Question

Equal Number of molecules of hydrogen and oxygen are contained in a vessel at one atmospheric pressure. The ratio of the collision frequency hydrogen molecules to the oxygen molecules on the container:
$A.$ $1:4$
$B.$ $4:1$
$C.$ $1:16$
$D.$ $16:1$

Answer 1

According to the kinetic molecular theory, the collision frequency is equal to the root mean square velocity of the molecules divided by their mean free path. Collision frequency is denoted by symbol $\left( Z \right)$
Formula used:
$Z = \dfrac{{{V_{rms}}}}{\sigma }$ .
Where $Z$ is the collision frequency, ${V_{rms}}$ is the root-mean square velocity of molecules, and $2\sigma$ is the mean free path.

Complete step by step answer:
In the given question it is given that equal numbers of molecules and oxygen are contained at one atmospheric pressure. We have to calculate the ratio of the collision frequency hydrogen molecules to the oxygen molecules on the container.
Let, the collision frequency of Hydrogen molecules ${Z_H}$ and collision frequency of oxygen molecules ${Z_O}$ have molar mass ${M_H}$ and ${M_O}$ of hydrogen and oxygen molecules respectively.
By using above formula of collision frequency we have the collision frequency of hydrogen is
${Z_H} = \dfrac{{{V_{rms}}}}{{2\sigma }}$
$\Rightarrow$ ${Z_H} = \dfrac{{\sqrt {\dfrac{{3RT}}{{{M_H}}}} }}{\sigma }$ $\left[ {\because {V_{rms}} = \sqrt {\dfrac{{3RT}}{M}} } \right]$
$\Rightarrow$ ${Z_H} = \dfrac{{\sqrt {3RT} \times \sigma }}{{\sqrt {{M_H}} }}$
Similarly, the collision frequency of Oxygen molecules is
${Z_O} = \dfrac{{\sqrt {3RT} \times \sigma }}{{\sqrt {{M_O}} }}$
Now the ratio of the collision frequency of hydrogen molecules to the oxygen molecules is
$\dfrac{{{Z_{^H}}}}{{{Z_O}}} = \dfrac{{\sqrt {\dfrac{{3RT}}{{{M_H}}}} }}{{\sqrt {\dfrac{{3RT}}{{{M_O}}}} }}$
$\Rightarrow$ $\dfrac{{{Z_H}}}{{{Z_O}}} = \sqrt {\dfrac{{{M_O}}}{{{M_H}}}}$
We know the molar mass of oxygen molecules is $16$ and molar mass of hydrogen is $1$.Put the value of molar mass of both the molecules in above equation, we get
$\dfrac{{{Z_H}}}{{{Z_O}}} = \dfrac{4}{1}$
Thus, the ratio of collision frequency of hydrogen molecules to oxygen molecules is $4:1$.

So, the Correct option is $B.$

Additional information:
Mean free path: Mean free path is the average distance between two collisions for a gas may be estimated from kinetic theory.

Note:
By above formula used for collision frequency $\left( {\dfrac{{{Z_H}}}{{{Z_O}}} = \sqrt {\dfrac{{{M_O}}}{{{M_H}}}} } \right)$. It may conclude that collision frequency is mainly depending on temperature and molar mass of gas present in the container. Collision frequency is inversely proportional to molar mass of gas.