Chemistry - Why is dU an exact differential and dq an inexact differential?

Solution 1:

As opposed to an exact differential, an inexact differential cannot be expressed as the differential of a function, i.e. while there exist a function $U$ such that $U = \int \mathrm{d} U$, there is no such functions for $\text{đ} q$ and $\text{đ} w$. And the same is, of course, true for any state function $a$ and any path function $b$ respectively: an infinitesimal change in a state function is represented by an exact differential $\mathrm{d} a$ and there is a function $a$ such that $a = \int \mathrm{d} a$, while an infinitesimal change in a path function $b$ is represented by an inexact differential $\text{đ} b$ and there is no function $b$ such that $b = \int \text{đ} b$.

Consequently, for a process in which a system goes from state $1$ to state $2$ a change in a state function $a$ can be evaluated simply as $$\int_{1}^{2} \mathrm{d} a = a_{2} - a_{1} \, ,$$ while a change in a path function $b$ can not be evaluated in such a simple way, $$\int_{1}^{2} \text{đ} b \neq b_{2} - b_{1} \, .$$ And for a state function $a$ in a thermodynamic cycle $$\oint \mathrm{d} a = 0 \, ,$$ while for a path function $b$ $$\oint \text{đ} b \neq 0 \, .$$ The last mathematical relations are important, for instance, for the first law of thermodynamics, because while $\oint \text{đ} q \neq 0$ and $\oint \text{đ} w \neq 0$ it was experimentally found that $\oint (\text{đ} q + \text{đ} w) = 0$ for a closed system, which implies that there exist a state function $U$ such that $\mathrm{d} U = \text{đ} q + \text{đ} w$.

Solution 2:

Not a complete answer, but the path is exactly what it sounds like. Say you are rolling a boulder up a hill. This increases its potential energy, which can be released by rolling the boulder down the hill. But the path you take to get to the top of the hill is irrelevant, only the height you raise the boulder matters for potential energy. So if you roll it up half way, let it fall back a quarter of the way, or any such combination of forward and back, none of this matters to the total change in potential energy (assuming perfect conditions, no friction, etc.).

With respect to Wildcat's great answer, this means for state functions the endpoints of your definite integral are all that matter: you could parameterize any path you want between the endpoints and the resulting (line) integral is the same.