Maths problem: deriving expressions for the critical temperature/volume/pressure

[Lisa...]

I have to show that for a van der Waals gas the critical temperature, volume and pressure are given by:

$$T_c= \frac{8a}{27bR}$$

$$V_c= 3nb$$

$$p_c= \frac{a}{27b^2}$$

I started off this way:

Van der Waals states that for a non ideal gas the pressure is:

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

with a and b constants...

The point of inflection of the (V,p) graph is the critical volume + temperature. Therefore this point is given by the conditions:

$$\frac{\delta p}{\delta V} =0$$

$$\frac{\delta ^2 p}{\delta V^2} =0$$

I came to the conclusion that these conditions are:

$$\frac{\delta p}{\delta V} = - \frac{nRT}{(V-nb)^2} + \frac{2an^2}{V^3}=0$$

$$\frac{\delta ^2 p}{\delta V^2} = \frac{2 nRT}{(V-nb)^3} - \frac{6an^2}{V^4}=0$$

Now how do I obtain V /Tfrom these two equations?
I believe I need to set the second equation equal to 0 and solve for V & T. Then substitute the answer in the first one and show that it also equals 0.
Only: how do I get an expression for V & T?! I've done the following:

$$\frac{2 nRT}{(V-nb)^3} - \frac{6an^2}{V^4}=0$$

$$\frac{2 nRT}{(V-nb)^3} = \frac{6an^2}{V^4}$$

$$\frac{nRT}{(V-nb)^3} = \frac{3an^2}{V^4}$$

What steps should I take next? Please help me

 

[Bystander]

Two equations and three unknowns? Think there might be a third relation laying around somewhere? See anything you can easily eliminate by rearranging your two derivatives?

 

[Lisa...]

I think I've managed to solve the problem:

I've used the second derivative to obtain an expression for T, then I've substituted it into the first one and solved for V. This expression matched the given one, so I knew I was on the right track. Next I've substituted the expression for V into the one I had for T and simplified, which also gave me the given expression. Finding p was really easy by then (with filled in values for V and T in the van der waals formula)...

Thanks anyways for trying to help :)

 

[Gokul43201]

Lisa!

You had :

$$\frac{nRT}{(V-nb)^2} = \frac{2an^2}{V^3}$$

and

$$\frac{2 nRT}{(V-nb)^3} = \frac{6an^2}{V^4}$$

Dividing one by the other directly gives you Vc = 3nb. Plug this into either of the two to find Tc. Plug Vc and Tc into the Van der Waals equation to find Pc.

 

[IITian]

These will be quite useful for people working with critical fluids, particularly for Carbon dioxide capture, transportation and storage. Needless to say that Van der Waals EoS is not valid for CO2!