Two gases-argon (atomic radius 0.07 ; nm,atom | vedclass

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(A) The mean free path $\lambda$ is given by $\lambda = \frac{1}{\sqrt{2} \pi n_{v} d^{2}}$,where $n_{v}$ is the number density and $d$ is the diamete

Vedclass · Accepted Answer

$1.09$

Vedclass · Answer

(A) The mean free path $\lambda$ is given by $\lambda = \frac{1}{\sqrt{2} \pi n_{v} d^{2}}$,where $n_{v}$ is the number density and $d$ is the diameter of the atom.The mean free time $	au$ is defined as $	au = \frac{\lambda}{v_{rms}}$,where $v_{rms} = \sqrt{\frac{3RT}{M}}$.Substituting $\lambda$ and $v_{rms}$ into the expression for $	au$:$	au = \frac{1}{\sqrt{2} \pi n_{v} d^{2}} \sqrt{\frac{M}{3RT}}$.Since $n_{v}$,$R$,and $T$ are the same for both gases,the ratio of the mean free times $	au_{Ar} / 	au_{Xe}$ is:$\frac{	au_{Ar}}{	au_{Xe}} = \sqrt{\frac{M_{Ar}}{M_{Xe}}} 	imes \left( \frac{d_{Xe}}{d_{Ar}} ight)^{2}$.Given $M_{Ar} = 40$,$M_{Xe} = 140$,$d_{Ar} = 2 	imes 0.07 \; nm$,and $d_{Xe} = 2 	imes 0.1 \; nm$:$\frac{	au_{Ar}}{	au_{Xe}} = \sqrt{\frac{40}{140}} 	imes \left( \frac{0.1}{0.07} ight)^{2} = \sqrt{\frac{2}{7}} 	imes \left( \frac{10}{7} ight)^{2} \approx 0.5345 	imes 2.0408 \approx 1.09$.

Two gases-argon (atomic radius $0.07 \; nm$,atomic weight $40$) and xenon (atomic radius $0.1 \; nm$,atomic weight $140$) have the same number density and are at the same temperature. The ratio of their respective mean free times is closest to:

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