# Change in Free Energy in terms of First Law

1. May 16, 2010

### jbunten

I understand why

dE=TdS-PdV ...[1]

at constant temperature and pressure

Also, I see that F=E-TS

and therefore

dF=dE-SdT-TdS ...[2]

and how combining [2] and [3] gives

dF=-SdT -PdV

QUESTION: shouldn't [1] be expressed as dE=TdS+SdT-PdV

i.e if you're going to substitute into equation [2] where T isn't kept constant, don't you need to start off with T not being kept constant in your expression for thermal energy? Why are we letting temperature change in equation 2 but not in equation 1?

This seemingly obvious question has me stumped.

2. May 16, 2010

### Mapes

Your intuition is correct; when we take the differential form of the energy equation

$$U=TS-PV+\mu N$$

(where $\mu$ is the chemical potential and $N$ is the number of particles), we get

$$dU=T\,dS+S\,dT-P\,dV-V\,dP+\mu\,dN+N\,d\mu$$

But the energy is also an expansion in terms of extensive variables; that is,

$$dU=\left(\frac{\partial U}{\partial S}\right)_{V,N}dS+\left(\frac{\partial U}{\partial V}\right)_{S,N}dV+\left(\frac{\partial U}{\partial N}\right)_{S,V}dN$$

from which we define $T\equiv(\partial U/\partial S)_{V,N}$, $P\equiv -(\partial U/\partial V)_{S,N}$, and $\mu\equiv(\partial U/\partial N)_{S,V}$. From this we must conclude that

$$S\,dT-V\,dP+N\,d\mu=0$$

which is known as the Gibbs-Duhem relation. Does this answer your question?

3. May 16, 2010

### jbunten

OK, so energy, being an extensive variable, gets expanded in terms of all the other extensive variables, giving you a relation SdT=VdP (assuming chemical potential = 0 for simplicity) which if you then substitute back into your differential equation for dE gives you the dE=TdS-PdV result.

I'm reasonably sure I understand this now. Is there a good intuitive way to understand this result? On the face of it the equation states that an increment in temperature or pressure doesn't affect free energy.

Thanks.

4. May 16, 2010

### Mapes

Why do you say that?

5. May 16, 2010

### jbunten

We have the expression:

dE=TdS-PdV

if increasing temperature or pressure affected internal energy the dT and dP terms would be necessary to state or otherwise to use partial derivatives and specify T and P as constant.

Am I thinking in the wrong way?

6. May 16, 2010

### Mapes

Yes. The condition of energy not being dependent on a variable $X$ is $dE/dX=0$. But we don't have that here, do we? Just because temperature and pressure don't appear in differential form doesn't mean that they don't affect internal energy.

7. May 16, 2010

### jbunten

OK. So I see all the maths works and now I'm trying to get an intuitive understanding of this.

Here's what I've reasoned out:

VdP doesn't need to be in this expression because an increase in pressure at a constant volume doesn't mean any work so don't include it.

Also, the expression TdS is essentially the small increment in heat, using ds=(sigma*q)/T so SdT isn't included in the equation.

I see that Temperature and Pressure are the intensive properties and therefore that's fundamentally why they aren't in the equation..is this right?

Now for instance if we have a gas kept at a constant volume in a cylinder and we heat it so that it's temperature increases by dT. How do we represent that change in internal energy using dE=TdS-PdV ? The volume has remained constant and so has entropy.

8. May 16, 2010

### Mapes

Pretty much, yes.

Not quite; to heat something means to transfer entropy, almost by definition. In agreement with this, heating a gas at constant volume means increasing its internal energy.

9. May 16, 2010

### nonequilibrium

The fact that SdT - VdP + Ndmu = 0 doesn't mean each term is zero. Mathematically it states that the change in T, P and mu are dependent of each other and that this combination happens to cancel each other. It is an interesting question why that exact combination happens to be zero, and I must say I don't know, but you must realise that the fundamental reason that they are zero, is because dE = TdS - PdV + mu dN. Why is this? It is called the thermodynamic identity. I don't know if there is a proof for it, but I can make it plausible: essentially it states your assumption that the energy of the system can be changed by either heat, work or adding particles, so dE = Q + W + mu dN. Now comes the dodgy part: because it is an infinitesimal change between equilibrium states, Q = TdS (per definition of S, somewhat). Also, if we ignore any other work and just allow mechanical work, then W = -PdV. This makes for dE = TdS - PdV + mu dN.

As for your question in bold: due to definition of heat capacity: nc dT = dQ and due to infinitesimal change: dQ = TdS. Since dV = 0, we get dE = dQ = nc dT

Integrating this, we get E = ncT. For an ideal mono-atomic gas, c = 3/2R, so E = 3/2 nRT. We've derived the expression for energy of an ideal gas, which seems to be only dependent of T.

10. May 16, 2010

### jbunten

Thanks to both of you this is now much more clear in my head, I really appreciate it!

11. May 16, 2010

### jbunten

So just to recap, in principle for the situation described in bold, In principle you could work out the increase in entropy dS for that system when heating it at constant volume, plug it into the thermodynamic identity and get the same result you would if you had used the heat capacity and temperature increment.

Sorry for being so pedantic but this seems like one of those things which are much more subtle than they appear and I want to really understand it.

12. May 16, 2010

### Mapes

You got it. In fact, the specific heat $C_V$ is defined as $T(\partial S/\partial T)_V$, so you should be able to work out the connection.