1
$\begingroup$

In one of my thermodynamics lectures, I came across something of the from $S = \int \frac{dU+pdV}{T}$ which I know to be a line integral in differential form. I saw that in a problem this was simplified to $S= \int\frac{dU}{T}+\int\frac{pdV}{T}$. I was under the impression that we cannot simply separate the terms like this for a line integral. Is this a consequence of the fact that $S$(entropy) is a proper differential?

$\endgroup$
1
  • 3
    $\begingroup$ You can always do that, because integrals are linear: $\int f(x) + g(x) \, dx = \int f(x) \, dx + \int g(x) \, dx$. $\endgroup$
    – knzhou
    Oct 26, 2020 at 17:01

2 Answers 2

4
$\begingroup$

$S$ is a function of state, so as long as we move from state $A=(U_0,V_0)$ to state $B=(U_1,V_1)$ in a reversible way, the change in entropy $\Delta S = S(B) - S(A)$ is independent of the path taken. So we can choose to go from $A$ to $B$ via $C=(U_1,V_0)$. As we go from $A$ to $C$ we keep $V=V_0$ constant, and as we go from $C$ to $B$ we keep $U=U_1$ constant. Then we can see that

$\displaystyle \Delta S = \int_A^B \frac{dU+p \space dV}{T} = \int_A^C \frac{dU+p\space dV}{T} +\int_C^B \frac{dU+p\space dV}{T} = \int_A^C \frac{dU}{T} +\int_C^B \frac{p\space dV}{T}$

$\endgroup$
0
$\begingroup$

If you wanna be technical about it, you can think of saying that U is some function of temperature, and the $ \frac{P}{T}$ ratio is some function of volume.

For an ideal gas, We can use the ideal gas law to turn the function which we are integrating into one of volume.

$$ \frac{P}{T} = \frac{nR}{V}$$

However, if you think of it from a purely mathematical standpoint, we are actually turning $dq$ into an exact differential via an integrating factor to arrive at a state function. Putting more precisely, if $S(U,V)$ then(*):

$$ (\frac{\partial S}{\partial U})_V = \frac{1}{T}$$

And,

$$ (\frac{\partial S}{\partial V})_U = \frac{P}{T}$$

The existence due to this condition(something about...curl?):

$$ \frac{ \partial }{\partial V} (\frac{\partial S}{\partial U})_U = \frac{ \partial}{\partial T}(\frac{\partial S}{\partial U})_V $$

Working out lhs:

$$ \big[ \frac{ \partial }{\partial V} (\frac{1}{T})\big]_U = \big[\frac{ \partial }{\partial V} (\frac{nC_v}{U}) \big]_U =0$$

And, RHS:

$$ \big[ \frac{ \partial}{\partial T}(\frac{\partial S}{\partial U})_V \big]= \big[ \frac{ \partial}{\partial T} \frac{P}{T} \big]_V = \big[\frac{ \partial}{\partial T} \frac{nR}{V} \big]_V = 0$$

Hence $dS$ is an exact differential and we must be able to find a state function of it.

Edit: A point to note maybe that when you have a process, the variables all change together that is a state maybe specified by the variables of $(P,V,T)$ so you can think of the integration as going from a state with state variables $(P,V,T)$ to one with say something like $(P',V',T')$


*: Those variables let us equate the partials when we take differential For proving that $ \frac{1}{T}$ is integrating factor: See this question I asked here

Might want to see gradient theorem and independence of path conditions wiki

About the point on curl, I had made a stack post about in math stack exchange here

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.