Integrating Differentials of Two Variables

Question

In a thermodynamics text I'm reading, the author says the following,

"The integral of the perfect differential $dz = y\,dx + (x+c) \,dy \; (\mathrm{where} \; z = xy + cy)$ from point (0,0) to pint $(a,b)$, over the path $(0,0)$ to $(0,b)$ to $(a,b)$ is

$$ c\cdot \int_0^b dy + b\cdot \int_0^adx = cb + ab. \quad (1)"$$

Later in the text - and presumably using the same integration rule - says the following (here I'm paraphrasing),

"Integrating $dS = \frac{3}{2} nR/T) dT + (nR/V) dV$ gives $$ S = nR \ln[(T/T_0)^{3/2}(V/V_0)] + S_0 \quad (2)$$

where $S = S_0$ when $T = T_0$ and $V = V_0.$"

My question is - where does this come from? I've taken two semesters of analysis and I haven't seen anything like this. In the case of (1), wouldn't one have to integrate the right-hand side with respect to both $x$ and $y$? If so, the resultant should have two terms, each of which being functions of both $x$ and $y$. Why can I integrate each term separately with respect to its variable? Where's the justification?

Answer 1

The integral of a one-form $Pdx + Qdy$ over a curve $\gamma(t) = (x(t), y(t))$, parameterized for $0 \leq t \leq 1$, is by definition $$ \int_\gamma (P dx + Q dy) = \int_0^1 \left(P(x(t),y(t)) x'(t) + Q(x(t),y(t)) y'(t) \right) dt $$ In other words, it is technically an integral over $t$, the parameter defining the curve. In your example (1), the curve $\gamma$ consists of two line segments which are horizontal and vertical. For the first part, from $(0,0)$ to $(0,b)$, you can take $\gamma(t) = (x(t), y(t)) = (0, t)$ for $0 \leq t \leq b$. In this case, $y(t) = t$, so the integral can be thought of as being over the variable "$y$". Similarly, the second segment is horizontal, so $x(t) = t$ in that case. Hopefully this answers your question:

Why can I integrate each term separately with respect to its variable? Where's the justification?

On a side note: computing this integral explicitly is unnecessary, since by the fundamental theorem, $$ \int_\gamma dz = z(a,b) - z(0,0) $$

Answer 2

The entropy $S$ of a system is a state function and so the differential $dS$ is an exact differential, given by:

$$ dS = \left( \frac{\partial S}{\partial T} \right)_V dT + \left( \frac{\partial S}{\partial V} \right)_T dV $$

Or, for ideal gases:

$$ dS = \frac{C_V}{T} dT + \frac{nR}{V} dV $$

We can verify it is an exact differential by the symmetry in second derivates, i.e. $\frac{\partial S}{\partial T \partial V} = \frac{\partial S}{\partial V \partial T} = 0$. Since $S$ is a state function it takes the form of differentiable scalar function $S(V,T)$ and so we may write:

$$ \Delta S = \int_i^f dS = \int_C \nabla S(\textbf{r}) \cdot d \textbf{r} = \int_C \nabla S(\textbf{r}(t)) \cdot \textbf{r}'(t) dt $$

Which is a line integral along some curve $C$ represented by $\textbf{r}(t)$. Now $\nabla S$ is conservative, as shown by the symmetry of second derivatives, and so the integral is path independent according to the gradient theorem. Since the integral is path independent we can integrate along any path we like, e.g. integrating $\frac{C_V}{T}$ at constant $V$ and $\frac{nR}{V}$ at constant $T$. This means we may write:

$$ \Delta S = \int_i^f \frac{C_V}{T} dT + \int_i^f \frac{nR}{V} dV $$

And so we conclude:

$$ \Delta S = C_V \ln \frac{T}{T_0} + nR \ln \frac{V}{V_0}$$

Or for monatomic gases:

$$ \Delta S = nR \ln \left( \left( \frac{T}{T_0} \right)^{\frac{3}{2}} \frac{V}{V_0} \right)$$