91Ó°ÊÓ

Because nitrogen is a diatomic gas, the thermal energy is calculated as follows:

$E=\frac{5}{2}N{K}_{B}T$

Now we know that , $pV=N{K}_{B}T$

From equation localid="1648381889785" $\left(1\right)$and$\left(2\right)$we get that

$E=\frac{5}{2}pV$

Given that$p=100\mathrm{atm}=1.01×{10}^7\mathrm{N}/{\mathrm{m}}^2$and $V=15000{\mathrm{cm}}^3=0.015{\mathrm{m}}^3,$, then substituting these to the formula:

localid="1648390595486" $E=\frac{5}{2}×1.01×{10}^7×0.015=379\mathrm{kJ}.$

The mean free path is given by

$\mathrm{λ}=\frac{1}{4\sqrt{2}\mathrm{π}(N/V)r^2}$

Now we have

$\frac{N}{V}=\frac{p}{K_{B}T}=\frac{1.01×{10}^7}{1.38×{10}^{−23}×293}=2.5×{10}^{27}$

and $r$for nitrogen is $r=1×{10}^{−10}$,then the mean free path will be:

$\begin{array}{r}\mathrm{λ}=\frac{1}{4\sqrt{2}\mathrm{π}×{10}^{27}×{\left(1×{10}^{−10}\right)}^2}\\ =2.25×{10}^{−9}\mathrm{m}\\ =2.25\mathrm{nm}.\end{array}$

But even though the formula shows that thermal energy is proportional to the number of gas molecules, degrees of freedom, and gas temperature, none of these factors change in this isothermal expansion.

As a result, the total amount of thermal energy will remain unchanged. Thermal energy will remain unchanged.