Van der Waals gas is not real gas?

by Outrageous
Tags: real, waals
 P: 375 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
Mentor
P: 12,037
(P+a/v^2)(v-b)=RT

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.

 But from the experiment, we get T,P,v
Are you sure your real gas is a perfect van-der-Waals gas?
P: 375
 Quote by mfb (P+a/v^2)(v-b)=RT Are you sure your real gas is a perfect van-der-Waals gas?
I though all real gas is van der Waals ? Then what do you mean by perfect van der Waals?

 PF Gold P: 962 Van der Waals gas is not real gas? 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.]

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