Question

N molecules each of mass m, of gas A and 2N molecules, each of mass 2m, of gas B are contained in the same vessel. Which is maintained at a temperature T. The mean square velocity of the molecules of type B is denoted by ${V^2}$ and the mean square velocity of the X component of the velocity of A type is denoted by ${\omega ^2}$, $\dfrac{{{\omega ^2}}}{{{V^2}}}$ is:
A) $2$
B) $1$
C) $\dfrac{1}{3}$
D) $\dfrac{2}{3}$

Answer 1

Hint: To solve this question we should know what the mean square velocity for any molecule is. We can find the mean square velocity for molecules of type $A$ in different directions and then use the one in $X$ direction. We simply have to find the ratio of the mean square velocity of the two given types of molecules.

Formulae used:
${V^2}_{rms} = \dfrac{{3kT}}{m}$
Where ${V_{rms}}$ is the mean square velocity, $k$ is the Boltzmann constant, $T$ is the temperature and $m$ is the molecular mass.

Complete step by step answer:
To solve this question we should know what the mean square velocity for any molecule is. Root mean square velocity of any molecule can be defined as the square root of the mean of squares of the velocity of a gas molecules which can be given by
${V^2}_{rms} = \dfrac{{3kT}}{m}$
Where ${V_{rms}}$ is the mean square velocity, $k$ is the Boltzmann constant, $T$ is the temperature and $m$ is the molecular mass.
Also
${V^2}_{rms} = v_x^2 + v_y^2 + v_z^2$
Here ${V_{rms}}$ is the mean square velocity, ${v_x}$ is the velocity along $X$ direction, ${v_y}$ is the velocity along $Y$ direction and ${v_z}$ is the velocity along $Z$ direction.
As ${v_x} = {v_y} = {v_z}$,
$ \Rightarrow {V^2}_{rms} = 3v_x^2$
$ \Rightarrow v_x^2 = \dfrac{1}{3}{V^2}_{rms}$
In the question, the velocity along $X$ component of $A$ type is denoted by ${\omega ^2}$
$ \Rightarrow {\omega ^2} = \dfrac{1}{3}{V^2}_{rms}$
Putting the value of ${V_{rms}}$ we get,
$ \Rightarrow {\omega ^2} = \dfrac{1}{3}\dfrac{{3kT}}{m} = \dfrac{{kT}}{m}$
$ \Rightarrow {\omega ^2} = \dfrac{{kT}}{m}$
Mean square velocity of the molecules of type $B$, denoted by ${V^2}$ will be
\[ \Rightarrow {V^2}_{rms} = {V^2} = \dfrac{{3kT}}{{2m}}\]
\[ \therefore {V^2} = \dfrac{{3kT}}{{2m}}\]
So the ratio between mean square velocity of the molecules of type $B$ and the mean square velocity of the $X$ component of the velocity of $A$ type is,
$ \Rightarrow \dfrac{{{\omega ^2}}}{{{V^2}}} = \dfrac{{\dfrac{{kT}}{m}}}{{\dfrac{{3kT}}{{2m}}}} = \dfrac{2}{3}$

So option (D) is the correct answer.

Note: While solving questions for root mean square velocity, always use the correct formulae. The root mean square velocity is different from average velocity as it is the square root of the mean of squares of the velocity of a gas molecule whereas average velocity is simply the arithmetic average of all the velocities.