91Ó°ÊÓ

Formula for quantum volume is:

${\mathrm{ν}}_Q={\left(\frac{h}{\sqrt{2\mathrm{Ï€³¾°ì°Õ}}}\right)}^3$

where,$T$is temperature,role="math" localid="1647248715772" $m$is molecule mass,$h$is Planck's constant,$k$is Boltzmann constant.

Mass of $N_2$ is calculated as:

Substitute variables in quantum volume formula:

$$\begin{gathered} {\mathrm{ν}}_Q={\left(\frac{6.63×{10}^{-34}J·s}{\sqrt{2\mathrm{π}(46.48×{10}^{-27}kg)(1.38×{10}^{-23}J/K)(300K)}}\right)}^3 \\ ={\left(\frac{6.63×{10}^{-34}}{\sqrt{347.91×{10}^{-50}}}\right)}^3 \\ =6.9×{10}^{-33}{m}^3 \end{gathered}$$

So, quantum volume of$N_2$molecule is$6.9×{10}^{-33}{m}^3$.

Ideal gas equation is $PV=NkT$

So, $\frac{V}{N}=\frac{kT}{P}$

Substitute Pressure,$P=1.013×{10}^{5}N/m^2$, Temperature,$T=300K$.

role="math" localid="1647253225593" $$\begin{gathered} \frac{V}{N}=\frac{(1.38×{10}^{-23}J/K)(300K)}{1.013×{10}^{5}N/m^2} \\ =4×{10}^{-26}{m}^3 \end{gathered}$$

In quantum statistics, $\frac{V}{N}≈{\mathrm{ν}}_Q$

$$\begin{gathered} \frac{kT}{P}≈{\left(\frac{h}{\sqrt{2\mathrm{Ï€³¾°ì°Õ}}}\right)}^3 \\ T.T^{\frac{3}{2}}≈\frac{h^3}{{\left(2\mathrm{Ï€³¾°­}\right)}^{{\displaystyle \frac{3}{2}}}}×\frac{P}{k} \\ {T}^{\frac{5}{2}}≈\frac{P{h}^3}{{\left(2\mathrm{Ï€³¾}\right)}^{{\displaystyle \frac{3}{2}}}{k}^{{\displaystyle \frac{5}{2}}}} \\ T≈\frac{1}{k}{\left[\frac{P^2{h}^6}{{\left(2\mathrm{Ï€³¾}\right)}^3}\right]}^{\frac{1}{5}} \end{gathered}$$

Substitute the values of variables in above equation:

$$\begin{gathered} T≈\frac{1}{1.38×{10}^{-23}J/K}{\left[\frac{{(1.013×{10}^{5}N/m^2)}^2{(6.63×{10}^{-34}J·s)}^6}{{\left(2\mathrm{π}\right(28×1.67×{10}^{-27}kg\left)\right)}^3}\right]}^{\frac{1}{5}} \\ T=5.681K \end{gathered}$$

So, temperature of the system is$5.681K$.