The definition of the Joule-Thomson effect is:

$$\mu=\left(\frac{\partial T}{\partial P}\right)_H$$

And this is defined in an isenthalpic process, i.e. $dH=0$.

My book shows the derivation of the isothermal Joule-Thomson coefficient ($\varphi$) using the cyclic rule:

$$\left(\frac{\partial T}{\partial P}\right)_H\times\left(\frac{\partial P}{\partial H}\right)_T\times\left(\frac{\partial H}{\partial T}\right)_P=-1$$

Using this cyclic rule,

$$\mu=\left(\frac{\partial T}{\partial P}\right)_H = -\frac{\left(\frac{\partial H}{\partial P}\right)_T}{\left(\frac{\partial H}{\partial T}\right)_P} = -\frac{\left(\frac{\partial H}{\partial P}\right)_T}{C_P}$$

And therefore

$$\left(\frac{\partial H}{\partial P}\right)_T = -\mu C_P = \varphi$$

I understand the mathematics of this derivation, but I don't understand it on a conceptual level.

$\mu$ was derived in an isenthalpic process, with $dH=0$. It owes it existence to the fact that there is no enthalpy change. So how can we now use this variable when $dH\neq 0$?

This question may be a more fundamental misunderstanding on my behalf of how thermodynamics equations are derived. Often I see an equation derived under the assumption that some variable is held constant, but then the equation applied when that variable is not constant any more. The above example is just one where I think this discrepancy is obvious.

  • $\begingroup$ "Often I see an equation derived under the assumption that some variable is held constant, but then the equation applied when that variable is not constant any more." If you give some examples..... $\endgroup$
    – Zenix
    Apr 8, 2020 at 12:47
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    $\begingroup$ @Zenix let's focus on the current question, and then maybe more examples will not be necessary. $\endgroup$
    – Gimelist
    Apr 8, 2020 at 12:48
  • $\begingroup$ Your last equation is correct but for a different experiment and that is one carried out at constant temperature in a calorimeter. The derivative is the heat absorbed per unit difference in pressure at constant T. This change in conditions is generally the case when moving between different forms of these equations. $\endgroup$
    – porphyrin
    Apr 8, 2020 at 15:33
  • $\begingroup$ Are you familiar with the equation $$dH=C_pdT+\left[V+T\left(\frac{\partial V}{\partial T}\right)_P\right]dP$$ $\endgroup$ Apr 8, 2020 at 19:48

1 Answer 1


When you write a total differential such as $$dH = \left(\frac{\partial H}{\partial T}\right)_p dT + \left(\frac{\partial H}{\partial p}\right)_T dp \tag{1}$$ you are applying the methods of differential geometry, so in reality the answer to your question lies in the applicability of these methods in thermodynamics (the rest being "math"), something which statements such as "so-and-so is a state function" implicitly justify. One can write a total or exact differential of a state function, as for the enthalpy in the equation above. This equation can be interpreted as follows: small (differential) changes in p and T, which are orthogonal dimensions (in the sense that they can be varied independently), additively cause a linearly proportional differential change in the function H. In the differential limit, the surface of H looks like a plane. The partial derivatives describe the slope of the plane in the orthogonal dimensions.

The cyclic rule can be derived from the above equation by taking the partial derivative wrt one of the independent variables while holding H constant.

$$ 0 = \left(\frac{\partial H}{\partial T}\right)_p \left(\frac{\partial T}{\partial p}\right)_H + \left(\frac{\partial H}{\partial p}\right)_T \tag{2}$$

What does it mean here to hold H constant? It means we are looking for an isenthalpic path on the enthalpy surface, from the initial point at which we computed the partial differentials of the surface wrt T and p, in direction $\left(\left(\frac{\partial T}{\partial p}\right)_H dp, dp\right)$, where the partial differential $\left(\frac{\partial T}{\partial p}\right)_H$ also happens to be given (thanks to the geometry of the problem) by

$$ \left(\frac{\partial T}{\partial p}\right)_H = -\frac{\left(\frac{\partial H}{\partial p}\right)_T }{\left(\frac{\partial H}{\partial T}\right)_p} \tag{3}$$

