Alternative proof Joule-Thomson process is isenthalpic I found in general as proof for the Joule Thomson experiment to be isenthalpic:
$$
 \Delta U = U_2 - U_1= \Delta W = - \int_{V_1}^{0} p_1  dV - \int_{0}^{V_2} p_2 dV = p_1 V_1 - p_2 V_2 
$$
and then
$$
U_2 - U_1 = p_1 V_1 - p_2 V_2 
$$
$$
U_2 +  p_2 V_2 = U_1  + p_1 V_1
$$
$$
H_1 = H_2
$$
But why can't I use the total derivative and say,
$$
dH = dU + pdV + Vdp = dQ - pdV + pdV + V dp 
$$
where the experiment is by constuction isobaric ($dp=0$) and adiabatic ($dQ=0$) and hence:
$$
d H = -pdV + pdV = 0 
$$
I'm sure this is wrong, otherwise someone would have used it in a book, why though?
Best!
 A: *

*We can always write $dU=\delta Q+\delta W$ and $dU=T\,dS-P\,dV$, but we can write $dU=\delta Q-P\,dV$ only for a reversible process. The Joule–Thompson process isn't reversible, so in general $dU\neq \delta Q-P\,dV$. (Put another way, entropy increases—is generated—in a manner independent of heating, so $dS\neq\frac{\delta Q}{T}$; as you note, $\delta Q=0$.)


*The pressure isn't constant in the Joule–Thompson process, so $dP=0$ doesn't hold.
This is why certain terms don't drop out as easily as they are proposed to do in the derivation above.
A: The proper application of the first law of thermodynamics to steady Joule Thomson flow is $$\Delta h=Q-W_S$$where h is the enthalpy per unit mass of the flowing fluid, Q is the heat added per unit mass of flowing fluid, and $W_S$ is the shaft work done per unit mass of flowing fluid.  In Joule Thomson flow through a valve or porous plug, Q = 0 and $W_S=0$, so $\Delta h = 0$.
A: The fact that the initial and final enthalpy is the same (i.e., $_{1}=_2$) does not necessarily mean the process is isenthalpic. An isenthalpic process is one in which the enthalpy is constant throughout the process. For that to be the case, the process needs to be reversible. The Joule-Thompson process is irreversible.
Hope this helps.
