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I remember an exact differential as:

$$A=M(x,y)dx+N(x,y)dy $$

and the condition for be exact is:

$$\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}. $$

Can I use that definition to proof that $dW=pdv$ is not an exact differential?

I was thinking in use $W=W(p,V)$ and calculate

$$dW=\frac{\partial W}{\partial p}dp+\frac{\partial W}{\partial V}dV$$

and try to find a way to refute the idea of an exact differential for $pdV$. Am I right?

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  • $\begingroup$ is W function of p and V though? You formula for exact differential works only if p and V are independent variables. $\endgroup$
    – Umaxo
    May 10, 2019 at 4:36
  • $\begingroup$ If it was the exact differential, cars, steam engines, refrigerators, heat pumps and air conditioning would not work. $\endgroup$
    – Poutnik
    May 10, 2019 at 15:51

4 Answers 4

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Work depends on the path between final and initial states, so by stating $W=W(P,V)$ you are ignoring that path dependence. Work isn't an exact differential because it's not only a function of variables; it's also a function of path.

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  • $\begingroup$ I think OP is going for a proof by contradiction — assuming $\mathrm{d}W$ is integrable, then working their way to a contradiction. $\endgroup$
    – Bob Knighton
    May 10, 2019 at 7:53
  • $\begingroup$ My 2 cents: it's difficult to take a physics definition of work and contort it such that it fits a random math definition. Math can be purely theoretical but physics is constrained by the fact that it is supposed to model the real world. $\endgroup$
    – David White
    May 10, 2019 at 15:42
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You could do that - you're basically right there. What would $M$ and $N$ be, and would they satisfy the condition you quote?

Here's an alternative way to look at it. If there were some function $F$ such that $dF = pdV$, then it would follow that the integral of $pdV$ along any contour in the $(p,V)$ plane would be

$$\int_A^B p dV = \int_A^B dF = F(B)-F(A)$$

and in particular, the integral of $pdV$ along any closed contour would be zero. Is that the case?

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  • $\begingroup$ @Umaxo If you aren't treating $p$ and $V$ as independent variables, then it doesn't make sense to talk about arbitrary curves in the $(p,V)$ plane. Imposing a particular functional relation $p=p(V)$ amounts to choosing a specific (open) curve, in which case obviously integrating from one end to the other and back again will yield zero, but only if the integral is zero for any closed curve (which it obviously isn't) is the quantity $pdV$ an exact differential of a state function $F=F(p,V)$. $\endgroup$
    – J. Murray
    May 10, 2019 at 5:43
  • $\begingroup$ how would you know the integral is zero? Usually p,V are not independent variables and if it would hold p=p(V), then such function F indeed exists. That would be the case for the system where internal energy does not depend on enthropy. So is there some reason why such system shouldnt exist? $\endgroup$
    – Umaxo
    May 10, 2019 at 5:46
  • $\begingroup$ Yes it doesnt. But p is defined as partial derivative of internal energy in respect to volume, when enthropy and all other state coordinates are fixed. So p is not an independent variables, unless you change your coordinates. But than, would W be still defined as pdV? I dont think so, because that would mean W=pV +f(p) (or not?) and somehow i dont think this is the most general form of W. $\endgroup$
    – Umaxo
    May 10, 2019 at 5:50
  • $\begingroup$ @Umaxo You are, as far as I can tell, arguing that it is not possible to change a system's pressure and volume independently. The existence of isobaric and isochoric processes would suggest that this is incorrect. $\endgroup$
    – J. Murray
    May 10, 2019 at 5:55
  • $\begingroup$ Existence of isochoric process is given by the fact, that internal energy depends on the enthropy. Without this dependence the process would be impossible. So the question is back to the first one, wheter it makes sense to have internal energy that is not enthropy dependent. But i guess not since enthropy is in TD and statistical physics so important...I think i ran too far though, i guess it is better i stopped discussing it here. $\endgroup$
    – Umaxo
    May 10, 2019 at 6:11
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In order for dW=PdV to be an exact differential, if you had two points in P,V space, $(P_1,V_1)$ and $(P_2,V_2)$, the integral of dW=PdV would have to be independent of the path between these two points. If you could think of more than just one single path where the integral differs, then dW could not be an exact differential. It is, of course, very easy to do this.

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In a common form we have that inner parameters of system $b_k$ defined by outer parameters and themperature: $$b_k=f_k(a_1, ... ,a_n; T), \qquad(1)$$ where $a_i$ - outer parameters; $T$-temperature. You can also write in this way so called thermal equations of state $$A_i=A_i(a_1, ... ,a_n, T),\qquad(2)$$ where $A_i$ - generalized forces conjugates with outer parameters $a_i$. In common expression for elementary work $$\delta W = \sum_{i}{A_i d{a_i}} \qquad(3)$$ we have no differential of temperature (as if coefficient before $d{T}$ equals to zero). So if expression (3) is full differential of someone function of state we will have $$\frac{\partial A_i}{\partial T}=\frac{\partial 0}{\partial a_i}=0. \qquad (compare\ with\ your \quad \frac{\partial M}{\partial y}=\frac{\partial N}{\partial x} ) $$ Last means that generalized forces (e.g. pressure) do not depend on temperature. And this contradicts base thermodynamic consumption that (2) exist.

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