Question

In the text I showed that for an Einstein solid with three oscillators and three units of energy, the chemical potential is $\mu =-ϵ$(where $ϵ$is the size of an energy unit and we treat each oscillator as a "particle"). Suppose instead that the solid has three oscillators and four units of energy. How does the chemical potential then compare to $-ϵ$ ? (Don't try to get an actual value for the chemical potential; just explain whether it is more or less than $-ϵ$.)

Short answer

$\mu <-\in$

Step-by-step solution

Explanation of Solution

Given:

For $\mathrm{N}=3$and $\mathrm{q}=3$the chemical potential is $\mu =-\in$

Calculation

The chemical potential formula is

$\mu ={\left(\frac{\Delta U}{\Delta N}\right)}_S$

The entropy for $\mathrm{N}=3$and $\mathrm{q}=3$is

$S=k\ln \Omega$

$S=k\ln \frac{\mid N+q-1}{\frac{|q-1|q}{\mid 3+3-1}}$

$S=k\ln \frac{|3-1|3}{\underset{¯}{\mid 5}}$

$S=k\ln \frac{\left|2\right|3}{\frac{\left|2\right|}{\left|2\right|}}$

$S=k\ln \left(10\right)$

The entropy must be constant throughout.

The entropy for $\mathrm{N}=3$and $\mathrm{q}=4$is,

$\begin{array}{r}S=k\ln \mathrm{\Omega }\\ S=k\ln \frac{\mid N+q-1}{\frac{|q-1|q}{\mid 3+3-1}}\\ S=k\ln \frac{|3-1|3}{\underset{\_}{\mid 5}}\\ S=k\ln \frac{\left|2\right|3}{\frac{\left|2\right|}{\left|2\right|}}\\ S=k\ln \left(10\right)\end{array}$

Thus, the entropy increases so to reduce the entropy to its original value.

Conclusion 

The chemical potential is lowered.

$\mu <-\in$