10.1: Exact Differentials

Last updated
Save as PDF

Page ID: 238798

\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\)

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.

Contributors and Attributions

Patrick E. Fleming (Department of Chemistry and Biochemistry; California State University, East Bay)