van der Waals gas is not real gas?

Outrageous

From van der Waals , (P+a/v^2)(v-b)=RT,
At critical temperature, I get (∂P/∂V)at constant temperature =0
and (∂^2P/∂V^2) at constant temperature ,T=0.
then critical pressure,P = a/(27b^2)--------1
critical volume,v=3b-----------2
critical temperature=8a/(27Rb)----------3
then simultaneous equation 1 and 3,
I get b=(RT/8P), b=(v/3) ------------------4
But from the experiment, we get T,P,v and then substitute into the two equation from 4,both b have different values. Why?

Thank you

mfb

(P+a/v^2)(v-b)=RT

Using your equations:
P = a/(27b^2)
v=3b
T=8a/(27Rb)

I get
(a/(27b^2)+a/(9b^2))(2b)=8Ra/(27Rb)
8/(27b) = 8/(27b)

Looks fine.

Are you sure your real gas is a perfect van-der-Waals gas?

Outrageous

I though all real gas is van der Waals ? Then what do you mean by perfect van der Waals?

Philip Wood

On other threads which you've started, it's been made clear (I think) that the V der W equation is a theoretical equation based on some quite crude assumptions. No actual gas obeys the V der W equation perfectly. [The confusion may be caused because 'real gas' is sometimes used to mean non-ideal gas, even a theoretical non-ideal gas, and not necessarily an actual gas.]