Why is w(max) = delta G?

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Viraj B 3A
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Joined: Fri Sep 26, 2014 2:02 pm

Why is w(max) = delta G?

Postby Viraj B 3A » Fri Jan 30, 2015 11:26 pm

In formulating the equation: G = -nFE, how can we assume that wmax = G under constant temperature and pressure? Thank you for any insight.

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Re: Why is w(max) = delta G?

Postby Chem_Mod » Sat Jan 31, 2015 1:06 am

G is actually defined as the maximum non-expansion work under constant T and P. Non-expansion means not related to a volume change, for example, electrical work, which is also what we call "useful work". This can be proven mathematically using differentials, obviously you won't be responsible for this!

1st Law, split the work into expansion and non-expansion
Expansion work is -PdV, assume from now that dw refers to non-expansion
Clausius inequality: dS > dq/T so TdS > dq
Move terms
We can arbitrarily add VdP and SdT terms. They are zero since T, P constant
Product rule for derivatives

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

Recall that work done by the system is negative, so if DeltaG < w, then DeltaG is the maximum work that may be done.

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