# Thermodynamics, Exact differentials

by manimaran1605
Tags: differentials, exact, thermodynamics
 Sci Advisor P: 2,210 Here's a long-winded description of the difference between exact differentials and inexact differentials: A generalized differential can be written in the form: $\sum_j Q_j dx_j$ where the $x_j$ are independent state variables, and $Q_j = Q_j(x_1, x_2, ...)$ are functions of those state variables. In contrast, an exact differential is the special case where there is a single state function $F(x_1, x_2, ...)$ and $Q_j = \dfrac{\partial F}{\partial x_j}$ In that special case, $\sum_j Q_j dx_j = dF$, an exact differential. How do you know whether there is such an $F$? Well, you can figure it out by using a special property of partial derivatives, which is that the order of differentiation doesn't matter: $\dfrac{\partial}{\partial x} \dfrac{\partial}{\partial y} F = \dfrac{\partial}{\partial y} \dfrac{\partial}{\partial x} F$ In terms of the $Q_j$, this means that: $\dfrac{\partial Q_j}{\partial x_k} = \dfrac{\partial Q_k}{\partial x_j}$ Now, relating all this back to the question about work: $dW = -P dV$ A complete set of independent state variables for a monoatomic gas is volume, temperature and number of particles: $V, T, N$. So to make an exact differential out of $-P dV$, you would have to add other terms, to get something like: $d ? = -P dV + Q_1 dT + Q_2 dN$ where the mixed derivatives work out: $\dfrac{\partial Q_1}{\partial V} = - \dfrac{\partial P}{\partial T}$ $\dfrac{\partial Q_2}{\partial V} = -\dfrac{\partial P}{\partial N}$ $\dfrac{\partial Q_1}{\partial N} = \dfrac{\partial Q_2}{\partial T}$ One set of choices that work out are: $Q_1 = S$ (entropy) $Q_2 = \mu$ (chemical potential) With these choices, our inexact differential $dW$ is turned into the exact differential, $dU = =Pdv + S dT + \mu dN$