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A gas that follows $P(V - nb)= nRT$ is subjected to Joule-Thomson expansion. Tell whether it cools or heats up.


$$\mu = {\partial T \over \partial P} = {\partial P(V - nb)/nR \over \partial P} = \frac1{nR}\left(V - nb + {\partial V \over \partial P}\right) = \frac1{nR}\left(V - nb - {nRT\over P^2}\right)$$

Now how do I determine whether $\mu >0 $ or $\mu < 0$ without knowing anything about temperature or anything else ?

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  • $\begingroup$ The equation you gave for $\mu$ is incorrect. That partial derivative is supposed to be at constant enthalpy H. Do you know the mathematical relationship between dH, dT, and dP? $\endgroup$ – Chet Miller Apr 2 '17 at 1:18
  • $\begingroup$ Yes if you mean $dH = -\mu C_p dP + C_p dT$. $\endgroup$ – A---B Apr 2 '17 at 1:20
  • $\begingroup$ I set $dH = 0$ then I get $\mu dP = dT$. $\endgroup$ – A---B Apr 2 '17 at 1:23
  • $\begingroup$ Not that part. The next part. Do you really think that that expression you wrote is equal to the partial derivative of T with respect to P at constant H? $\endgroup$ – Chet Miller Apr 2 '17 at 1:36
  • $\begingroup$ There are some details in this answer chemistry.stackexchange.com/questions/71543/… . You should find that the coefficient for your gas is $-B/C_p$. $\endgroup$ – porphyrin Apr 2 '17 at 9:03
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The equation for dH is: $$dH=C_pdT+\left[V-T\left(\frac{\partial V}{\partial T}\right)_P\right]dP$$

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