Alternately, consider the line resulting from intersection of a horizontal isenthalpic plane $c(T,p)=H_0=H(T_0,p_0)$ and the plane $s(T,p)$ tangential to the surface $H$ at the point $(T_0,p_0,H_0)$, the tangential plane given by $$s(T,p) = H_0 + C_p \Delta T + \varphi \Delta p$$


$$C_p=\left[\left(\frac{\partial H}{\partial T}\right)_p\right]_{(T_0,p_0)}$$ $$\varphi=\left[\left(\frac{\partial H}{\partial p}\right)_T\right]_{(T_0,p_0)}$$

are the partial derivatives of $H$ evaluated at $(T_0,p_0)$, and $\Delta T = T-T_0,~ \Delta p = p-p_0$. Solving for the intersection line by setting $s(T,p)=c(T,p)$ gives

$$T = -\frac{\varphi}{C_p} p + T_0 + \frac{\varphi}{C_p} p_0 +\frac{c-H_0}{C_p}$$

The slope of the intersection line can be recognized to be the same as $\left(\frac{\partial T}{\partial p}\right)_H$ given by Eq. (3). The geometric nature of the problem and relation between the different derivatives should then be clear.

So how can we now use this variable when dH≠0?

A Joule-Thompson coefficient $\mu$, like any other thermodynamic (or state) properties, is strictly valid at the conditions under which it is determined (it may have a broader useful range depending on how much it varies with T and p and the tolerated error). This is different from the question of the mathematical accuracy of the relationships used to derive the properties. $\mu$ is derived at a specific state defined for a pure substance by a specific point (T,p) and as such is a fixed property of the substance at that point.

  • $\begingroup$ Ok, so the fact that $\mu=\left(\frac{\partial T}{\partial P}\right)_H$ was found at constant enthalpy does not require for it to be valid only at constant enthalpy. We used only constant enthalpy to simply the equation so we can practically find what $\left(\frac{\partial T}{\partial P}\right)_H$ is. A simpler analogy would be finding the intercept in something like $y=2x+c$. You put $x=0$ and $y=3$ (e.g., some experimentally measured value, just like $\mu$ is) to find $c=3$. But then this is correct for all $x$, not just $x=3$. Is this the general idea? $\endgroup$
    – Gimelist
    Apr 8, 2020 at 23:48
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    $\begingroup$ Yes, a geometric interpretation in terms of intersecting planes is possible. I added this to the answer. $\endgroup$
    – Buck Thorn
    Apr 9, 2020 at 8:16
  • $\begingroup$ Thanks, I now understand. Regarding your last paragraph on the validity, so it is correct to say that it's strictly valid at the conditions for which it was derived (eg $dH=0$), and an approximation for all other conditions, yet often the approximation is good enough? For example, in the derivation of the $PVT$ relations in adiabatic expansion, we often use $C_p$ and $C_v$. Yet, $dP \neq 0$ and $dV \neq 0$. So how does this work out? Or should this be a new question? $\endgroup$
    – Gimelist
    Apr 9, 2020 at 10:42
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    $\begingroup$ The thing is that $dH$ is not a property at a particular point, H is (or rather H relative to its value at some reference state). dH is just a small change in H at that point, and is for all purposes so small as to not affect the properties (state) of the system. $\endgroup$
    – Buck Thorn
    Apr 9, 2020 at 10:51
  • $\begingroup$ Ok, so even if $dH \neq 0$, because it is a differential, $H$ itself is infinitesimally close to whatever new $H$ we have due to $dH \neq 0$, so for all practical purposes it doesn't matter. This can be generalised to my other question, for example with adiabatic expansion. Even though $C_p$ was derived with $dP = 0$, the "new" $P$ when $dP \neq 0$ is close enough to the "old" $P$, therefore it works. $\endgroup$
    – Gimelist
    Apr 9, 2020 at 10:56

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