Nk,\tag{1.7}\] and so \[\left(\frac{\partial V}{\partial T}\right)_P = \frac{Nk}{P - aN^2/V^2} \approx \frac{Nk}{P}\left(1 + \frac{aN^2}{PV^2}\right).\tag{1.8}\] To leading order in $N/V$ we may replace $V \approx NkT/P$ in the correction term only, giving \[\left(\frac{\partial V}{\partial T}\right)_P \approx \frac{Nk}{P} + \frac{aN^2}{kT^2 V}.\tag{1.9}\] Substituting into the expression for $\eta_{\mathrm{JT}}$ and using $V \approx NkT/P + Nb - aN/kT$ to the same order, \[T\left(\frac{\partial V}{\partial T}\right)_P - V = \frac{NkT}{P} + \frac{aN}{kT} - \frac{NkT}{P} - Nb + \frac{aN}{kT} = \frac{2aN}{kT} - Nb.\tag{1.10}\] Hence, \[\eta_{\mathrm{JT}} = \frac{N}{C_P}\left(\frac{2a}{kT} - b\right). \tag{1.11}\] The gas can be cooled ($\eta_{\mathrm{JT}}>0$) only below the inversion temperature, \[T < T_{\mathrm{inv}} \equiv \frac{2a}{kb}.\tag{1.12}\]