91影视

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

The quantum volume expression is:

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

$h=$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,

$P_0=\frac{kT{Z}_{\text{inm}}{e}^{渭/TT}}{v_{\mathrm{鈩�}}}$

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

$渭=$chemical potential

$P_0=$pressure at fixed temperature.

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

Rearrange the terms,

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

The Bose-Einstein distribution's expression is:

${\stackrel{炉}{n}}_{\mathrm{BE}}=\frac{1}{e^{(x鈭�渭)k{T}_{鈭�1}}}$

$蔚=$energy at any state and

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

The Fermi-Dirac distribution's expression is:

${\stackrel{炉}{n}}_{\mathrm{FD}}=\frac{1}{e^{(k-渭)kT}+1}$

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

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

$\begin{array}{r}\frac{{\stackrel{炉}{n}}_{\mathrm{BE}}}{{\stackrel{炉}{n}}_{\mathrm{FD}}}=\frac{e^{(x鈭�渭)dT}+1}{e^{(x鈭�渭)XT}鈭�1}\\ =\frac{1+e^{鈭�(c鈭�渭)kT}}{1鈭�e^{鈭�(c鈭�渭)kT}}\\ 鈮�1+2e^{鈭�(蔚鈭�渭)/kT}\end{array}$

Here, $(蔚鈭�渭)/kT鈮�1$so, the above approximation is valid.

Calculation:

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

The value of $蔚$cannot be negative.

So, $-渭/kT$must be greater than $5.3$

Substitute localid="1651131552143" $1.38脳{10}^{-23}\mathrm{J}/\mathrm{K}=\mathrm{k}$

$300K=T$

$4.65脳{10}^{鈭�26}\mathrm{kg}=m$

localid="1651131671206" $6.63脳{10}^{鈭�34}\mathrm{J}.\mathrm{s}$

$50=Z_{int}$

${10}^{5}Pa=P$in equation (I).

$\begin{array}{r}鈭�渭/kT=\ln 鈦�\left[\begin{array}{c}\frac{\left(1.38脳{10}^{鈭�23}\mathrm{J}/\mathrm{K}\right)\left(300\mathrm{K}\right)\left(50\right)}{{10}^5\mathrm{Pa}}\\ {\left(\frac{\sqrt{2蟺\left(4.65脳{10}^{鈭�26}\mathrm{kg}\right)\left(1.38脳{10}^{鈭�23}\mathrm{J}/\mathrm{K}\right)\left(300\mathrm{K}\right)}}{6.63脳{10}^{鈭�34}\mathrm{J}鈰�\mathrm{s}}\right)}^3\end{array}\right]\\ =\ln 鈦�\left(3脳{10}^8\right)\\ =\ln 鈦�3+8\ln 鈦�10\\ =19.52\end{array}$

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.