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Van der Waals gas is not real gas? 
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#1
Nov1312, 09:54 PM

P: 375

From van der Waals , (P+a/v^2)(vb)=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=3b2 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 


#2
Nov1412, 08:13 AM

Mentor
P: 11,601

(P+a/v^2)(vb)=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. 


#3
Nov1412, 09:10 AM

P: 375




#4
Nov1412, 09:52 AM

PF Gold
P: 941

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 nonideal gas, even a theoretical nonideal gas, and not necessarily an actual gas.]



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