Enthalpy in adiabatic process


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> A gas adiabatically expanded from $\pu{32atm}$ and $\pu{273K}$ to $\pu{1atm}$ and $\pu{251K}$, Calculate Joule-Thomson coefficient $\mu$ at $\pu{273K}$.
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**Answer:**


$$\mu = {\Delta T \over \Delta p} = {-22 \over -31} = 0.7$$




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I think the answer uses the definition of $\mu$ that is $\displaystyle\left({\partial T \over \partial p}\right)\_H$, but this definition assumes the process to have constant enthalpy.


But in adiabatic process $dH = dU + Vdp + w\_{ad}$, so for $dH = 0$ we need to have $dU + Vdp + w\_{ad} = 0$ which is not possible since pressure is clearly changing and so $Vdp$ is not zero. 


My questions are 


1. Is $dH = 0$ for a adiabatic and I am missing something ?
2. If not then how did the author used $\displaystyle\left({\partial T \over \partial p}\right)\_H$ to get the answer ?




The Joule-Thomson experiments occurs with no change in enthalpy. 


Suppose that at the left of a porous plug there is a pressure $p\_1$ and temperature $T\_1$ and $p\_2,T\_2$ to the right of the plug, as $p\_1>p\_2$ the gas moves left to right. The experimental configuration must ensure that pressures remain constant and that the experiment is performed under adiabatic conditions when $q=0$. 


If a volume $V\_1$ of gas moves from the left to right the work done/mole is $W=p\_1V\_1-p\_2V\_2$. This is the difference between the work of compression on the left of the plug and work recovered on expansion on the right. If the gas were ideal then $w=0$, but real gases are not. The gas expansion is also adiabatic so that no heat leaves or enters then $q=0$ and the change in internal energy $\Delta U$ is equal to the net work $$\Delta U =U\_2-U\_1 = p\_1V\_1-p\_2V\_2$$ therefore $$U\_2+p\_2V\_2=U\_1 + p\_1V\_1$$
As $H=U+pV$, then $$\Delta H = H\_2-H\_1=U\_2 +p\_2V\_2-U\_1 -p\_1V\_1 =0$$


The Joule-Thompson coefficient $\mu$ is defined, as you write, $\left ( \partial T/\partial P \right)\_H$ and this measures how much the intermolecular interactions make the gas differ from an perfect gas. Most gases cool when passing from high to low pressures at room temperature.


Notes:


The coefficient can be rewritten in other forms using
$$ \left ( \frac{\partial T}{\partial p} \right)\_H \left ( \frac{\partial H}{\partial T} \right)\_p \left ( \frac{\partial p}{\partial H} \right)\_T =-1$$
then
$$ \mu C\_p= -\left (\frac{\partial H}{\partial p} \right)\_T $$


as
$$ \frac{dH}{dP} = C\_p\frac{dT}{dp} +V-T\left (\frac{\partial V}{\partial T} \right)\_p$$
then if $\alpha=(1/V)(\partial V/\partial T)\_p$ is the coefficient of expansion then at constant *T*, $(\partial H/\partial p)\_T= V(1-\alpha T)$ which means that the Joule-Thompson coefficient can be written as $\mu=(V/C\_p)(\alpha T-1)$ (This last equation has been used as a way of measuring absolute temperature because $V,\mu$ and $ \alpha $ are all measurable quantities.


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