Calculate the value of the mean free path λ for oxygen molecules at temperature 27°C and pressure 1.01×10^5 Pa. Assume the molecular diameter 0.3 nm and the gas is ideal. k=1.38×10^(-23) J K^(-1)

Correct answer: (D) 102 nm. λ=RT/(√2πd^2 N_AP) λ=(8.314×300)/(√2×3.14×0.3×10-9^2×6.023×10^23×1.01×10^5) λ=102 nm

Q70 / 90 · JEE Main 2021 (16 Mar Shift 2) · Physics

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

$λ = \frac{R T}{\sqrt{2} π d^{2} N_{A} P}$

$λ = \frac{8.314 \times 300}{\sqrt{2} \times 3.14 \times {(0.3 \times 10^{- 9})}^{2} \times 6.023 \times 10^{23} \times 1.01 \times 10^{5}}$

$λ = 102 nm$

Hence, the correct answer is option: 'D'.