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Thermodynamics Part 2

Section: Physics

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Question : 7 of 100

Marks: +1, -0

Calculate the value of mean free path

(\lambda)

for oxygen molecules at temperature

27^{\circ}\,\mathrm{C}

1.01 \times 10^{5}\,\text{Pa}

. Assume the molecular diameter

0.3\,\text{nm}

and the gas is ideal.

(k=1.38 \times 10^{-23}\,\mathrm{J\,K}^{-1})

[16 Mar 2021 Shift 2]

$58\,\text{nm}$
$32\,\text{nm}$
$86\,\text{nm}$
$102\,\text{nm}$

Solution:

Given,

Pressure of the gas,

p=1.01 \times 10^{5}\,\text{Pa}

Absolute temperature of gas,

T=27^{\circ}\,\mathrm{C}

=27+273=300\,\text{K}

Molecular diameter,

d=0.3\,\text{nm}=0.3 \times 10^{-9}\,\text{m}

Boltzmann constant,

k_{B}=1.38 \times 10^{-23}\,\mathrm{J/K}

Mean free path,

\lambda = \frac{k_{B} T}{\sqrt{2}\,\pi d^{2} p}

....(i)

Substituting the given values in Eq. (i), we get

\lambda = \frac{1.38 \times 10^{-23} \times 300}{\sqrt{2} \times 3.14 \times (0.3 \times 10^{-9})^{2} \times 1.01 \times 10^{5}}

=102\,\text{nm}

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