# Finding Partial Derivatives

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. $$P=\frac{nRT}{V-nb}-\frac{n^{2}a}{V^{2}}$$

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

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Do you know how to take normal derivatives? Because a partial derivative is exactly a normal derivative except every variable is a constant except for the variable that you are taking the derivative to respect to. in the dP/dV case, V is the only variable that is not a constant...This is pretty much an issue of product rule (f = g/h, f' = ((h)(g') - (g)(h'))/h^2 – mathguy Oct 5 '12 at 14:28
@mathguy So would this be what you said? $P=\frac{nRT}{V-nb}-\frac{n^{2}a}{V^{2}}$ $$\frac{\partial P}{\partial V}= \frac{nRT}{-nb}-\frac{n^{2}a}{2V}$$ – John Oct 6 '12 at 4:59
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) – mathguy Oct 7 '12 at 5:21

You can try to find $\partial P / \partial T$ and $\partial P / \partial V$ such that: $$\frac{\frac{\partial P}{\partial T}}{\frac{\partial P}{\partial V}}=\frac{\partial V}{\partial T}$$
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 $P(T)$ – M. L. Oct 7 '12 at 9:49