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Heat and Thermodynamics question

2021 · 16 Mar · Shift 2 · Q54
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Heat and Thermodynamics question

2021 · 16 Mar · Shift 2 · Q54

JEE MainPhysicsHeat and ThermodynamicsMCQ+4 / −1
Calculate the value of mean free path (λ\lambdaλ) 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)
  1. A
    32 nm
  2. B
    58 nm
  3. C
    86 nm
  4. D
    102 nm
View written solutionFree

Correct answer: D

  1. Formula for mean free path

For an ideal gas, the mean free path is

λ=kT2 πd2P\lambda = \frac{kT}{\sqrt{2}\,\pi d^2 P}λ=2​πd2PkT​

where:

  • k=1.38×10−23 J K−1k = 1.38 \times 10^{-23}\,\text{J K}^{-1}k=1.38×10−23J K−1
  • T=27∘C=300 KT = 27^\circ \text{C} = 300\,\text{K}T=27∘C=300K
  • P=1.01×105 PaP = 1.01 \times 10^5\,\text{Pa}P=1.01×105Pa
  • d=0.3 nm=0.3×10−9 md = 0.3\,\text{nm} = 0.3 \times 10^{-9}\,\text{m}d=0.3nm=0.3×10−9m
  1. Substitute the values

First, calculate d2d^2d2:

d2=(0.3×10−9)2=0.09×10−18=9×10−20 m2d^2 = (0.3 \times 10^{-9})^2 = 0.09 \times 10^{-18} = 9 \times 10^{-20}\,\text{m}^2d2=(0.3×10−9)2=0.09×10−18=9×10−20m2

Now numerator:

kT=1.38×10−23×300=4.14×10−21kT = 1.38 \times 10^{-23} \times 300 = 4.14 \times 10^{-21}kT=1.38×10−23×300=4.14×10−21

Now denominator:

2πd2P=(1.414)(3.1416)(9×10−20)(1.01×105)\sqrt{2}\pi d^2 P = (1.414)(3.1416)(9 \times 10^{-20})(1.01 \times 10^5)2​πd2P=(1.414)(3.1416)(9×10−20)(1.01×105) 2π≈4.442\sqrt{2}\pi \approx 4.4422​π≈4.442

So,

denominator=4.442×9×10−20×1.01×105\text{denominator} = 4.442 \times 9 \times 10^{-20} \times 1.01 \times 10^5denominator=4.442×9×10−20×1.01×105 =39.978×10−20×1.01×105= 39.978 \times 10^{-20} \times 1.01 \times 10^5=39.978×10−20×1.01×105 ≈40.378×10−15=4.0378×10−14\approx 40.378 \times 10^{-15} = 4.0378 \times 10^{-14}≈40.378×10−15=4.0378×10−14
  1. Compute λ\lambdaλ
λ=4.14×10−214.0378×10−14\lambda = \frac{4.14 \times 10^{-21}}{4.0378 \times 10^{-14}}λ=4.0378×10−144.14×10−21​ λ≈1.025×10−7 m\lambda \approx 1.025 \times 10^{-7}\,\text{m}λ≈1.025×10−7m

Convert to nm:

1.025×10−7 m=102.5 nm1.025 \times 10^{-7}\,\text{m} = 102.5\,\text{nm}1.025×10−7m=102.5nm
  1. Match with the options

The closest option is:

102 nm\boxed{102\,\text{nm}}102nm​

So, Option D is correct.

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