Partial Derivative of Van der Waals Equation

[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~

 

[SammyS]

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 ?

Sincerely thanks~

What have you tried?

Where are you stuck ?

 

[Scharles]

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

 

[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 ?

 

[jackmell]

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

If I had an equation:

y(x)+x=k

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

y'(x)+1=0

Ok, not too bad.

Suppose I had:

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

How about:

(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.