Gibbs free energy doesn't increase (constant T and P) - proof doesn't seem right


by Silversonic
Tags: constant, energy, free, gibbs, increase, proof
Silversonic
Silversonic is offline
#1
May7-12, 10:45 AM
P: 126
First my notes discover that for an isothermal transformation;

ΔW ≥ ΔF

Where W is the work done and F is the Helmholtz Free Energy, F = E - TS.


Then it defines the Gibbs free Energy;

G = F + PV

"For a system at constant temperature and pressure, G never increases";

So ΔG = ΔF + Δ(PV) = ΔF + PΔV = ΔF + ΔW


Then it says "We already know ΔW ≤ -ΔF and therefore ΔW + ΔF = ΔG ≤ 0".

But how can we assume ΔW ≤ -ΔF from the fact that ΔW ≥ ΔF? If ΔW is positive and |ΔW| ≥ |ΔF| then this assumption does not work, and also I have nothing that indicates to me the change in the work done is either positive or negative. I can't seem to show how this proof was meant to work otherwise though, any help?
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jfizzix
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#2
May12-12, 01:01 AM
P: 219
Quote Quote by Silversonic View Post
First my notes discover that for an isothermal transformation;

ΔW ≥ ΔF

Where W is the work done and F is the Helmholtz Free Energy, F = E - TS.


Then it defines the Gibbs free Energy;

G = F + PV

"For a system at constant temperature and pressure, G never increases";

So ΔG = ΔF + Δ(PV) = ΔF + PΔV = ΔF + ΔW


Then it says "We already know ΔW ≤ -ΔF and therefore ΔW + ΔF = ΔG ≤ 0".

But how can we assume ΔW ≤ -ΔF from the fact that ΔW ≥ ΔF? If ΔW is positive and |ΔW| ≥ |ΔF| then this assumption does not work, and also I have nothing that indicates to me the change in the work done is either positive or negative. I can't seem to show how this proof was meant to work otherwise though, any help?


From what I can see...

[itex] dW = -PdV + \mu dN[/itex]
[itex] dE = dW + dQ = dW + T dS[/itex]

At constant temperature...

[itex]d F = d E - T dS = dW[/itex]
so
[itex]ΔF = ΔW[/itex].

I would be interested in knowing how you show
[itex]ΔF ≤ ΔW[/itex]



Also, there is an interesting way (which may be along the same lines as your notes) to show that the Gibbs free energy never increases in processes at constant temperature and pressure.

We note that in all processes where a system is in contact with a pressure and temperature reservoir, the system will evolve toward a state which maximizes the entropy of both system plus reservoir.

[itex]dS^{s+r} = dS^{s} + dS^{r} ≥0[/itex]
[itex]dS^{s} = (1/T^{s})*(dU^{s} + P^{s} dV^{s} - \mu^{s} dN^{s})[/itex]

With these two expressions and knowing that the temperature and pressure are the same for both system and reservoir, you can show that
[itex]dS^{s+r} = -(1/T^{s})*dG^{s}[/itex]
so that since the system inexorably tends to maximize the entropy of the universe (universe = system plus reservoir), it must also inexorably tend to minimize its Gibbs free energy for processes at constant temperature and pressure.

Anywhoo, that's why I'd say the Gibbs free energy of the system cannot increase for processes at constant temperature and pressure.

-James


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