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Multiple choice physics behaviour of perfect gas and kinetic theory of gases mean free path law of equipartition of energy and mean free path behavior of perfect gas and kinetic theory

There are two vessels of same consisting same no of moles of two different gases at same temperature . One of the gas is $CH _{4}$ & the other is unknown X. Assuming that all the molecules of X are under random motion whereas in $CH _{4}$ except one all are stationary. Calculate $Z _{1}$ for X in terms of $Z _{1}$ of $CH _{4}$. Given that the collision diameter for both gases are same & $\displaystyle (U _{rms}) _{x}=\frac{1}{\sqrt{6}}(Uav) _{CH _{4}}$.

  1. $\displaystyle \frac{2\sqrt{2}}{3\sqrt{\pi }}Z _{1}$

  2. $\displaystyle \frac{3\sqrt{2}}{2\sqrt{\pi }}Z _{1}$

  3. $\displaystyle \frac{2\sqrt{3}}{2\sqrt{\pi }}Z _{1}$

  4. $\displaystyle \frac{4\sqrt{2}}{3\sqrt{\pi }}Z _{1}$
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A Correct answer
Explanation

V, n, T $\rightarrow  same$(25) so $P\rightarrow $ also same ( P  5  25)
$\displaystyle \sigma \rightarrow same (25)$
given

$\displaystyle (v {rms})\times

x=\dfrac{1}{\sqrt{6}}(v _{avg.}) _{CH _{4}}$ &

$v _{rms}=\sqrt{\dfrac{3\pi }{8}}(v _{avg.})$ so
$\displaystyle \sqrt{\dfrac{3\pi }{8}}(v _{avg.}) _{CH _{4}}$
$\displaystyle \dfrac{(v _{avg.})x}{(v _{avg.}) _CH _{4}}=\sqrt{\dfrac{8}{3\pi }}.\frac{1}{\sqrt{6}}=\dfrac{2}{3\sqrt{\pi }}$
For X (9< ) : $\displaystyle Z _{1}=\sqrt{2}\pi \sigma ^{2}(v _{avg.}) _{x}N^{\ast }$
For CH{4} (9< ) : $\displaystyle Z _{1}=\pi \sigma ^{2}(v _{avg.}) _{CH _{4}}N^{\ast }$
Since T, P, v, n are same, $N\ast $ will also be same.
$\displaystyle



\frac{Z _{1}X}{Z _{1}(CH _{4})}=\sqrt{2}\frac{(v _{avg.}) _{x}}{(v _{avg.}) _{CH _{4}}}=\sqrt{2}.\frac{2}{3\sqrt{\pi

}}$
$\displaystyle Z _{1}(X)=Z _{1}(CH _{4}).\frac{2\sqrt{2}}{3\sqrt{\pi }}$