Mean free path - class-XI

Initial momentum, $P _1=mvcos 30$
and final momentum, $P _2 = mvcos30$
change in momentum
$\Delta P = -2mv cos30$
$\Delta P =-\sqrt 3 mv$
Force on wall-1
$F _1=\frac {2mv}{\Delta t}$
Force on wall-2
$F _2=\frac {\sqrt 3mv}{\Delta t}$, so $F _1 > F _2$

The mean free path or average distance between collisions for a gas molecule may be estimated from kinetic theory.

The collision frequency $(f)$ is the ratio of rms velocity $(v _{rms})$ to mean free path $(\lambda)$.

The mean-free path of molecule is the distance traveled by a molecule in two consecutive collision. If pressure is reduced and there are less particle then a molecule will travel longer distance before collision, so mean free path is increased.

As gas pressure increases mean free path of the gas decreases. 
Mean free path is the distance traveled by a gas molecule between two successive collisions.
So, as pressure increases number of collisions increase. Hence, mean free path decreases.   

The expression for mean free path $\lambda =\dfrac { RT }{ \sqrt { 2 } \pi { d }^{ 2 }NP } $ mean free path is directly proportional to Temperature

The concept of mean free path is popular in all branches of physics and it has a number of applications including estimation of resistivity of material, designing a chemical apparatus, and optics and acoustics, etc.

The mean free path formula is lambda = 1 / (sqrt(2) * pi * n * d^2). Given n = 2.5 * 10^6 m^-3 and d = 0.24 * 10^-9 m, calculating this yields approximately 1.56 * 10^12 m.

Using the ideal gas law to find number density n = P/kT, then applying the mean free path formula lambda = 1 / (sqrt(2) * pi * n * d^2) with d = 2 * 10^-10 m, we get approximately 1.12 * 10^-7 m.

The mean free path in 3D is lambda = 1 / (sqrt(2) * pi * n * d^2). When considering motion along a single coordinate axis, the effective mean free path is reduced by a factor of sqrt(3) due to the projection of the velocity components.

Mean free path $\lambda=1.2 \times 10^{-7}$

The mean free path formula lambda = 1 / (sqrt(2) * pi * n * d^2) shows that it depends on both the molecular diameter (d) and the molecular number density (n).

Mean free path is the average distance between collisions for a gas and is given as $\lambda=\dfrac{RT}{\sqrt2\pi d^2N _AP}$

As atomicity of gas increases, its mean path decreases, also as attraction force between gas molecules increases, mean free path decreases so order is
$He > H _2 > O _2 > N _2 > CO _2$

$\displaystyle \lambda =2.6\times 10^{-5}:m, :\sigma =0.26:nm=2.6\times 10^{-10}m$
$\displaystyle T=300:K$
$\displaystyle \lambda =\frac{1}{\sqrt{2}\pi \sigma ^{2}N^{\ast }}$
$\displaystyle 2.6\times 10^{-5}=\frac{1}{\sqrt{2}\times 3.14\times (2.6\times 10^{-10})^{2}\times N^{\ast }}$
$\displaystyle N^{\ast }=1.281\times 10^{23}m^{-3}$
$\displaystyle N^{\ast }=\frac{P}{KT}$
$\displaystyle P=1.281\times 10^{23}\times 1.38\times 10^{-23}\times 300$
$\displaystyle P=530.3:Pa$

Collision Frequency is the number of times a molecule of a gas collides with other molecules. 

The mean free path estimated by the kinetic theory of gases is given by

The mean free path is the average distance traveled by a moving particle (such as an atom , a molecule, a photon) between successive impacts (collisions), which modify its direction or energy or other particle properties.

The mean free path of a molecule is given by the formula ($\lambda $)  $\cfrac{1}{\sqrt{2\pi d^2 n}}$ 

The mean free path is given by, $\lambda=\dfrac{1}{\pi \sigma n}$

Given,

Partition has hole of diameter $d$, Mean pressure between both sections is equal.

Boltzmann constant ${{K} _{B}}$

At constant pressure, Mean Free path $\lambda \ \alpha \ \sqrt{{{K} _{B}}T}$

Mean free path in ${{1}^{st}}$ section ${{\lambda } _{I}}=\sqrt{{{K} _{B}}\times 150}$

Mean free path in ${{2}^{nd}}$ section ${{\lambda } _{II}}=\sqrt{{{K} _{B}}\times 300}$

$\dfrac{{{\lambda } _{I}}}{{{\lambda } _{II}}}=\dfrac{\sqrt{{{K} _{B}}\times 150}}{\sqrt{{{K} _{B}}\times 300}}=0.707$

Hence, $\dfrac{{{\lambda } _{I}}}{{{\lambda } _{II}}}\cong 0.7$ 

Using the formula lambda = 1 / (sqrt(2) * pi * n * d^2) with n = 2.7 * 10^25 m^-3 and d = 2 * 10^-10 m, the result is approximately 2.81 * 10^-7 m.

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

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