# 10.1: Exact Differentials


In general, if a differential can be expressed as

$df(x,y) = X\,dx + Y\,dy$

the differential will be an exact differential if it follows the Euler relation

$\left( \dfrac{\partial X}{\partial y} \right)_x = \left( \dfrac{\partial Y}{\partial x} \right)_y \label{euler}$

In order to illustrate this concept, consider $$P(\overline{V}, T)$$ using the ideal gas law.

$P= \dfrac{RT}{\overline{V}}$

The total differential of $$P$$ can be written

$dP = \left( - \dfrac{RT}{\overline{V}^2} \right) dV + \left( \dfrac{R}{\overline{V}} \right) dT \label{Eq10}$

Example $$\PageIndex{1}$$: Euler Relation

Does Equation \ref{Eq10} follow the Euler relation (Equation \ref{euler})?

Solution

Let’s confirm!

\begin{align*} \left[ \dfrac{1}{\partial T} \left( - \dfrac{RT}{\overline{V}^2} \right) \right]_\overline{V} &\stackrel{?}{=} \left[ \dfrac{1}{\partial \overline{V}} \left( \dfrac{R}{\overline{V}} \right) \right]_T \\[4pt] \left( - \dfrac{R}{\overline{V}^2} \right) &\stackrel{\checkmark }{=} \left( - \dfrac{R}{\overline{V}^2} \right) \end{align*}

$$dP$$ is, in fact, an exact differential.

The differentials of all of the thermodynamic functions that are state functions will be exact. Heat and work, which are path functions, are not exact differential and $$dw$$ and $$dq$$ are called inexact differentials instead.