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

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

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 ?
• Do you not understand how adiabatic flow through a porous plug or valve, from high pressure upstream to low pressure downstream, gives no change for the enthalpy change per mole? Are you aware of the equation: $$dH=C_pdT+\left[V-T\left(\frac{\partial V}{\partial T}\right)_P\right]dP$$ – Chet Miller Mar 30 '17 at 3:13
• No I only know $\displaystyle dH = -\mu C_pdp + CpdT$. – A---B Mar 30 '17 at 10:58
• If you combine these two equations, you get a relationship for calculating $\mu$ from knowledge of the heat capacity and the equation of state. – Chet Miller Mar 30 '17 at 11:16
• Ok I subtract my equation from our equation, $$0 = \left[V-T\left(\frac{\partial V}{\partial T}\right)_P +\mu C_p\right]dP$$ How can I get rid of $C_p$ ? – A---B Mar 30 '17 at 11:22
• You can't. But, in any event, you did the algebra incorrectly. What happened to the dT. – Chet Miller Mar 30 '17 at 11:28

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 $$\Delta H = H_2-H_1=U_2 +p_2V_2-U_1 -p_2V_2 =0$$

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.

• Most gases cool when passing from high to low pressures at room temperature? Not according to Wikipedia: en.wikipedia.org/wiki/Joule%E2%80%93Thomson_effect – Chet Miller Mar 30 '17 at 11:32
• perhaps 'most' was loose, 'many' would be better, or best only those with positive coefficients at the prevailing temperature will cool, i.e. $\alpha T>1$ – porphyrin Mar 30 '17 at 11:56