Calculate heat capacity at constant pressure $C_p$ for the following equation of state $$P(v-b)=RT$$ Where $b$ is a constant. Now I did calculate it, but I think this is wrong!

$$C_p=T \left( \frac{\partial s}{\partial T} \right)_p=T \left( \frac{\partial s}{\partial v} \right)_p \left( \frac{\partial v}{\partial T} \right)_p=T \left( \frac{\partial P}{\partial T} \right)_v \left( \frac{\partial v}{\partial T} \right)_p=T \frac{R}{v-b}\frac{R}{P}=R$$

I searched a little bit and I saw in a book that for the same equation of state a similar approach gives $C_p-C_v=R$ and if my reasoning is true it gives $C_v=0$. Where did I go wrong?

First thing that comes to my mind is that I wrote $$\left( \frac{\partial s}{\partial T} \right)_p= \left( \frac{\partial s}{\partial v} \right)_p \left( \frac{\partial v}{\partial T} \right)_p$$ I used this method of adding a new variable in derivation and It always worked.

  • $\begingroup$ $C_{p}-C_{v}=R$ is for an ideal gas. Your equation of state is not for an ideal gas. $\endgroup$ – Bob D Jul 9 at 9:57
  • $\begingroup$ See example 9.9 in books.google.com/… $\endgroup$ – Ghartal Jul 9 at 10:00
  • $\begingroup$ OK, example 9.9 only proves that if $v$ is changed by a constant in the ideal gas equation, you still have $C_{p}-C_{v}=R$ which means the equation still applies to an ideal gas because $C_{p}-C_{v}=R$ ONLY applies to an ideal gas. So clearly you did go wrong to conclude that $C_{p}=R$. I'll see if I can figure out why starting with the basic definition of $C_{p}$ in terms of enthalpy. $\endgroup$ – Bob D Jul 9 at 12:10
  • $\begingroup$ Thanks for your help. I'll wait for your response :) $\endgroup$ – Ghartal Jul 9 at 15:19
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    $\begingroup$ The Maxwell relationship for $(dS/dV)_p$ should be $(dp/dT)_S$. $\endgroup$ – Jeffrey J Weimer Jul 9 at 15:55

The starting point for this analysis should be $$dH=C_pdT+\left[v-T\left(\frac{\partial v}{\partial T}\right)_P\right]dP$$What does that give you for the term in brackets for your equation of state? Is the result for the term in brackets a function of T?

  • $\begingroup$ It gives $dh=C_p \ dT+ b \ dP$. How to calculate $C_p$ from this? $\endgroup$ – Ghartal Jul 9 at 16:19
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    $\begingroup$ It means that Cp is a function only of T and not P. So Cp is the same as the ideal gas heat capacity. $\endgroup$ – Chet Miller Jul 9 at 22:40
  • $\begingroup$ I derived it, its absolutely right. It gives $$\left( \frac{\partial C_p}{\partial P} \right)_T=0$$ But It still does not explain how to get the value of $C_p$. I wonder why my calculation is wrong, which gives $C_p=R$. $\endgroup$ – Ghartal Jul 10 at 19:02
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    $\begingroup$ It can’t be derived from the PvT equation of state. For that one would need the equation for, say, G vs T and P. Otherwise, we would need to measure the Cp in the ideal gas limit or know the number of degrees of freedom. Once we know the ideal gas Cp and the PvT equation of state, we can get Cp at any other higher pressure. $\endgroup$ – Chet Miller Jul 10 at 19:32

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