Calculate the value of mean free path ( λ) for oxygen molecules at temperature 27 ° C and pressure 1.01 × 105 Pa. Assume the molecular diameter 0.3 nm and the gas is ideal. (k = 1.38 × 10 - 23 JK ⁻¹) (a) 32 nm (b) 58 nm (c) 86 nm (d) 102 nm

Answer: (d). λ = (kT)/(√(2) π d² P) with T = 300 K, d = 0.3×10⁻⁹ m. Numerator: 1.38×10⁻²³×300 = 4.14×10⁻²¹ Denominator: 1.414×3.14×9×10⁻²⁰×1.01×10⁵ ≈ 4.04×10⁻¹⁴ λ ≈ 1.02×10⁻⁷ m = 102 nm So the answer is (d) 102 nm.

Calculate the value of mean free path ( λ) for oxygen molecules at… | JEE Main 2021 (16 Mar, Shift 2)

Q14JEE Main 2021 · 16 MAR · SHIFT 2Medium

Mean Free Path & Real Gases

Calculate the value of mean free path ( $\lambda$ ) for oxygen molecules at temperature 27 $^\circ$ C and pressure 1.01 $\times$ 105 Pa. Assume the molecular diameter 0.3 nm and the gas is ideal. (k = 1.38 $\times$ 10 $-$ 23 JK ${}^{-1}$ )

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Answer: (d)

$\lambda = \dfrac{kT}{\sqrt{2}\,\pi d^2 P}$ with $T = 300$ K, $d = 0.3\times10^{-9}$ m.
Numerator: $1.38\times10^{-23}\times300 = 4.14\times10^{-21}$
Denominator: $1.414\times3.14\times9\times10^{-20}\times1.01\times10^{5} \approx 4.04\times10^{-14}$
$\lambda \approx 1.02\times10^{-7}$ m $= 102$ nm
So the answer is (d) 102 nm.

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Mean Free Path & Real Gases question from JEE Main 2021 (16 Mar, Shift 2)

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