Q.7.14 - An Introduction to Thermal Physics

Question

For a system of particles at room temperature, how large must $ϵ-\mu$ be before the Fermi-Dirac, Bose-Einstein, and Boltzmann distributions agree within $1\%$ ? Is this condition ever violated for the gases in our atmosphere? Explain.

Step-by-Step Solution

Answer

As far as gases in the atmosphere are concerned, the condition was never violated.

Step 1: Given Information

The Fermi-Dirac, Bose-Einstein and Boltzmann distribution lies within $1\%$ of $1.$

Step 2: Explanation

The quantum volume expression is:

$v_{\mathrm{Q}}={\left(\frac{h}{\sqrt{2\pi mkT}}\right)}^3$

$h=$ Planck's constant,

$m=$mass of gas molecule,

$k=$gas constant

$T=$temperature

$v_Q=$quantum volume.

The pressure of gas molecule will be denoted by,

$P_0=\frac{kT{Z}_{\text{inm }}{e}^{\mu /TT}}{v_{\ell }}$

$Z_{\text{int }}=$partition function,

$\mu =$chemical potential

$P_0=$pressure at fixed temperature.

 Substitute $2{\left(\frac{h}{\sqrt{2\pi mkT}}\right)}^3$for $v_Q$ in above expression,

Rearrange the terms,

$-\mu /kT=\ln \left[\frac{kT{Z}_{\text{int }}}{P_0}{\left(\frac{\sqrt{2\pi mkT}}{h}\right)}^3\right]\dots \dots ..\left(1\right)$

Step 3: Explanation

The Bose-Einstein distribution's expression is:

${\overline{n}}_{\mathrm{BE}}=\frac{1}{e^{(x-\mu )k{T}_{-1}}}$

$\epsilon =$ energy at any state and

${\overline{n}}_{\mathrm{BE}}=$Bose-Einstein distribution.

The Fermi-Dirac distribution's expression is:

${\overline{n}}_{\mathrm{FD}}=\frac{1}{e^{(k-\mu )kT}+1}$

${\overline{n}}_{\mathrm{FD}}$Bose-Einstein distribution.

Equation (I) divided by equation (II) is:

$\begin{array}{r}\frac{{\overline{n}}_{\mathrm{BE}}}{{\overline{n}}_{\mathrm{FD}}}=\frac{e^{(x-\mu )dT}+1}{e^{(x-\mu )XT}-1}\\ =\frac{1+e^{-(c-\mu )kT}}{1-e^{-(c-\mu )kT}}\\ \approx 1+2e^{-(\epsilon -\mu )/kT}\end{array}$

Here, $(\epsilon -\mu )/kT\gg 1$so, the above approximation is valid.

Step 4: Explanation

Calculation:

The value of $e^{-(\epsilon -\mu )/kT}$ Should be less than $1/200$for the ratio to be within $1\%$ of 1 ; it means the value $(\epsilon -\mu )/kT>\ln 200=5.3$.

The value of $\epsilon$ cannot be negative.

So, $-\mu /kT$must be greater than $5.3$

Substitute $1.38\times {10}^{-23} \mathrm{J}/\mathrm{K}=\mathrm{k}$ 

$300K=T$

$4.65\times {10}^{-26}\mathrm{kg}=m$

$6.63\times {10}^{-34}\mathrm{J}.\mathrm{s}$

$50=Z_{int}$

${10}^{5}Pa=P$ in equation (I).

$\begin{array}{r}-\mu /kT=\ln \left[\begin{array}{c}\frac{\left(1.38\times {10}^{-23}\mathrm{J}/\mathrm{K}\right)\left(300\mathrm{K}\right)\left(50\right)}{{10}^5\mathrm{Pa}}\\ {\left(\frac{\sqrt{2\pi \left(4.65\times {10}^{-26}\mathrm{kg}\right)\left(1.38\times {10}^{-23}\mathrm{J}/\mathrm{K}\right)\left(300\mathrm{K}\right)}}{6.63\times {10}^{-34}\mathrm{J}\cdot \mathrm{s}}\right)}^3\end{array}\right]\\ =\ln \left(3\times {10}^8\right)\\ =\ln 3+8\ln 10\\ =19.52\end{array}$

This indicates that the condition is valid since $-\mathrm{mu}/\mathrm{kT}$ is greater than $5.3.$

No violation of the condition occurred as far as atmospheric gases are concerned.