Inversion Curve for a gas obeying Dieterici's equation of state

AI Thread Summary
The discussion revolves around deriving the inversion curve for a gas described by Dieterici's equation of state. Participants focus on the condition for the inversion curve, which involves the derivative (dV/dT) at constant pressure. There are challenges in performing implicit differentiation and eliminating variables to reach the target equation. Eventually, one participant confirms their differentiation approach and successfully simplifies the problem, leading to the conclusion that the maximum inversion temperature can be determined by setting pressure (P) to zero, resulting in T = 2a/bR. The conversation highlights the complexities of the derivation and the eventual clarity achieved through collaboration.
ncholland
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Homework Statement



For a gas obeying Dieterici's equation of state:

P(V-b) = RTexp(-a/RTV)

for one mole, prove that the equation of the inversion curve is

P = ((2a/b^2) - (RT/b)) * exp((1/2) - (a/(RTb)))

and hence find the maximum inversion temperature.

Homework Equations



N/A

The Attempt at a Solution



So, I know that for the inversion curve, the condition is (dV/dT) = V/T (where the derivative is evaluated at constant pressure). But this would need implicit differentiation to find dV/dt ... and it seems completely intractable - is there something I'm missing?
 
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Do the partial differentiation.
To arrive at the target equation, you need to eliminate ∂V/∂T (for which you have an equation) and V (for which you have the original equation).
If you're still stuck, please post your working.
 
Hi, thanks for the help!

I've tried the partial differentiation, but when I try and eliminate V and ∂V/∂T between the equation I obtain and the equation of state I just get a horrible mess - I'm not sure if I'm doing the differentiation right...

I rearranged the equation to get:

\frac{-a}{RTV} = ln(p(V-b)) - ln(RT)

So, differentiating wrt T:

\frac{a}{RT^{2}V} + \frac{a}{RTV^{2}}*\frac{∂V}{∂T} = \frac{1}{V-b}*\frac{∂V}{∂T} - \frac{1}{T}

There's not really any point me posting any of the further work / manipulation I've done - I've tried a load of different things and nothing gets anywhere...

Is the initial differentiation correct?

Cheers!
 
ncholland said:
\frac{a}{RT^{2}V} + \frac{a}{RTV^{2}}*\frac{∂V}{∂T} = \frac{1}{V-b}*\frac{∂V}{∂T} - \frac{1}{T}
Yes, that looks good. Substitute for ∂V/∂T and get it into the form V = ...
 
Awesome, thanks! Got it now, much less nasty than I'd thought :-) (I was just messing up the cancellation of terms - which made me doubt I'd got the differentiation right in the first place because the whole thing looked such a mess!)
 
And the maximum inversion temperature is just found by setting P = 0, right? So that gives T = 2a/bR ?
 
ncholland said:
And the maximum inversion temperature is just found by setting P = 0, right?
If you say so. I know nothing about this subject matter.
 
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