Question

Thermodynamic Square Derivative Help

Answer 1

so I am given the relations:

dU = -pdV + TdS
dH = VdP + TdS
dG = Vdp - SdT
dA = -pdV - SdT

which can be seen through the diagonals of the thermodynamic square, however, I am slightly confused about the sign convention vs direction of the diagonals and how to remember/derive them :S

Answer 2

@sillybilly123 may I have some assistance?

Answer 3

Voca, I know some maths and physics but virtually no chemistry.

i watched a vid on Youtube that matches your thermo square and, frankly, concluded very quickly that this is no help whatsoever - would be way easier just to memorise the relationships. None of it, including the square itself.

**however**, Wiki has a Thermodynamic square that is really well written/ set out, that I can follow quite easily, without really knowing what some of the potentials really mean.

https://en.wikipedia.org/wiki/Thermodynamic_square

......and, with a knowledge of Legendre transforms, can see this way forward:

If we start with the fundamentals, 1.01 on any physics Thermo course:

1st Law: \(dU = - dW_{\text{by}} + dQ_{in}\)

Definition (classical) entropy: \(ds = \dfrac{dQ}{T}\)

Work done by expanding gas: \(dW = p dV\)

Then we can see that:

\(dU = - p dV + T ds\)

And that's what the Thermo square table predicts.

Next we [randomly] reach for some Legendre transforms:

We can try this way:

\(dU = - p dV + \color{red}{T} d\color{red}{s}\)

\(d(U - \color{red}{sT}) = - p dV + T ds - s dT - Tds = - p dV - s dT\)

Well that just happens to be \(dF\) from the square

Next we use the transforms:

\(dU = - \color{red}{p} d\color{red}{V} + T ds\)

\(d(U + \color{red}{pV}) = - p dV + T ds + p dV + V dp = T ds + V dp\)

And that just happens to be \(dH\) from the square

Next we'll do a double Legendre:


\(d(U - sT + pV) = - p dV + T ds - s dT - T ds + p dV + V dp = - s dT + V dp \)

And that just happens to be \(dG\)


All from flailing around, knowing a little calculus and very little chemistry. ***But being careful to pick a plus or minus sign inside the differential so that some stuff cancels out.***


There is so much info packed into this stuff. Simple example: \(dW = p dV\) really means the infinitessinal amount of work done by gas in an infinitessimal expansion so \(p \approx \text{const} \)

IOW: Haven't thought about turning these into PDE's - the Maxwell relations - but I'm guessing that's something one could also bodge just using math.

Bit waffly, and late, but hope it helps :)

Answer 4

so all those meduls doshed out in Physics :)

funny that !