# For what temperature is the Joule-Thomson coefficient for a gas zero?

Sep 9, 2017

For what kind of gas?

The Joule-Thomson experiment essentially pushes a gas through a porous plug at constant enthalpy (insulated so that no heat is exchanged with the environment). This is done in such a way that all of the expansion work goes into changing the internal energy:

Any gas is then described by the Joule-Thomson coefficient

${\mu}_{J T} = {\left(\frac{\partial T}{\partial P}\right)}_{H}$,

that is, it is the change in temperature brought about by a change in pressure at constant enthalpy.

(This may be weird... but if you can perform the expansion slowly enough, the pressure can re-equilibrate and constant enthalpy is then possible with perfect insulation.)

A Joule-Thomson coefficient for any gas is given by:

${\mu}_{J T} = {\left(\frac{\partial T}{\partial P}\right)}_{H} = \frac{V}{C} _ P \left(\alpha T - 1\right)$

An ideal gas always has a Joule-Thomson coefficient ${\mu}_{J T}$ of zero. Note that for ideal gases:

$\alpha = \frac{1}{V} {\left(\frac{\partial V}{\partial T}\right)}_{P} = \frac{n R}{P V} = \frac{1}{T}$

And thus, $\alpha T = 1$ and ${\mu}_{J T} = 0$ for an ideal gas, regardless of temperature (as long as $T \ne \text{0 K}$).

For real gases, when ${\mu}_{J T}$ is zero, it is known to be at the inversion temperature ${T}_{\text{inv}}$:

$0 = \frac{V}{C} _ P \left(\alpha {T}_{\text{inv}} - 1\right)$

Since the volume of any gas is never zero and since assuming we don't need to reach $T = \text{0 K}$ for the sign change to occur, the heat capacity does not reach zero:

$\textcolor{b l u e}{{T}_{\text{inv}} = \frac{1}{\alpha}}$

DISCLAIMER: LONG DERIVATION BELOW!

To derive this, begin at the cyclic rule of partial derivatives:

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

By definition, the constant-pressure heat capacity is:

${C}_{P} = {\left(\frac{\partial H}{\partial T}\right)}_{P}$, $\text{ "" } \boldsymbol{\left(2\right)}$

which is an experimentally-determined quantity. Next, note that, as with all derivatives:

${\left(\frac{\partial P}{\partial H}\right)}_{T} = \frac{1}{\frac{\partial H}{\partial P}} _ T$ $\text{ "" } \boldsymbol{\left(3\right)}$

The Gibbs' free energy is a function of $T$ and $P$, so consider the ubiquitous relation in a constant-temperature system:

$\mathrm{dG} = \mathrm{dH} - T \mathrm{dS}$ $\text{ "" } \boldsymbol{\left(4\right)}$

By taking the partial derivative with respect to $P$ at constant $T$ of $\left(4\right)$, we get:

${\left(\frac{\partial G}{\partial P}\right)}_{T} = {\left(\frac{\partial H}{\partial P}\right)}_{T} - T {\left(\frac{\partial S}{\partial P}\right)}_{T}$ $\text{ "" } \boldsymbol{\left(5\right)}$

For a similar reason, we then consider the Gibbs' free energy Maxwell Relation:

$\mathrm{dG} = - S \mathrm{dT} + V \mathrm{dP}$ $\text{ "" } \boldsymbol{\left(6\right)}$

Since for state functions, cross-derivatives are equal, we have:

${\left(\frac{\partial S}{\partial P}\right)}_{T} = - {\left(\frac{\partial V}{\partial T}\right)}_{P}$ $\text{ "" } \boldsymbol{\left(7\right)}$

By definition, the coefficient of thermal expansion is:

$\alpha = \frac{1}{V} {\left(\frac{\partial V}{\partial T}\right)}_{P}$ $\text{ "" } \boldsymbol{\left(8\right)}$

which is experimentally-determined. So, using $\left(7\right)$ and $\left(8\right)$,

${\left(\frac{\partial S}{\partial P}\right)}_{T} = - V \alpha$. $\text{ "" } \boldsymbol{\left(9\right)}$

Next, from the Maxwell Relation $\left(6\right)$,

${\left(\frac{\partial G}{\partial P}\right)}_{T} = V$. $\text{ "" } \boldsymbol{\left(10\right)}$

As a result, we so far have, by combining $\left(9\right)$ and $\left(10\right)$ into $\left(5\right)$:

$V = {\left(\frac{\partial H}{\partial P}\right)}_{T} + T \cdot V \alpha$

Therefore:

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

Lastly, plug this back into the cyclic rule $\left(1\right)$ to find:

${\left(\frac{\partial T}{\partial P}\right)}_{H} {C}_{P} / \left(- V \left(\alpha T - 1\right)\right) = - 1$

And so, the Joule-Thomson coefficient is found as:

$\textcolor{b l u e}{\overline{\underline{|}} \stackrel{\text{ ")(" "mu_(JT) = ((delT)/(delP))_H = V/C_P(alphaT - 1)" }}{|}}$