I'm getting frustrated finding my mistake, because I get the exact same answer as my professor, except for a constant. The problem is the following:
For a system, we know that $c_P = c + \frac{2aP}{RT^2}$ and $Pv = RT + > P(b-\frac{a}{RT})$, and that it undergoes a Joule-Kelvin process where the temperature halves. Find the change in entropy.
In order to find the change in entropy, which can be calculated as an integral in terms of $c_P(T,P)$ and $\alpha(T,P)$, I need to know the change in pressure, which I could find out by establishing that $\Delta h = 0$. So, since I need to know $h(P,T)$ I can write its differential: $$dh = \Big{(}\frac{\partial h}{\partial T}\Big{)}_P dT + \Big{(}\frac{\partial h}{\partial P}\Big{)}_T dP$$ And, knowing that $dh = Tds + vdP$, it follows: $$dh = \Bigg{(} T\Big{(}\frac{\partial s}{\partial P}\Big{)}_T + v\Bigg{)}dP + T\Big{(}\frac{\partial s}{\partial T}\Big{)}_P dT$$
From Gibbs' potential we can find an adequate Maxwell relation: $$\Big{(}\frac{\partial s}{\partial P}\Big{)}_T = -\Big{(}\frac{\partial v}{\partial T}\Big{)}_P$$ And therefore: $$dh = \Bigg{(} -T\Big{(}\frac{\partial v}{\partial T}\Big{)}_P + v\Bigg{)}dP + T\Big{(}\frac{\partial s}{\partial T}\Big{)}_P dT$$ $$dh = (v-Tv\alpha)dP + c_P dT$$
I have proceeded directly to calculate the integral: $$h=\int dh = \int \Big{(} c + \frac{2aP}{RT^2} \Big{)} dT + \int\Big{(} \frac{RT}{P}+b-\frac{a}{RT} - T(\frac{R}{P} + \frac{a}{RT^2}) \Big{)}dP$$
And my result is: $$h = cT +(b-\frac{4a}{RT})P + C$$ While my professor's is: $$h = cT +(b-\frac{2a}{RT})P + C$$
Where did I go wrong?
