Understanding Free Energy in Fuel Cell Reactions

AI Thread Summary
In fuel cell reactions, ΔG represents the maximum electrical work obtainable from the system, which is why ΔG equals W(electrical) alone, excluding Q, as Q is not considered useful work in this context. The confusion arises from the distinction between electrical work and heat, where waste heat is calculated using the relationship Q = ΔG - ΔH. This indicates that waste heat is derived from the difference between the Gibbs free energy change and the enthalpy change. Understanding these relationships clarifies the roles of energy transformations in fuel cells. The discussion highlights the importance of accurately applying thermodynamic principles to fuel cell efficiency calculations.
lustrog
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I'm given a fuel cell that uses methane as fuel:

(CH4) + 2(O2) --> 2(H2O) + (CO2)

And I'm asked to find (a) the electrical work I can get out for each mole of methane, and (b) the amount of waste heat produced.

I understand that ΔG is defined as the amount of useful non-mechanical work that can be obtained from the system. So my first thought was that ΔG = Q + W(electrical). However, I'm told it's actually the case that ΔG = W(electrical) only. So why exactly is that? Isn't Q a form of non-mechanical work? Secondly, when I'm asked to compute the amount of waste heat, I go ahead and accept the confusing notion that ΔG = W(electrical) and attempt to compute the waste heat simply as ΔH = ΔU + PΔV = ΔU - W = Q. However, I'm told the waste heat is actually Q = ΔG - ΔH. Can anyone explain these two areas of confusion for me?
 
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Just read the sticky. Perhaps this should have been in the HW section. But it is a conceptual question. I just pulled a relevant textbook question to demonstrate my areas of confusion about free energy.
 
Thread 'Question about pressure of a liquid'
I am looking at pressure in liquids and I am testing my idea. The vertical tube is 100m, the contraption is filled with water. The vertical tube is very thin(maybe 1mm^2 cross section). The area of the base is ~100m^2. Will he top half be launched in the air if suddenly it cracked?- assuming its light enough. I want to test my idea that if I had a thin long ruber tube that I lifted up, then the pressure at "red lines" will be high and that the $force = pressure * area$ would be massive...
I feel it should be solvable we just need to find a perfect pattern, and there will be a general pattern since the forces acting are based on a single function, so..... you can't actually say it is unsolvable right? Cause imaging 3 bodies actually existed somwhere in this universe then nature isn't gonna wait till we predict it! And yea I have checked in many places that tiny changes cause large changes so it becomes chaos........ but still I just can't accept that it is impossible to solve...

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