Is Gibbs Energy Truly Equivalent to Work?

AI Thread Summary
The discussion centers around the relationship between Gibbs free energy and work in thermodynamics. The original poster attempts to derive that Gibbs energy equals negative work through various equations, including the ideal gas law and the definitions of enthalpy and internal energy. However, other participants point out errors in the equations, specifically regarding the assumptions of constant temperature and the mixing of temperature and entropy in the Gibbs free energy equation. Clarifications are made about the correct formulation of the equations used. The conversation highlights the importance of precise definitions and assumptions in thermodynamic calculations.
freek_g
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Hi,

I'm preparing for my exams in a few weeks, of which one covers Thermodynamics.

I was trying to solve a question, where I noticed the Gibb's free energy had to equal the (negative) work. I kind of came to an answer, but was not sure if I did it the right way. All steps are reversible.

First, from the ideal gas law : $$ pV = nRT$$
Then, if I differentiate both sides: $$d(pV) = d(nRT)$$
$$ dpV = -pdV$$

So, let's hold that thought. Now for the Gibb's free energy (with T = constant): $$ dG = dH - TdS$$
And because ##H = U + pV##: $$ dG = dU + d(pV) - TdS$$
Now, ## U = q + w = q - pdV##: $$dG = q - pdV + dpV + pdV - TdS$$

Because ##q = q_{rev} = TdS##: $$ dG = q_{rev} + dpV - q_{rev} = dpV$$

Then, from the beginning: $$dG = dpV = -pdV = - w_{rev}$$

So my question, am I right with this reasoning?

Thanks in advance :)
 
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Your third equation (which would be more clearly written ##V\,dp=-p\,dV##) already assumes constant temperature, right? Otherwise, I don't know how you're getting it.

Also, your first equation for Gibbs free energy is incorrect; you've mixed up ##T## and ##S##.
 
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