# Thermodynamic Identities Proof - Gibbs and Helmholtz

#### TFM

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

$$S = -(\frac{\partial F}{\partial T})_V$$ and $$S = -(\frac{\partial F}{\partial T})_P$$

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

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

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

2. Relevant equations

Helmholtz:

$$F = U - TS$$

Gibbs:

$$G = H - TS = U - TS + PV$$

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?

$$F(T) = V - TS$$

$$\frac{\partial F}{\partial T} = -S$$

insert:

$$U = -T^2 \left(\frac{-S}{\partial T} \right)_V$$

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:

$$U = -2T \left(-S} \right)_V$$

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

Any ideas?

Many thanks,

TFM

#### Mapes

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

$$\left(\frac{\partial (F/T)}{\partial T}\right)_V$$

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

#### TFM

Okay, so using chain rule:

$$\frac{\partial \frac{F(T)}{T}}{dT}$$

Would the quotient rule not be best:

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

giving:

$$\frac{TF'(T) - 1F(T)}{T^2}$$

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

$$\frac{T(-S) - U - TS}{T^2}$$

$$\frac{-TS - U - TS}{T^2}$$

$$\frac{-2TS - U}{T^2}$$

Reinsert:

$$U = -T^2 \left(\frac{-2TS - U}{T^2} \right)_V$$

This cancels to:

$$U = -2TS - U$$

$$2U = -2TS$$

$$U = -TS$$

I don't think this is right...?

#### Vuldoraq

Okay, so using chain rule:

$$\frac{\partial \frac{F(T)}{T}}{dT}$$

Would the quotient rule not be best:

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

giving:

$$\frac{TF'(T) - 1F(T)}{T^2}$$

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

$$\frac{T(-S) - U - TS}{T^2}$$

$$\frac{-TS - U - TS}{T^2}$$

$$\frac{-2TS - U}{T^2}$$

Reinsert:

$$U = -T^2 \left(\frac{-2TS - U}{T^2} \right)_V$$

This cancels to:

$$U = -2TS - U$$

$$2U = -2TS$$

$$U = -TS$$

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

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

#### Vuldoraq

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

Okay, so

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

F/T = U/T - S

put into equation:

$$U = -T^2 \left(\frac{\partial (\fracd{U}{T} - S)}{\partial T} \right)_V$$

differentiating this gives:

$$U = -T^2 \left(U)_V$$

?

#### Vuldoraq

Okay, so

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

F/T = U/T - S

put into equation:

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

$$\frac{\partial (F/T)}{\partial T}=\frac{\partial (\frac{U}{T} - S)}{\partial T}$$

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

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#### Mapes

Homework Helper
Gold Member
You made an arithmetic error in post #3. You're almost there.

#### Vuldoraq

That is strange we posted at exactly the same time. Are you my long lost twin brother?

#### TFM

What you want is this;

$$\frac{\partial (F/T)}{\partial T}=\frac{\partial (\frac{U}{T} - S)}{\partial T}$$

Then think what is d(U/T)/dT?
Would it be U/T^2 ?

#### Vuldoraq

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

Okay, so using the quotient rule:

U = U ()
V = T

du/dt = du/dt

dvdt = 1

thus:

$$= \frac{U - Tdu/dT}{T^2}$$

Look okay now?

#### Vuldoraq

Okay, so using the quotient rule:

U = U ()
V = T

du/dt = du/dt

dvdt = 1

thus:

$$= \frac{U - Tdu/dT}{T^2}$$

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

$$(\frac{u}{v})'=\frac{u'*v-u*v'}{v^{2}}$$ 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

Okay, so it should have been:

$$\frac{dU/dT T - U}{T^2}$$

so now the thermodynamic identity:

dU = Tds - pdv

so now I have to rearrange

$$\frac{dU/dT T - U}{T^2}$$

into dU = , then replace du with the thermo identity, and solve?

#### Vuldoraq

Okay, so it should have been:

$$\frac{dU/dT T - U}{T^2}$$

so now the thermodynamic identity:

dU = Tds - pdv

so now I have to rearrange

$$\frac{dU/dT T - U}{T^2}$$

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

so,

dU = Tds + pdV

sub into:

$$\frac{dU/dT T - U}{T^2}$$

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

this will give:

$$\frac{T^2 ds/dT + pT dV/dT - U}{T^2}$$

???

#### Vuldoraq

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

#### TFM

Okay so:

$$\frac{T^2 ds/dT + pT dV/dT - U}{T^2}$$

break it down into:

$$\frac{T^2 ds/dT}{T^2} + \frac{PT dV/dT}{T^2} - \frac{u}{T^2}$$

$$\frac{ds}{dT} + \frac{P dV/dT}{T} - \frac{u}{T^2}$$

$$\frac{ds}{dT} + \frac{P dV/dT}{T} = \frac{u}{T^2}$$

Okay so far?

#### Vuldoraq

Okay so:

$$\frac{T^2 ds/dT + pT dV/dT - U}{T^2}$$

break it down into:

$$\frac{T^2 ds/dT}{T^2} + \frac{PT dV/dT}{T^2} - \frac{u}{T^2}$$

$$\frac{ds}{dT} + \frac{P dV/dT}{T} - \frac{u}{T^2}$$

$$\frac{ds}{dT} + \frac{P dV/dT}{T} = \frac{u}{T^2}$$

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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