# Partial Derivative of Van der Waals Equation

1. Dec 20, 2011

### Scharles

Given that the Van Der Waals equation is (p + (an^2)/v^2)(v-nb)=nRT where n,a,R and b are constants...

How to we find the derivative of p wrt v ?

How to find the derivative of p wrt T without further differentiation ??

Can anyone teach me how to do this question ?

Sincerly thanks~

2. Dec 20, 2011

### SammyS

Staff Emeritus
What have you tried?

Where are you stuck ?

3. Dec 21, 2011

### Scharles

i have no idea on how to solving this...
please kindly teach me how to start on solving this sort of question...

4. Dec 21, 2011

### dextercioby

Do you know the difference between an implicit function and an explicit one ? What do you know about the derivatives for explicit functions ? How about implicit ones ?

5. Dec 21, 2011

### jackmell

$$y(x)+x=k$$

and I wanted to take the derivative of y with respect to x, I'd get:

$$y'(x)+1=0$$

$$y(x)+\frac{1}{x^2}=k$$

still not too bad if I want the derivative of y with respect to x. That's:

$$y'-2x^{-3}=0$$

$$(y(x)+\frac{c}{x^2})(x-k)=a$$

That's still not too bad cus' I'd use the chain rule this time:

$$(y(x)+\frac{c}{x^2}) \frac{d}{dx} (x-k)+(x-k)\frac{d}{dx}(y(x)+\frac{c}{x^2})=0$$

and that's:

$$(y(x)+\frac{c}{x^2})(1)+(x-k)(y'(x)-2cx^{-3})=0$$

ok, now you do one but instead of y(x), I'll say:

$$(p(v)+\frac{k}{v^2})(v-c)=a$$

and I want to take the derivative of p with respect to v. Do that one, then do yours with all the other parameters.

Last edited: Dec 21, 2011