Question

Calculate the number of collisions per second of one hydrogen molecule at \[{24^ \circ }C\] and $2.00bar$. The diameter of a hydrogen molecule is $270pm$?

Answer 1

Hint: In this question, we need to find out the number of collisions per second. This is known as the collision frequency. Collision frequency is the ratio of the root mean square speed to the mean free path. So, first we will calculate the mean free path, ten we will calculate the root mean square speed and the ratio of the two will give us the final answer.


Complete step by step solution:
The mean free path of a molecule is given by the expression,
$\lambda = \dfrac{1}{{\sqrt 2 \pi {d^2}P}}m$
The number of collisions per second $ = \dfrac{{{v_{rms}}}}{\lambda }$
$\lambda = \dfrac{1}{{\sqrt 2 \pi {d^2}P}}m.......(1)$
In this question,
$d = 270pm$
On converting it into metre, we get,
$d = 2.7 \times {10^{ - 10}}m$
$P = 2.00bar$
On converting it into Pascal,
$P = 2 \times {10^5}Pa$
$T = {24^o}C$
On converting it into Kelvin scale, we get,
$T = 24 + 273$
$T = 297K$
We know that the value of the universal gas constant is,
$R = 8.314\dfrac{J}{K}$
On putting all these values in equation (1), we get,
$\lambda = \dfrac{1}{{\sqrt 2 \pi \times {{(2.7 \times {{10}^{ - 10}})}^2} \times 2 \times {{10}^5}}}m$
On simplifying the above expression, we get,
$\lambda = 5.52 \times {10^{ - 8}}m$
We know that,
The number of collisions per second $ = \dfrac{{{v_{rms}}}}{\lambda }......(2)$
So, in order to find the value of the number of collisions per second, first we need to find the value of root mean square speed.
The expression for the root mean square speed is,
${v_{rms}} = \sqrt {\dfrac{{3RT}}{M}} $
${v_{rms}} = \sqrt {\dfrac{{3 \times 8.314 \times 297}}{{2 \times {{10}^{ - 3}}}}} $
On further solving this, we get,
${v_{rms}} = 1917\dfrac{m}{s}$
On putting all these values in the equation (2), we get,
The number of collisions per second $ = \dfrac{{1917}}{{5.52 \times {{10}^{ - 8}}}}$
The number of collisions per second $ = 3.46 \times {10^{10}}{s^{ - 1}}$
So, the number of collisions per second of one hydrogen molecule at \[{24^ \circ }C\] and $2.00bar$ is $3.46 \times {10^{10}}{s^{ - 1}}$.

Note:
The mean free path is defined as the average distance which is travelled by a moving particle between two successive collisions, which modify its direction. This particle can be any particle which carries mass such as atom, molecule and photon.