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Studiot
#4
Jan17-13, 06:47 PM
P: 5,462
Confused about thermodynamic quantities and their minimums.

Not quite sure of your symbols here. E is not normally used for enthalpy. Assuming you are using F for what is normally called the Helmoltz free energy (or work function) and H for enthalpy, equation 38 should read.

G = H - TS = F + PV.

since there is some doubt as to the meaning of E I shall avoid it and use U for internal energy.

So for a change at constant temperature along a reversible path

dU = TdS + dwrev (Gibbs equation)

At constant temp TdS = d(TS) so we can rewrite the above as

d(U-TS) = dwrev (at constant T)

Let F = A-TS then dF = dwrev (at const T)

Now the expression wrev refers to all possible kinds of work.

If only pressure/volume work is envisaged

dwrev = PdV

At const volume dV = 0 so dF = 0 so F is a minimum, at equilibrium.

So the condition for equilibrium at constant temp and volume is that the Helmholtz free energy is a minimum and dF = 0.

Now to develop a similar discussion for constant pressure Note that

H = U + PV

dH = dU + PdV + VdP

Substitute from the above

dH = TdS + dwrev + PdV + VdP

or at constant T and P

d(H-TS) = PdV + dwrev

This new function is called the Gibbs free energy and is G = H - TS = F+PV

So dG = PdV + dwrev

Again restricting this to PV work dwrev, dwrev = -PdV

So at const T and P dG = 0, Thus G is a minimum, at equilibrium.

Now applying all that to my melting ice example.

dG= 0 = d(H-TS) = dH - d(TS) = dH - TdS

dS = dH / T

The entropy of melting (fusion) can be calculated by dividing the enthalpy (=latent heat) by the melting point.

You might like to look at this thread and in particular post#11 (Which no longer contains a typo)

http://www.physicsforums.com/showthr...lmholtz&page=2

does this help?