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Question

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

Solution

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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:\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:}

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

$\textnormal{So, we also know from chapter 4.A4.A that for one component system or to be more exact, for a pure substance, chemical potential is equal to molar Gibbs energy.}$

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

dGm=dμdG_m = d\mu

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

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

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

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

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

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

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

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

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