# Heat capacity $C_p$ for Clausius equation of state $P(v-b)=RT$

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.

• $C_{p}-C_{v}=R$ is for an ideal gas. Your equation of state is not for an ideal gas. Jul 9 '19 at 9:57
• See example 9.9 in books.google.com/… Jul 9 '19 at 10:00
• 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. Jul 9 '19 at 12:10
• Thanks for your help. I'll wait for your response :) Jul 9 '19 at 15:19
• The Maxwell relationship for $(dS/dV)_p$ should be $(dp/dT)_S$. Jul 9 '19 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?
• It gives $dh=C_p \ dT+ b \ dP$. How to calculate $C_p$ from this? Jul 9 '19 at 16:19
• 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 $C_p=R$. Jul 10 '19 at 19:02