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Van der Waals gas is not real gas?

by Outrageous
Tags: real, waals
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Outrageous
#1
Nov13-12, 09:54 PM
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
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mfb
#2
Nov14-12, 08:13 AM
Mentor
P: 11,928
(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.

But from the experiment, we get T,P,v
Are you sure your real gas is a perfect van-der-Waals gas?
Outrageous
#3
Nov14-12, 09:10 AM
P: 375
Quote Quote by mfb View Post
(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?

Philip Wood
#4
Nov14-12, 09:52 AM
PF Gold
P: 956
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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