Calculate the value of mean free path ($\lambda$) for oxygen molecules at temperature $27^{\circ} \mathrm{C}$ and pressure $1.01 \times 10^{5} \mathrm{~Pa}$. Assume the molecular diameter 0.3 nm and the gas is ideal. $(\mathrm{k}=1.38 \times 10^{-23} \mathrm{JK}^{-1})$

JEE Main · 2021 · 16 Mar 2021 · Shift 2 · Heat & Thermodynamics

Calculate the value of mean free path (

λ

) for oxygen molecules at temperature

2 7^{\circ} C

and pressure

1.01 \times 1 0^{5} Pa

. Assume the molecular diameter 0.3 nm and the gas is ideal.

(k = 1.38 \times 1 0^{- 23} JK^{- 1})

Answer: (D)

For an ideal gas of molecular diameter

d

at pressure

P

and temperature

T

the mean free path is

λ = \frac{k T}{2 π d ^{2} P}

Putting

T = 2 7^{\circ} C = 300 K

P = 1.01 \times 1 0^{5} Pa

d = 0.3 \times 1 0^{- 9} m

and

k = 1.38 \times 1 0^{- 23} J K^{- 1}

λ = \frac{1.38 \times 1 0 ^{- 23} \times 300}{1.414 \times 3.14 \times ( 0.3 \times 1 0 ^{- 9} ) ^{2} \times 1.01 \times 1 0 ^{5}} = \frac{4.14 \times 1 0 ^{- 21}}{4.04 \times 1 0 ^{- 14}}

which is

1.02 \times 1 0^{- 7} m

, so the mean free path of the oxygen molecules is about

102 nm

JEE Main · 2021 · 16 Mar 2021 · Shift 2 · Single correct option