## Related questions with answers

What is the physical reason for the increase of the chemical potential of a pure substance as the pressure is raised?

Solution

Verified$\textnormal{In order to determine the change of chemical potential in respect to pressure, we have to remind ourselves of expression for the change in Gibbs energy which states:}$

$\begin{align*}dG = Vdp - SdT\tag{1}\end{align*}$

$\textnormal{So, we also know from chapter $4.A$ that for one component system or to be more exact, for a pure substance, chemical potential is equal to molar Gibbs energy.}$

$\textnormal{\textit{Note:}}$ $G_m =\frac{G}{n}$ $\textnormal{where $n$ is a number of moles of that particular substance}$

$dG_m = d\mu$

$\textnormal{So if we rewrite equation (1), i.e. divide everything with n we get:}$

$\begin{align*}d\mu = V_mdp - S_mdT\tag{2}\end{align*}$

$\textnormal{In the first case, if we arrange conditions of constant temperature, expression from equation (2) will get this shape:}$

$\begin{align*}d\mu = V_mdp\tag{3}\end{align*}$

$T=const.$, $dT = 0$

$\textnormal{In conditions of constant pressure, expression from equation (2) will have this kind a shape:}$

$\begin{align*}d\mu = - S_mdT\tag{4}\end{align*}$

$p=const.$, $dp = 0$

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