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?

  • 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


$S$ 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

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


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(*):

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


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

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


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