# Joule-Thomson Expression for dT

In H. Callen's "Thermodynamics and an Introduction to Thermostatistics" in the chapter about the Joule-Thomson process, they say:

"If the change in pressure in a throttling process is sufficiently small we can employ the usual differential analysis" $$dT = \left(\frac{\partial T}{\partial P}\right)\bigg|_{H, N_1, N_2, ...}dP$$

How does one get to this equation for $$dT$$?

I can't find any references to this (while skimming through other chapters) and I'm not entirely sure what this relationship is called. As far as I know, you can write $$T(S,P,N)$$ as an equation of state using the enthalpy equation, but writing $$dT$$ gives

$$dT = \frac{\partial T}{\partial S}\bigg|_{P,N}dS + \frac{\partial T}{\partial P}\bigg|_{S,N}dP + \frac{\partial T}{\partial N}\bigg|_{S,P}dN$$

I can understand if $$dN = 0$$, but is it possible to also have $$dS = 0$$ for this process?

One can show $$dH = 0$$, and since it appears to be held constant, I figure this is the right track, but this is not in the equation above.

Any insight is appreciated.

• eliminate $S$ and replace it with $H$, then $T=T(H,P,N)$ – hyportnex Nov 10 '18 at 20:53
• Thanks for the comment, just not sure how one would go about doing that. $H \ne S$ right? Can you just replace it, or is it an application of the chain rule? – Alexander Julian Ty Nov 10 '18 at 21:42
• since $dH=TdS+Vdp+\mu dN$ you can immediately turn this around for $dS$ that can be integrated (solved) for $S=S(H,V,N)$, and if you already have a function $T=T_1(S,V,N)$ then you can have an other function $T=T_2(H,V,N)$ – hyportnex Nov 10 '18 at 21:57

## 1 Answer

The equation to start with is $$dH=C_pdT+\left[V-T\left(\frac{\partial V}{\partial T}\right)_P\right]dP$$which is derived directly from $$dH=TdS+VdP$$