Integral Help: Van der Waals Equation of State

[Peregrine]

I'm trying to integrate the Van der Waals equation of state for an isothermal problem, but based on my results I think I'm doing a bit of simple calculus wrong and hope someone here can help.

$$P = \int \frac{RT}{v-b} dv$$
where R, b and T are known constants.

I tried to do a u-substitution for 1/(v-b) with u=(v-b)
so, du = 1.

Thus. $$P = \int \frac{RT}{u} du$$

So, P = RT ln(u) = RT ln(v/b)

Any ideas where I went wrong? Thanks.

 

[quasar987]

ln(v-b) does not equal ln(v/b) to the best of my knowledge. The rule is ln(A/B)=ln(A)-ln(B)

The result makes physical sense too. b is a measure of the intermolecular repulsion force. As v-->b, P-->oo.

 

[Gib Z]

Ahh sorry I didn't help earlier lol, as soon as I read the first sentence, I thought I didn't know the physics involved so i couldn't help. As soon as you told me they were all constants it made it much easier.

$$P=RT \int \frac{1}{v-b} dv= RT \ln (v-b)$$, which can not be simplified. I thinking quasar987 also pointed that out, just clarifying or sumthin..

 

[Peregrine]

Ah, I see. Thanks for the help!

(of course, since you are familiar with the physics, Gib z, you may see I'm trying use the relation of your namesake to fund du, and I could do the attraction potion which is the ultimately simple a/v^2.

Again, I greatly appreciate the help!

 

[ssd]

I do not have good idea about Physics, but mathematically we shall add the constant of integration with what GibZ has shown.