# When can we separate the terms of a line integral?

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?

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

$$S$$ 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(*):

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

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