# 10.3: The Joule-Thomson Experiment


The experiment is also known as the Joule-Kelvin experiment. William Thomson was created Lord Kelvin. The experiment is also known as the porous plug experiment.

In the Joule-Thomson experiment a constant flow of gas was maintained along a tube which was divided into two compartments separated by a porous plug, such that the pressure and molar volume on the upstream side were P1, V1, and the pressure and molar volume on the downstream side were P2, V2. Under such circumstances the net work done on a mole of gas in passing from one compartment to the other is P1V1P2V2. (Imagine, for example, that a piston pushes a mole of gas towards the plug from the upstream side, through a distance x1 ; if A is the crosssectional area of the tube, the work done on the gas is P1Ax1 = P1V1. Imagine also that the gas on the downstream side pushes a piston away from the plug through a distance x2. The work done by the gas is P2Ax2 = P2V2. Therefore the net external work done on the gas is P1V1P2V2.) If no heat is supplied to or lost from the system, the increase in internal energy of this gas is just equal to this work done on it:

$$U_{2}-U_{1}=P_{1} V_{1}-P_{2} V_{2},$$

or

$U_{1}+P_{1} V_{1}=U_{2}+P_{2} V_{2}.$

That is, there is no change in enthalpy. Therefore, we want to find $$\left(\frac{\partial T}{\partial P}\right)_{H}$$, which is the Joule-Thomson coefficient, for which I shall be using the symbol µ.

In the experiment we are discussing, we are interested in how temperature varies with pressure in an experiment in which the enthalpy is constant. We shall therefore choose H as our state function and P and T as our independent state variables. That is we shall write H = H(P,T), so that

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

The second of these partial derivatives is CP, and therefore

$\left(\frac{\partial T}{\partial P}\right)_{H}=-\frac{1}{C_{P}}\left(\frac{\partial H}{\partial P}\right)_{T}.$

Now

$d H=T d S+V d P.$

That is,

$d S=\frac{1}{T}[d H-V d P]=\frac{1}{T}\left[\left(\frac{\partial H}{\partial P}\right)_{T} d P+\left(\frac{\partial H}{\partial T}\right)_{P} d T-V d P\right].$

$d S=\frac{1}{T}\left[\left(\frac{\partial H}{\partial P}\right)_{T}-V\right] d P+\frac{1}{T}\left(\frac{\partial H}{\partial T}\right)_{P} d T.$

But we also have

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

Therefore

$\left(\frac{\partial S}{\partial P}\right)_{T}=\frac{1}{T}\left[\left(\frac{\partial H}{\partial P}\right)_{T}-V\right]$

and

$\left(\frac{\partial S}{\partial T}\right)_{P}=\frac{1}{T}\left(\frac{\partial H}{\partial T}\right)_{p}.$

The mixed second derivatives are

$\frac{\partial^{2} S}{\partial T \partial P}=-\frac{1}{T^{2}}\left[\left(\frac{\partial H}{\partial P}\right)_{T}-V\right]+\frac{1}{T}\left[\frac{\partial^{2} H}{\partial T \partial P}-\left(\frac{\partial V}{\partial T}\right)_{P}\right]$

and

$\frac{\partial^{2} S}{\partial P \partial T}=\frac{1}{T} \frac{\partial^{2} H}{\partial P \partial T}.$

But entropy is a function of state and dS is an exact differential, so the mixed second derivatives are equal. Whence, after simplification:

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

Hence, returning to equation 10.3.3, we obtain, for the Joule-Thomson coefficient,

$\mu=\left(\frac{\partial T}{\partial P}\right)_{H}=\frac{1}{C_{p}}\left[T\left(\frac{\partial V}{\partial T}\right)_{P}-V\right].$

Trivial Exercise: Show that, for an ideal gas, the Joule-Thomson coefficient is zero, and also that, for an ideal gas,

$\left(\frac{\partial H}{\partial P}\right)_{T}=0.$

This is analogous to equation 8.1.4 for an ideal gas, namely $$\left(\frac{\partial U}{\partial V}\right)_{T}=0$$.

Exercise. Show that, for a van der Waals gas, the Joule-Thomson coefficient is

$\left(\frac{\partial T}{\partial P}\right)_{H}=-\frac{V}{C_{p}} \cdot \frac{\left(R T V^{2} b-2 a(V-b)^{2}\right)}{R T V^{3}-2 a(V-b)^{2}}.$

(Verify the dimensions of this expression.) Hint: It is difficult to calculate (∂V/∂T)P directly, because it is difficult to express V explicitly as a function of P and T. It is not actually impossible to do it algebraically, because van der Waals' equation is a cubic equation in V, and a cubic equation does have an algebraic solution. It is easier, however, to calculate (∂V/∂T)P from $$\left(\frac{\partial V}{\partial T}\right)_{P}=-\left(\frac{\partial P}{\partial T}\right)_{V} /\left(\frac{\partial P}{\partial V}\right)_{T}$$, or from $$\left(\frac{\partial V}{\partial T}\right)_{P}=1 /\left(\frac{\partial T}{\partial V}\right)_{P}$$.

Note also that the Joule-Thomson coefficient may be negative or positive; i.e., it may result in cooling or heating. It will result in heating if you start above a certain temperature called the inversion temperature, and cooling if you start below the inversion temperature. The Joule-Thomson effect is used in the Linde method for cooling and ultimately liquefying gases. For most gases, the inversion temperature is higher than room temperature, so that cooling starts immediately. But for hydrogen, the inversion temperature is about −80 oC, and hydrogen must be cooled below this temperature before the Joule-Thomson effect can be used to cool it further and to liquefy it. You can see from equation 10.3.14 that the inversion temperature for a van der Waals gas is equal to $$\frac{2 a(V-b)^{2}}{R V^{2} b} \approx \frac{2 a}{R b}$$. Here V is the molar volume.

Summary:

Joule coefficient

$\eta=\left(\frac{\partial T}{\partial V}\right)_{U}=\frac{1}{C_{V}}\left[P-T\left(\frac{\partial P}{\partial T}\right)_{V}\right]$

Joule-Thomson coefficient

$\mu=\left(\frac{\partial T}{\partial P}\right)_{H}=\frac{1}{C_{P}}\left[T\left(\frac{\partial V}{\partial T}\right)_{P}-V\right].$

This page titled 10.3: The Joule-Thomson Experiment is shared under a CC BY-NC license and was authored, remixed, and/or curated by Jeremy Tatum.