Solving the Berthelot equation for pressure

[jbowers9]

1. The problem statement, all variabl
The attempt at a solution

 

[chemisttree]

The answer is obviously 'c'.

 

[jbowers9]

I'm sorry. I had some problems using the Latex editor. It's my first time using it. I did post earlier about proton affinity though. I haven't gotten a heads up from anyone about it but I think my reasoning is correct. I really wish that there was something like PF when I was in school.
Oh, and the correct answer to the above problem is c.

 

[jbowers9]

What I was trying to ask was how does the author Klotz get from a to b?
a
(P + a/Vm^2)( Vm - b) = RT

b
P Vm = RT [1 + 9/128 P/Pc * Tc / T (1 - 6 T^2c / T^2)

 

[jbowers9]

See below please

 

[jbowers9]

(P + a/V2m)(Vm - b) = RT

P = RT/( Vm - b) - a/Vm^2

∂Vm/∂ Pc = 0 = -RTc/( Vmc - b)^2 + 2a/TcVmc^3
∂Vm2/∂2 Pc = 0 = 2RTc/( Vmc - b)^3 - 6a/TcVmc^4


b = Vmc /3 ; 2a = 9RTc/4Vmc^2 * TcVmc^3

a = 9/8 RTc^2 Vmc = 3PcVmc^2 Tc


PcTcVmc^2 = 3/2R Tc^2 Vmc - a

4 PcTcVmc^2 = 3/2R Tc^2 Vmc

3 R Tc/8Pc = Vmc = 3b
b = R Tc/8Pc
4/3 a = 3/2 RTc^2 Vmc
a = 3/4 * 3/2 * 3/8 R2 Tc^3 / Pc
a = 27/64 R^2 Tc^3 / Pc ; b = R Tc/8Pc ; R = 8Pc Vmc/3 Tc
PVmc = RT [1+1/8 PTc /PcT - 9/8 Tc Vmc /T^2 Vm + 3/8 Tc^2 Vmc^2 /T^2 Vm^2 ]

How do I get Vm and Vmc in terms of Pc & Tc? Solve a cubic in terms of Vm?

P Vm = RT[1 + 9/128 P/Pc * Tc/T (1 - 6 Tc^2 /T^2)]

 

[Gokul43201]

(a) is the van der Waals equation and (b) is the Berthelot equation. They arise from two different models, and I don't believe they are functionally identical (I could be wrong here, I haven't tried to check). Is there a reason that you believe one is derivable from the other?

 

[jbowers9]

Hello sir. Thank you for your time and attention on these matters. Weell, a should actually be

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

which is a form of the Berthelot equation. The development and derivations for a, b, R and the form of the eq. which contains Vmc and Vm are the same as the original post.

P = RT/(Vm - b) - a/TVm^2

∂Vm/∂ Pc = 0 = -RTc/( Vmc - b)^2 + 2a/TcVmc^3
∂Vm2/∂2 Pc = 0 = 2RTc/( Vmc - b)^3 - 6a/TcVmc^4


b = Vmc /3 ;
2a = 9RTc/4Vmc^2 * TcVmc^3

a = 9/8 RTc^2 Vmc = 3PcVmc^2 Tc

The constant a differs from the van der Waals constant value by a factor of Tc.

PcTcVmc^2 = 3/2R Tc^2 Vmc - a

4 PcTcVmc^2 = 3/2R Tc^2 Vmc
3 R Tc/8Pc = Vmc = 3b

b = R Tc/8Pc
which is the same as the van der Waals value.

4/3 a = 3/2 RTc^2 Vmc
a = 3/4 * 3/2 * 3/8 R2 Tc^3 / Pc

a = 27/64 R^2 Tc^3 / Pc ; b = R Tc/8Pc ; R = 8Pc Vmc/3 Tc

R and b are the same as for the van der Waals eq. The constant a varies only by a factor of Tc.


PVmc = RT [1+1/8 PTc /PcT - 9/8 Tc Vmc /T^2 Vm + 3/8 Tc^2 Vmc^2 /T^2 Vm^2 ]

So how do I get Vm and Vmc in terms of Pc & Tc? Solve a cubic in terms of Vm?

P Vm = RT[1 + 9/128 P/Pc * Tc/T (1 - 6 Tc^2 /T^2)]