Thermodynamic Identities Proof - Gibbs and Helmholtz

TFM

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1. The problem statement, all variables and given/known data

Use the definitions of Helmholtz free energy, F, and Gibbs free energy, H, together with the thermodynamic identity, to show that

[tex] S = -(\frac{\partial F}{\partial T})_V [/tex] and [tex] S = -(\frac{\partial F}{\partial T})_P [/tex]

Then use those definitions again, and the expressions above, to show that

[tex] U = -T^2 \left(\frac{\partial (F/T)}{\partial T} \right)_V [/tex] and [tex] H = -T^2 \left(\frac{\partial (G/T)}{\partial T} \right)_P [/tex]

These equations are known as the Gibbs-Helmholtz equations and are useful in chemical thermodynamics.

2. Relevant equations

Helmholtz:

[tex] F = U - TS [/tex]

Gibbs:

[tex] G = H - TS = U - TS + PV [/tex]

3. The attempt at a solution

I have succesfully done the first part (showing S = F/T...) But I am stuck on the second.

I asked the ;ectuirer, whop said I should first work backwards, and do df/DT, and insert into the equation. This has given me?

[tex] F(T) = V - TS [/tex]

[tex] \frac{\partial F}{\partial T} = -S [/tex]

insert:

[tex] U = -T^2 \left(\frac{-S}{\partial T} \right)_V [/tex]

now the thermodynamic identity is:

dU = Tds - pdv

I do I get rid of the squared part?

because if I just differentiate T squared, I get 2T,

so this could leave me with:

[tex] U = -2T \left(-S} \right)_V [/tex]

but this doesn't make the S a ds? and I have a factor of 2?

Any ideas?

Many thanks,

TFM
 

Mapes

Science Advisor
Homework Helper
Gold Member
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I would use the chain rule to evaluate

[tex]\left(\frac{\partial (F/T)}{\partial T}\right)_V[/tex]

which will give you two terms. Then plug in [itex]F=U-TS[/itex], which leads to the desired first expression. The same approach works for the H-G pair.
 

TFM

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Okay, so using chain rule:

[tex] \frac{\partial \frac{F(T)}{T}}{dT} [/tex]

Would the quotient rule not be best:

[tex] u = F(T), v = T, du/DT = F'(t), dv/dT = 1 [/tex]

giving:

[tex] \frac{TF'(T) - 1F(T)}{T^2} [/tex]

Plug in F = U - TS, F' = -S


[tex] \frac{T(-S) - U - TS}{T^2} [/tex]

[tex] \frac{-TS - U - TS}{T^2} [/tex]

[tex] \frac{-2TS - U}{T^2} [/tex]

Reinsert:

[tex] U = -T^2 \left(\frac{-2TS - U}{T^2} \right)_V [/tex]

This cancels to:


[tex] U = -2TS - U [/tex]

[tex] 2U = -2TS [/tex]

[tex] U = -TS [/tex]

I don't think this is right...?
 
Okay, so using chain rule:

[tex] \frac{\partial \frac{F(T)}{T}}{dT} [/tex]

Would the quotient rule not be best:

[tex] u = F(T), v = T, du/DT = F'(t), dv/dT = 1 [/tex]

giving:

[tex] \frac{TF'(T) - 1F(T)}{T^2} [/tex]

Plug in F = U - TS, F' = -S


[tex] \frac{T(-S) - U - TS}{T^2} [/tex]

[tex] \frac{-TS - U - TS}{T^2} [/tex]

[tex] \frac{-2TS - U}{T^2} [/tex]

Reinsert:

[tex] U = -T^2 \left(\frac{-2TS - U}{T^2} \right)_V [/tex]

This cancels to:


[tex] U = -2TS - U [/tex]

[tex] 2U = -2TS [/tex]

[tex] U = -TS [/tex]

I don't think this is right...?
Try this approach,

First divide F by T. Then differentiate the answer. Then plug in the thermodynamic identity for dU. Rearrange, and let N and V be constant. This should give you the right answer.
 

TFM

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So when you say divide F by T, would thuis be F/T, or do we use F = U - TS, giving U/T - S?
 
So when you say divide F by T, would thuis be F/T, or do we use F = U - TS, giving U/T - S?
You want both of those.
 

TFM

1,026
0
Okay, so

[tex] U = -T^2 \left(\frac{\partial (F/T)}{\partial T} \right)_V [/tex]

F/T = U/T - S

put into equation:

[tex] U = -T^2 \left(\frac{\partial (\fracd{U}{T} - S)}{\partial T} \right)_V [/tex]

differentiating this gives:

[tex] U = -T^2 \left(U)_V [/tex]

?
 
