Thermodynamics: Partial derivatives

[WWCY]

Hi all, I have had the following question in my head for quite a while:

Thermodynamic potentials written in differential form look like
$$dU = TdS - PdV$$
and we can obtain equations for say, temperature by doing the following partial
$$T = \frac {\partial U}{\partial S} |_V$$
Does this mean that this equation for temperature is only valid in a system where volume is held fixed? That is to say, can I only use the equation obtained in relation to experiments whereby volume doesn't change throughout?

My intuition tells me that it's not the case, and that it's simply stating that we are doing a partial derivative of $U$ with respect to $S$, while leaving the $V$ variable untouched. However I keep getting a niggly feeling that I'm wrong.

Assistance is greatly appreciated.

 

[Charles Link]

When $U=U(S,V)$ , for small changes: $\\$ $\Delta U=(\frac{\partial{U}}{\partial{S}})_V \Delta S+ (\frac{\partial{U}}{\partial{V}})_S \Delta V=T \Delta S+(\frac{\partial{U}}{\partial{V}})_S\Delta V$. $\\$
When the volume is not fixed, this last equation applies. $\\$
Your question is somewhat puzzling, but maybe this helps answer it.
Also note $(\frac{\partial{U}}{\partial{V}})_S=-P$. $\\$ $(\frac{\partial{U}}{\partial{S}})_V=T$ always, but it may be impossible to isolate this if you are looking at changes in internal energy $\Delta U$ w.r.t. a change in entropy $\Delta S$ (not readily measured, but let's assume we can somehow monitor it and/or make it change by a specified small increment) and the volume is changing. You would also need to measure the pressure $P$ and the change in volume $\Delta V$.

 

[WWCY]

Thank you for the reply,

however, I don't get what you mean by this statement. Do you mind elaborating?

$\\$
When the volume is not fixed, this last equation applies. $\\$

Also, let me attempt to rephrase my question. We can obtain pressure by two partial derivatives $p = -\Big(\frac{\partial U}{\partial V} \Big)_S$ and $p = -\Big(\frac{\partial F}{\partial V} \Big)_T$. If I have some partition function, and attempt to get an expression for pressure using these two derivatives, will I get the same equations for pressure?

And finally, suppose that I am considering a system with fixed temperature, and I wish to compute pressure using $p = -\Big(\frac{\partial U}{\partial V} \Big)_S$ rather than $p = -\Big(\frac{\partial F}{\partial V} \Big)_T$, will I get the right expression? Which is to say, if I evaluate a partial derivative with some thermodynamic potential held fixed, does it mean that I have to be considering an actual system in which that thermodynamic potential is fixed too?

Apologies if I sound extremely confused, thank you for your patience.

 

[Charles Link]

If I remember correctly $F=-kT \ln{Z}$. The second equation for the pressure is often easy to compute from this.
The first equation for the pressure says we do a reversible expansion with no heat flow into or out of the system. For an ideal gas, the temperature will drop, and thereby $U$ decreases. The pressure can be computed as $p=-\frac{\Delta U}{\Delta V}$. It is basically being computed by the work the gas does on the surroundings, and thereby the energy it loses.
I don't recall how $U$ would be computed from the partition function $Z$ though, especially if it needs to be a function of $S$ and $V$. I think that calculation would be clumsy and difficult. $\\$
@Chestermiller Might you have any additional inputs?

 

[Chestermiller]

We have that $$dF=-SdT-PdV$$ so we know that $$P=-\left(\frac{\partial F}{\partial V}\right)_T$$ and $$S=-\left(\frac{\partial F}{\partial T}\right)_V$$From this, it follows that $$-\frac{\partial ^2 F}{\partial V\partial T}=-\frac{\partial ^2 F}{\partial T\partial V}=\left(\frac{\partial S}{\partial V}\right)_T=\left(\frac{\partial P}{\partial T}\right)_V$$The latter equality is a so-called Maxwell relationship, from which it can be shown that $$dU=C_vdT-\left[P-T\left(\frac{\partial P}{\partial T}\right)_V\right]dV$$This is essentially the relationship that @Lord Jestocost posted in another thread initiated by the OP.