Question

Estimate the number of collisions per second suffered by a molecule in a sample of hydrogen at STP. The mean free path (average distance covered by a molecule between successive collisions) = 1.38 × 10−5 cm.

Answer 1

The number of collisions per second suffered by a molecule in a sample of hydrogen at STP = 1.23 × 10^8

We will first find the mean velocity of the gas molecules , Vmean = √(8RT/πM)

T - temperature of the gas

M - mass of the gas = 2/1000 Kg = 0.002 Kg

R - gas constant

At STP - T = 273K

Mean velocity = Vmean = √((8 × 8.31 × 273)/π × 0.002)

= 1698.96 m/s

Mean free path = 1.38 × 10^−5 cm

Number of collisions of molecule per second = Vmean / mean free path

= 1698.96/1.38 × 10−5

= 1.23 × 10^8

Answer 2

The number of collisions per second suffered by a molecule in a sample of hydrogen at STP is $1.23 \times 10^{11}$

Explanation:

Step 1:

Given data,

$\lambda=1.38 \times 10^{-8} \mathrm{m}$

$T=273 K$

$\mathrm{M}=2 \times 10^{-3} \mathrm{kg}$

Step 2:

The average  speed of the H molecules is given by

$V_{a v g}=\sqrt{\frac{8 R T}{\pi M}}$

Step 3:

On substituting the values,

$=\sqrt{\frac{8 \times 8.31 \times 273}{3.14 \times 2 \times 10^{-3}}}$

$=\sqrt{\frac{66.48 \times 273}{6.28 \times 10^{-3}}}$

$=\sqrt{\frac{18149.04}{6.28 \times 10^{-3}}}$

$=\sqrt{2889.974522 \times 10^{3}}$

$=1699.99 \approx 1700 \mathrm{m} / \mathrm{s}$

The time between two crashes is given by

$t=\frac{\lambda}{V_{a v g}}$

$t=\frac{1.38 \times 10^{-8}}{1700}$

$t=8.11 \times 10^{-4} \times 10^{-8}$

$t=8.11 \times 10^{-12} \mathrm{s}$

Collision count in $1 \mathrm{s}=\frac{1}{8.11 \times 10^{-12} \mathrm{s}}=1.23 \times 10^{11}$