Question

Ultra-high-vacuum experiments are routinely performed at a total pressure of 1.0 x 10-10 torr. Calculate the mean free path of n2 molecules at 350 k under these conditions. The collision diameter of n2 is 3.75

Answer 1

Answer : The mean free path is, $5.82\times 10^5m$

Explanation :

Formula used for mean free path is :

$\text{Mean free path}=\frac{K_BT}{\sqrt{2}\pi d^2P}$

where,

d = diameter of molecule = $3.75\AA=3.75\times 10^{-10}m$

conversion used : $1\AA=10^{-10}m$

$K_B$ = Boltzmann constant = $1.38\times 10^{-23}J/K$

P = total pressure = $1.0\times 10^{-10}torr=1.33\times 10^{-8}Pa$

conversion used : 1 torr = 133.322 Pa

T = temperature = 350 K

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

$\text{Mean free path}=\frac{(1.38\times 10^{-23}J/K)\times (350K)}{\sqrt{2}\times (3.14) (3.75\times 10^{-10}m)^2\times (1.33\times 10^{-8}Pa)}$

$\text{Mean free path}=5.82\times 10^5m$

Therefore, the mean free path is, $5.82\times 10^5m$