Question

For two gases, A and B with molar masses ${{M}_{A}}$ and ${{M}_{B}}$, it is observed that at a certain temperature T, the mean velocity of A is equal to the root mean square velocity of B. Thus, the mean velocity of A can be made equal to the mean velocity of B, if:
A. A is at temperature T and B at T’, T’>T.
B. A is lowered to a temperature ${{T}_{2}}$ C. Both A and B are raised to a higher temperature.
D. Both A and B are placed at lower temperature.

Answer 1

Hint Try to think about the Kinetic theory of gases. According to the data given in the question we need to have an equation that can be used to relate the root mean square speed and the average speed of gas molecules.

Complete answer:
All the gas laws that we have discussed the Boyle's law, Charles Law Avogadro's Law are merely based on experimental evidence. There was no theoretical background to justify them So, the scientists were curious to know why the gases behave in a peculiar manner under certain not of condition. From Charles Law we got to know that the gases expand on heating But there was no theory to give the reason for such a tad So, there was a need for some theory which could tell about the happenings at the molecular level and so could answer the questions arising regarding the behaviour of gases. Later a theory was given called kinetic molecular theory of gases to provide a sound theoretical basis for the various gas laws.
The root-mean-square speed is the measure of the speed of particles in a gas, defined as the square root of the average velocity-squared of the molecules in a gas. The root-mean-square speed takes into account both molecular weight and temperature, two factors that directly affect the kinetic energy of a material. Whereas the average speed at a given temperature is the arithmetic mean of the speeds of different molecules of the gas.
Now,${{v}_{avg}}=\sqrt{\dfrac{8RT}{\pi M}}$ and ${{v}_{rms}}=\sqrt{\dfrac{3RT}{M}}$, at the temperature T, both velocities are equal. So,
$\sqrt{\dfrac{8RT}{\pi {{M}_{A}}}}=\sqrt{\dfrac{3RT}{{{M}_{B}}}}$,${{M}_{A}}\,and\,{{M}_{B}}$ are the molar masses.
So,$\dfrac{{{M}_{B}}}{{{M}_{A}}}=3\times \dfrac{\pi }{8};{{M}_{B}}>{{M}_{A}}$. That means, when they have same mean velocities, we have:
\[\begin{align}
 & \sqrt{\dfrac{8RT'}{\pi {{M}_{B}}}}=\sqrt{\dfrac{8RT}{\pi {{M}_{A}}}} \\
& Taking\,square,we\,have: \\
& \dfrac{8RT}{\pi {{M}_{B}}}=\dfrac{8RT}{\pi {{M}_{A}}} \\
& \dfrac{T'}{T}=\dfrac{{{M}_{B}}}{{{M}_{A}}}\,\,\,So,\,T'>T \\
\end{align}\]
So, option A is correct. We can also write that:
$\begin{align}
& \dfrac{8RT}{\pi {{M}_{B}}}=\dfrac{8R{{T}_{2}}}{\pi {{M}_{A}}} \\
& So,\, \\
& \dfrac{T}{{{T}_{2}}}=\dfrac{{{M}_{B}}}{{{M}_{A}}};\,T>{{T}_{2}} \\
\end{align}$
So, option B is also correct. So, mean velocity of A increases when heat energy is supplied to it.

So, options A and B are the correct options.

NOTE: The order of the 3 types of velocities can be remembered by recalling the word “RAM” where ‘R’ is r.m.s, ‘A’ means average and ‘M’ means most probable. ${{V}_{rms}}>{{V}_{average}}>{{V}_{most\,probable}}$.