2
$\begingroup$

For the equation below, of Van der Waal form: $\left(P+\frac{n^{2}a}{V^{2}}\right)(V-nb)=nRT$

Determine the partial derivatives; $\Bigl(\frac{\partial V}{\partial T}\Bigr)_{P,n} \text{and } \Bigl(\frac{\partial P}{\partial V}\Bigr)_{T,n}$

Where $a,b, n, R$ are constants.

This is what I've done so far for one of the partial derivatives. I dont know what else to do from here, sorry. $P=\frac{nRT}{V-nb}-\frac{n^{2}a}{V^{2}}$

$\frac{\partial P}{\partial V}=-\frac{nRT}{(V-nb)^2}+\frac{2n^{2}a}{V^{3}}$

  • 0
    No, I think you are taking that derivative wrong. I did it by hand and you should end up with what's on your question (the last expression)2012-10-07

1 Answers 1

1

You can try to find $\partial P / \partial T$ and $\partial P / \partial V$ such that: $\frac{\frac{\partial P}{\partial T}}{\frac{\partial P}{\partial V}}=\frac{\partial V}{\partial T}$

I hope this will help you

  • 0
    You can see $P(V)$ as $\alpha\cdot(V-\beta)^{-1}-\gamma V^{-2}$ where $\alpha=nRT,\beta=nb,\gamma=n^2 a$ are constants so taking the derivative: $\frac{\partial P}{\partial V}=\frac{\partial}{\partial V}[\alpha\cdot(V-\beta)^{-1}]-\frac{\partial}{\partial V}[\gamma V^{-2}]$ $\frac{\partial P}{\partial V}=(-1)\alpha\cdot(V-\beta)^{-2}-(-2)\gamma V^{-3}=-\frac{\alpha}{(V-\beta)^2} + \frac{2\gamma}{V^3} =-\frac{nRT}{(V-nb)^2} + \frac{2n^2 a}{V^3}$ You can do the same for $P(T)$2012-10-07