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 ?
  • $\begingroup$ 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: $$dH=C_pdT+\left[V-T\left(\frac{\partial V}{\partial T}\right)_P\right]dP$$ $\endgroup$ – Chet Miller Mar 30 '17 at 3:13
  • $\begingroup$ No I only know $\displaystyle dH = -\mu C_pdp + CpdT$. $\endgroup$ – A---B Mar 30 '17 at 10:58
  • $\begingroup$ If you combine these two equations, you get a relationship for calculating $\mu$ from knowledge of the heat capacity and the equation of state. $\endgroup$ – Chet Miller Mar 30 '17 at 11:16
  • $\begingroup$ 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 $C_p$ ? $\endgroup$ – A---B Mar 30 '17 at 11:22
  • $\begingroup$ You can't. But, in any event, you did the algebra incorrectly. What happened to the dT. $\endgroup$ – 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 $$\Delta H = H_2-H_1=U_2 +p_2V_2-U_1 -p_1V_1 =0$$

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.


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.

| improve this answer | |
  • $\begingroup$ 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 $\endgroup$ – Chet Miller Mar 30 '17 at 11:32
  • 1
    $\begingroup$ perhaps 'most' was loose, 'many' would be better, or best only those with positive coefficients at the prevailing temperature will cool, i.e. $\alpha T>1$ $\endgroup$ – porphyrin Mar 30 '17 at 11:56
  • $\begingroup$ @porphyrin What is the $p$ in the partial derivative? $p_1$ or $p_2$? Why for ideal gas $w=0$? Thanks in advance. $\endgroup$ – ado sar Feb 20 at 20:44
  • $\begingroup$ @adosar corrected typo in $\Delta H$ eqn. The last equations are general for any pressure. $\endgroup$ – porphyrin Feb 21 at 12:01
  • $\begingroup$ @porphyrin So we can measure either the change of $T_1$ with respect to $p_1$ or $T_2$ with respect to $p_2$? $\endgroup$ – ado sar Feb 21 at 13:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.