Question

In a gas at 1 atm pressure and 27°C a molecule
has diameter of 5Å find mean free path of
molecules:-
(1) 3.8 x 10-8 m (2) 2 x 10-8 m
(3) 8.3 x 10-9 m (4) 3.2 x 10-6 m​

Answer 1

Given that,

Pressure = 1 atm

Temperature = 27°C = 300 K

Diameter $d=5\ \AA$

We need to calculate the mean free path of  molecules

Using formula of mean free path

$\lambda=\dfrac{k_{b}T}{\sqrt{2}\pi d^2P}$

Where, T = temperature

P = pressure

d = diameter

k = boltzmann constant

Put the value into the formula

$\lambda=\dfrac{1.38\times10^{-23}\times300}{\sqrt{2}\times\pi\times1.013\times10^{5}\times(5\times10^{-10})^2}$

$\lambda=3.8\times10^{-8}\ m$

Hence, The mean free path of  molecules is $3.8\times10^{-8}\ m$

(1) is correct option.

Answer 2

Answer:

a. Is ANS

Explanation:

Mean free path is ,

= kbT/√2πd^2P

=3.8×10^-8