Joule–Thomson coefficient is defined as

$$\mu_\mathrm{JT}=\left(\frac{\partial T}{\partial P}\right)_H. \tag{1}$$

Using the cyclic rule for partial derivatives it can be written as $$\mu_\mathrm{JT}=-\frac{\left(\frac{\partial H}{\partial p}\right)_T}{\left(\frac{\partial H}{\partial T}\right)_p}, \tag{2}$$

which is equivalent to

$$\mu_\mathrm{JT}=-\frac{\left(\frac{\partial H}{\partial p}\right)_T}{C_p}. \tag{3}$$

Can we analogously write

$$\mu_\mathrm{JT}=-\frac{\left(\frac{\partial U}{\partial p}\right)_T}{C_V}? \tag{4}$$


1 Answer 1


No. You are mixing four variables in the last equation when applying the Euler's chain rule. In your first equation we have \begin{equation} \mu_\mathrm{JT} = -\frac{\left(\dfrac{\partial H}{\partial p}\right)_T} {\left(\dfrac{\partial H}{\partial T}\right)_p} \tag{1} \end{equation} and we count in Eq. (1) three variables: $H$, $p$, and $T$.

However, in your fourth equation \begin{equation} \mu_\mathrm{JT} = -\frac{\left(\dfrac{\partial U}{\partial p}\right)_T} {\left(\dfrac{\partial U}{\partial T}\right)_V} \tag{2} \end{equation} and we count in Eq. (2) four variables: $U$, $p$, $T$, and $V$. As you want to write something similar to Eq. (1) with the internal energy $U$ in terms of $p$ and $T$, the equation should be \begin{equation} \mu_\mathrm{AS} \equiv \left(\frac{\partial T}{\partial p}\right)_U = - \frac{\left(\dfrac{\partial U}{\partial p}\right)_T} {\left(\dfrac{\partial U}{\partial T}\right)_p} \tag{3} \end{equation} where we defined the Aarush Saharan's coefficient in the honour of the person that was interested in this magnitude. It is the rate of change of the temperature when the pressure increases as the internal energy is held constant. Since you asked in the other answer what this can mean, to illustrate we study its behaviour in hydrogen for the van der Waals' equation of state.

Eqs. (2) and (3) need to be put in terms of the practical variables $p$, $T$, and $V$. Skipping the derivations it possible to prove that \begin{align} \mu_\mathrm{JT} &= -\frac{1}{C_p}\left[ V - T\left(\frac{\partial V}{\partial T}\right)_p\right] \tag{4} \\ \mu_\mathrm{AS} &= - \frac{-T\left(\dfrac{\partial V}{\partial T}\right)_p - p\left(\dfrac{\partial V}{\partial p}\right)_T} {C_p - p\left(\dfrac{\partial V}{\partial T}\right)_p} \tag{5} \end{align} Now we need to obtain all the partial derivatives for the vdW equation of state. This will unnecessary flood the post with big expressions for no purposes. It is enough for us to know that with some effort we can calculate them via the implicit function theorem because we do not have a manageable expression of $V$ in terms of $p$ and $T$.

The results for Eq. (4) are shown below: enter image description here The JT coefficient shows the typical behaviour. We added a discontinuous gray line to divide $\mu_\mathrm{JT} > 0$ and $\mu_\mathrm{JT} < 0$ for clarity. For low pressures, the coefficient is first negative (heating region), then it is positive (cooling region), and then it is negative again (heating region). At high pressures outside the inversion curve, the temperature always decreases as the pressure increases for a constant enthalpy evolution, which we illustrated in the red curve.

The results for Eq. (5) are shown below: enter image description here The AS coefficient shows a similar behaviour to the JT coefficient. However, it is always positive. Therefore, at least for the hydrogen gas at these ranges, when the pressure increases at constant internal energy, the temperature always increases.

The van der Waals's equation of state is $$ p = \frac{RT}{V - b} - \frac{a}{V^2} \tag{6} $$ and the parameters utilised for this post were: \begin{array}{|c|c|} a \; (\pu{atm dm^6 mol^-2}) & 0.2420 \\ b \; (\pu{dm^3 mol^-1}) & 0.0265 \\ C_p \; (\pu{J mol^-1 K-1}) & \approx 5R/2 \end{array}


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