Question

In an interstellar gas cloud at 42.6 K, the pressure is 1.06 x 10-8 Pa. Assuming that the molecular diameters of the gases in the cloud are all 19.2 nm, what is their mean free path?

Answer 1

Answer : The mean free path is, 33.88 m

Explanation :

First we have to calculate the volume of for mole of gas by using ideal gas equation.

$PV=nRT$

where,

P = pressure of gas = $1.06\times 10^{-8}Pa$

V = volume of gas = ?

T = temperature of gas = 42.6 K

n = number of moles of gas = 1 mole

R = gas constant = $8.314Pa.m^3/mole.K$

Now put all the given values in the ideal gas equation, we get:

$(1.06\times 10^{-8}Pa)\times (V)=1mole\times (8.314Pa.m^3/mole.K)\times (42.6K)$

$V=3.34\times 10^{10}m^3$

Now we have to calculate the number of molecules per unit volume.

$\frac{N}{V}=\frac{n\times N_A}{V}=\frac{(1mole)\times (6.022\times 10^{23})}{3.34\times 10^{10}m^3}=1.803\times 10^{13}molecule/m^3$

Now we have to calculate the mean free path.

Formula used :

$\text{Mean free path}=\frac{1}{\sqrt{2}\pi d^2(\frac{N}{V})}$

where,

d = diameter pf molecule = $19.2nm=19.2\times 10^{-9}m$

conversion used : $1nm=10^{-9}m$

$\frac{N}{V}$ = number of molecules per unit volume

Now put all the given values in this formula, we get:

$\text{Mean free path}=\frac{1}{\sqrt{2}\times (3.14) (19.2\times 10^{-9})^2\times (1.803\times 10^{13})}$

$\text{Mean free path}=33.88m$

Therefore, the mean free path is, 33.88 m