Enthalpy derivation differential equation

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The discussion revolves around the challenges faced in deriving the enthalpy differential equation. The user attempted to apply the chain rule and Maxwell relations but found them ineffective. It is suggested that using the gas law would be a more straightforward approach to solve the problem. The conversation emphasizes the importance of simplifying the method rather than relying on complex relations. Overall, the focus is on finding a clearer path to derive enthalpy without unnecessary complications.
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Homework Statement
A state equation for a certain gas is ##(P + b)v = RT## and its internal energy is ##u = aT +bv +u_o##

show that $$(\frac {\partial H} {\partial v})_P = \frac {C_p T} {v}$$
Relevant Equations
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Good evening,
unfortunately I'm pretty lost in this problem.

I tried to use the chain rule $$(\frac {\partial H} {\partial v})_P = (\frac {\partial H} {\partial T})_P (\frac {\partial T} {\partial v})_P$$ and using some Maxwell relations but it doesn't work very well.
I know that $$T = (\frac {\partial H} {\partial S})_P$$ but I don't know how I would get to the answer.

I would be very happy if somone helped me.

Thanks.
 
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All you need to do is to use the gas law. The Maxwell relation is unnecessary for this problem.
 
I want to find the solution to the integral ##\theta = \int_0^{\theta}\frac{du}{\sqrt{(c-u^2 +2u^3)}}## I can see that ##\frac{d^2u}{d\theta^2} = A +Bu+Cu^2## is a Weierstrass elliptic function, which can be generated from ##\Large(\normalsize\frac{du}{d\theta}\Large)\normalsize^2 = c-u^2 +2u^3## (A = 0, B=-1, C=3) So does this make my integral an elliptic integral? I haven't been able to find a table of integrals anywhere which contains an integral of this form so I'm a bit stuck. TerryW

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