Question

For the gases A and B with molecular weights ${{{M}}_{{A}}}$ and ${{{M}}_{{B}}}$, respectively, it is observed that at certain temperature T, the mean velocity of A is equal to the ${{{V}}_{{{rms}}}}$ of B. Thus, the mean velocity A can be made equal to the mean velocity of B, if:
A. A is at temperature T and B is at temperature ${{{T}}^{{'}}}$ such that ${{T > }}{{{T}}^{{'}}}$
B. Temperature of A is lowered to ${{{T}}_{{2}}}$ while B is at T such that ${{{T}}_{{2}}}{{ < T}}$
C. Both A and B are raised to a higher temperature
D. Heat energy supplied to A

Answer 1

Hint: Usually, gas molecules produce a pressure when they collide with each other. The motion of molecules is related with pressure. Also, when we increase the temperature, the velocity of gas molecules also increases. This is explained in kinetic molecular theory.

Complete step by step answer:
From the given data at a certain temperature T, the mean velocity of A is equal to the value of ${{{V}}_{{{rms}}}}$of B.
 hus, it can be written as $\sqrt {\dfrac{{{{8RT}}}}{{{{\Pi }}{{{M}}_{{A}}}}}} {{ = }}\sqrt {\dfrac{{{{3RT}}}}{{{{{M}}_{{B}}}}}} $ where ${{{M}}_{{A}}}$ and ${{{M}}_{{B}}}$ are the molecular weights of gases A and B.
We have the equation,
Average velocity, ${{{\mu }}_{{{av}}}}{{ = }}\sqrt {\dfrac{{{{8RT}}}}{{{{\Pi M}}}}} $ and root mean square velocity, ${{{\mu }}_{{{rms}}}}{{ = }}\sqrt {\dfrac{{{{3RT}}}}{{{M}}}} $, where ${\mu _{{{av}}}}$ is the average velocity, ${{R}}$ is the gas constant, ${{T}}$ is the temperature, ${{M}}$ is the molecular weight of gas.
It is given that at temperature T, mean velocity of A is equal to ${{{V}}_{{{rms}}}}$of B
$\sqrt {\dfrac{{{{8RT}}}}{{{{\Pi }}{{{M}}_{{A}}}}}} {{ = }}\sqrt {\dfrac{{{{3RT}}}}{{{{{M}}_{{B}}}}}} $, where ${{{M}}_{{A}}}$ is the molecular weight of A and ${{{M}}_{{B}}}$ is the molecular weight of B.
Squaring on both sides, we get
$\dfrac{{{{8RT}}}}{{\pi {{{M}}_{{A}}}}}{{ = }}\dfrac{{{{3RT}}}}{{{{{M}}_{{B}}}}} \Leftrightarrow {{{M}}_{{B}}} \times {{8RT}} = {{3RT}} \times \pi {{{M}}_{{A}}}$
Cancelling common terms, we get
$\dfrac{{{{{M}}_{{B}}}}}{{{{{M}}_{{A}}}}} = 3\dfrac{\pi }{8}$
Thus when mean velocity of A= mean velocity of B
We have
\[\sqrt {\dfrac{{{{8R}}{{{T}}^{{'}}}}}{{{{\Pi }}{{{M}}_{{B}}}}}} {{ = }}\sqrt {\dfrac{{{{8RT}}}}{{{{\Pi }}{{{M}}_{{A}}}}}} \]
On squaring both sides
$\dfrac{{{{8R}}{{{T}}^{{'}}}}}{{{{\Pi }}{{{M}}_{{B}}}}}{{ = }}\dfrac{{{{8RT}}}}{{{{\Pi }}{{{M}}_{{A}}}}}$
$\dfrac{{{{{T}}^{{'}}}}}{{{T}}}{{ = }}\dfrac{{{{{M}}_{{B}}}}}{{{{{M}}_{{A}}}}}$ where ${{T < }}{{{T}}^{{'}}}$
Thus the first option is incorrect
$\dfrac{{{{8RT}}}}{{{{\Pi }}{{{M}}_{{B}}}}}{{ = }}\dfrac{{{{8R}}{{{T}}_{{2}}}}}{{{{\Pi }}{{{M}}_{{A}}}}}$
$\dfrac{{{T}}}{{{{{T}}_{{2}}}}}{{ = }}\dfrac{{{{{M}}_{{B}}}}}{{{{{M}}_{{A}}}}}$ where ${{{T}}_{{2}}}{{ < T}}$

So, the correct answer is Option B.

Additional Information:
Both temperature and molecular weight is taken into consideration for the root mean square speed. By taking the root of square of the average velocities it is possible to obtain the average speed of gaseous particles. We know that gases are in continuous random motion, all the particles have different speeds, they collide and change direction.

Note: Velocity is the term that is used to describe the movement of gas particles, thus considering both speed and direction. The gas particles move in all the directions with random speed. The average velocity will be zero while solving for the average velocity of gas particles, by assuming that all the particles are moving in different directions.