Okay, so

[tex] U = -T^2 \left(\frac{\partial (F/T)}{\partial T} \right)_V [/tex]

F/T = U/T - S

put into equation:

[tex] U = -T^2 \left(\frac{\partial (\fracd{U}{T} - S)}{\partial T} \right)_V [/tex]
What you want is this;

[tex]\frac{\partial (F/T)}{\partial T}=\frac{\partial (\frac{U}{T} - S)}{\partial T}[/tex]

Then think what is d(U/T)/dT?
 
Last edited:

Mapes

Science Advisor
Homework Helper
Gold Member
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You made an arithmetic error in post #3. You're almost there.
 
That is strange we posted at exactly the same time. Are you my long lost twin brother? :wink:
 

TFM

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What you want is this;

[tex]\frac{\partial (F/T)}{\partial T}=\frac{\partial (\frac{U}{T} - S)}{\partial T}[/tex]

Then think what is d(U/T)/dT?
Would it be U/T^2 ?
 
Would it be U/T^2 ?
That wold be part of it, although I think it should be negative U/T^2. Use the quotient rule and remember that you can't actually differentiate U so you have to leave it as dU/dT.
 

TFM

1,026
0
Okay, so using the quotient rule:

U = U (:rolleyes:)
V = T

du/dt = du/dt

dvdt = 1

thus:

[tex] = \frac{U - Tdu/dT}{T^2} [/tex]

Look okay now?
 
Okay, so using the quotient rule:

U = U (:rolleyes:)
V = T

du/dt = du/dt

dvdt = 1

thus:

[tex] = \frac{U - Tdu/dT}{T^2} [/tex]

Look okay now?
YOur quotient rule is back to front I believe. It should be (the following are for the general case);

[tex](\frac{u}{v})'=\frac{u'*v-u*v'}{v^{2}}[/tex] Note where the primes are, differentiate top times bottom, then take away the derivative of the bottom times the top and divide the whole lot by the bottom squared.

Once you've changed that let dU=identity and the rest is just rearranging.

Edit: I don't mean dU=1, I just mean the thermodynamic identity for dU, which you must have used earlier.
 
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TFM

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Okay, so it should have been:

[tex] \frac{dU/dT T - U}{T^2} [/tex]

so now the thermodynamic identity:

dU = Tds - pdv

so now I have to rearrange

[tex] \frac{dU/dT T - U}{T^2} [/tex]

into dU = , then replace du with the thermo identity, and solve?
 
Okay, so it should have been:

[tex] \frac{dU/dT T - U}{T^2} [/tex]

so now the thermodynamic identity:

dU = Tds - pdv

so now I have to rearrange

[tex] \frac{dU/dT T - U}{T^2} [/tex]

into dU = , then replace du with the thermo identity, and solve?
It would be easier if you just substituted for dU in the differential eqution and solved. Don't forget the -dS/dT term from earlier when you put everything back together.
 

TFM

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so,

dU = Tds + pdV

sub into:

[tex] \frac{dU/dT T - U}{T^2} [/tex]


[tex] \frac{(Tds + pdV/dT) T - U}{T^2} [/tex]

this will give:

[tex] \frac{T^2 ds/dT + pT dV/dT - U}{T^2} [/tex]

???
 
Thats right. Now all you have to do is simplify and you will have the right answer.
 

TFM

1,026
0
Okay so:

[tex] \frac{T^2 ds/dT + pT dV/dT - U}{T^2} [/tex]

break it down into:

[tex] \frac{T^2 ds/dT}{T^2} + \frac{PT dV/dT}{T^2} - \frac{u}{T^2} [/tex]

[tex] \frac{ds}{dT} + \frac{P dV/dT}{T} - \frac{u}{T^2} [/tex]

[tex] \frac{ds}{dT} + \frac{P dV/dT}{T} = \frac{u}{T^2} [/tex]

Okay so far?
 
Okay so:

[tex] \frac{T^2 ds/dT + pT dV/dT - U}{T^2} [/tex]

break it down into:

[tex] \frac{T^2 ds/dT}{T^2} + \frac{PT dV/dT}{T^2} - \frac{u}{T^2} [/tex]

[tex] \frac{ds}{dT} + \frac{P dV/dT}{T} - \frac{u}{T^2} [/tex]

[tex] \frac{ds}{dT} + \frac{P dV/dT}{T} = \frac{u}{T^2} [/tex]

Okay so far?
Not sure about the equals sign here, don't forget this is part of the rhs of the equation,

d(F/T)/dT=d/dT(U/T-S)
 

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