1
$\begingroup$

In thermodynamics I have come across a peculiar expression whose mathematical nature eludes me. Maybe one of you can help me finding out. First of all let us assume that in this scenario all premises of the Poincaré lemma are met.

We discussed the first theorem of thermodynamics and some applications: $$dU=\delta Q +\delta W.$$ Where '$d$' denotes an exact differential form and $'\delta'$ denotes an inexact one. Using the state equation $pV = nRT$ of an ideal gas and the definition of the work differential for ideal gases $\delta W = -pdV$ we derived this thing: $$\frac{\delta W}{dp}=-dV.$$

Physically speaking I understand what this equation means. However, mathematically I do not understand what the left-hand side actually means.

Also, given a differential form like $\delta Q=c_V\cdot dT$ one can calculate $\frac{\delta Q}{dp}=c_V\cdot\left(\frac{\partial T}{\partial p}\right)_V$.

Our professor said that this quotient is not a derivative but a difference quotient. But then I don't understand how in the second case this difference quotient comes out to be a partial derivative anyway.

I suppose it has to do with the fact that path integrals of inexact differential forms are path-dependent. However, as opposed to difference quotients of exact differentials I cannot imagine how to define a difference quotient of an inexact differential since those don't have an anti-derivative.

  • 0
    First of all, we should have $\delta W = p\,dV$ (I'm not sure where that sign came from). Second of all, the second displayed equation makes no sense. I think you meant just to have $\delta W/p$. Which quotient? I don't like writing expressions like $\delta Q/dp$. We should stay in the land of differential $1$-forms. – 2017-01-13

0 Answers 0