Show that equation 5.40 is in agreement with the explicit formula for the chemical potential of a monatomic ideal gas derived in Section 3.5. Show how to calculate $\mu °$for a monatomic ideal gas.

The explicit formula for a monoatomic ideal gas's chemical potential, as well as the value of ideal chemical potential.

The chemical potential of the gas expression is given below.

$\mu ={\left(\frac{\partial G}{\partial N}\right)}_{T,P}$

where

role="math" localid="1648236173152" $$\begin{gathered} \mu \text{is the chemical potential,}G\text{is the Gibbs energy and} \\ N\text{is the number of gas molecules.} \end{gathered}$$

When more particles are added to a system while the temperature and pressure remain constant, the system's Gibbs energy increases by $\mathit{\mu }$

Write the expression for chemical potential in terms of standard chemical potential at atmospheric pressure.

$\mu (T,P)={\mu }^0\left(T\right)+kT\ln \left(P/P^0\right)\dots \dots \dots \left(1\right)$

Here, ${\mu }^0\left(T\right)$is the standard chemical potential, $k$is Boltzmann constant, $T$is the absolute temperature and $P^0$is the atmospheric pressure.

The ideal gas equation is

$PV=NkT$

Here, $P$is the pressure of the gas,

$V$is the volume,

$k$is Boltzmann constant and

$T$is the absolute temperature.

Now, the chemical potential of an ideal gas expression is given below:

$\mu =-kT\ln \left[\frac{V}{N}{\left(\frac{2\pi mkT}{h^2}\right)}^{3/2}\right]\dots \dots ..\left(2\right)$

$\text{Substitute}\left(\frac{KT}{P}\right)\text{for}\left(\frac{V}{N}\right)\text{in expression (2).}$

$$\begin{gathered} \mu =-kT\ln \left[\frac{kT}{P}{\left(\frac{2\pi mkT}{h^2}\right)}^{3/2}\right] \\ =-kT\ln \left[\frac{{\displaystyle kT{p}^0}}{{\displaystyle P{P}^0}}{\left(\frac{{\displaystyle 2\pi mkT}}{{\displaystyle h^2}}\right)}^{3/2}\right] \\ =-kT\ln \left[\frac{kT}{p^0}{\left(\frac{2\pi mkT}{h^2}\right)}^{3/2}\right]-kT\ln \left(\frac{p^0}{P}\right) \end{gathered}$$

Simplify the above expression

$\mu =-kT\ln \left[\frac{kT}{p^0}{\left(\frac{2\pi mkT}{h^2}\right)}^{3/2}\right]+kT\ln \left(\frac{p}{p^0}\right)\dots \dots ..\left(3\right)$

The equation (1) and (3) are similar to each other.

${\mu }^0=-kT\ln \left[\frac{kT}{p^0}{\left(\frac{2\pi mkT}{h^2}\right)}^{3/2}\right]$

Substitute the above expression in (3)

$\mu ={\mu }^0+kT\ln \left(\frac{P}{p^0}\right)$

Therefore explicit formula for the chemical potential of a monoatomic ideal gas is

$\mu =-kT\ln \left[\frac{kT}{p^0}{\left(\frac{2\pi mkT}{h^2}\right)}^{3/2}\right]+kT\ln \left(\frac{p}{p^0}\right)$and the value of$u^0\text{is}\left(-kT\ln \left[\frac{kT}{p^0}{\left(\frac{2\pi mkT}{h^2}\right)}^{3/2}\right]\right)$