Question

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)?

Short answer

The quantum volume for an $N_2$molecule at room temperature is $6.9\times {10}^{-33}{m}^3$. The temperature of system at which quantum statistics become relevant is $5.681K$.

Step-by-step solution

Step 1. Formula quantum volume

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.

Step 2. Calculation quantum volume

Mass of $N_2$ is calculated as:

role="math" localid="1647249031293" $$\begin{gathered} m=(28u)\left(\frac{1.67\times {10}^{-27}kg}{1u}\right) \\ =46.48\times {10}^{-27}kg \end{gathered}$$

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$6.9\times {10}^{-33}{m}^3$.

Step 3. Calculation ideal gas equation

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

Step 4. Calculation temperature

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$5.681K$.