Compute the quantum volume for an $N_2$molecule at room temperature, and argue that a gas of such molecules at atmospheric pressure can be

treated using Boltzmann statistics. At about what temperature would quantum statistics become relevant for this system (keeping the density constant and pretending that the gas does not liquefy)?

Formula for quantum volume is:

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

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:

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

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

Ideal gas equation is $PV=NkT$

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

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

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

In quantum statistics, $\frac{V}{N}\approx {\nu }_Q$

$$\begin{gathered} \frac{kT}{P}\approx {\left(\frac{h}{\sqrt{2\mathrm{\pi mkT}}}\right)}^3 \\ T.T^{\frac{3}{2}}\approx \frac{h^3}{{\left(2\mathrm{\pi mK}\right)}^{{\displaystyle \frac{3}{2}}}}\times \frac{P}{k} \\ {T}^{\frac{5}{2}}\approx \frac{P{h}^3}{{\left(2\mathrm{\pi m}\right)}^{{\displaystyle \frac{3}{2}}}{k}^{{\displaystyle \frac{5}{2}}}} \\ T\approx \frac{1}{k}{\left[\frac{P^2{h}^6}{{\left(2\mathrm{\pi m}\right)}^3}\right]}^{\frac{1}{5}} \end{gathered}$$

Substitute the values of variables in above equation:

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

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