Question

The diameter of a gas molecule is $2.4×10^{−10}$ m. The mean free path at NTP is ….. (Given Boltzmann constant k=1.38×10-23 J/K)
(A) $1.46 \times {10}^{-7}$ m
(B) $2.46 \times {10}^{-6}$ m
(C) $1.46 \times {10}^{-6}$ m
(D) $2.46 \times {10}^{-7}$ m

Answer 1

Hint
The mean free path is the distance that a molecule travels between collisions. The mean free path is determined by the criterion that there is one molecule within the "collision tube" that is swept out by a molecular trajectory. The criterion is: $\lambda \left( {\dfrac{N}{V}} \right)\pi {r^2} = 1$ , where r is the radius of a molecule.

Complete step by step answer
According to the question given that,
Diameter of the gas molecule (d) = 2.4 × 10-10 m
There is no mention of volume in the question so we using the 2nd law of mean free path,
$\lambda = \dfrac{{{K_B}T}}{{\sqrt 2 \pi {d^2}P}}$ [ KB = constant, p = pressure, T = temperature]......(1)
At NTP the pressure is P = 1.01 × 105 N/m2
The temperature is T = 273 K
Now, put the value in equation 1 we get,
$\lambda = \dfrac{{1.30 \times {{10}^{ - 23}} \times 273}}{{1.41 \times 3.14 \times {{(2.4 \times {{10}^{ - 10}})}^2} \times 1.01 \times {{10}^5}}}$
$\therefore \lambda = 1.46 \times {10^{ - 7}}m$ .
Option (A) is correct.

Note
Ludwig Boltzmann (1844–1906) The Boltzmann constant (k) relates temperature to energy. It is an indispensable tool in thermodynamics, the study of heat and its relationship to other types of energy. It's named for Austrian physicist Ludwig Boltzmann (1844–1906), one of the pioneers of statistical mechanics.