Question

The molecular velocity of any gas is:
(A)- Inversely proportional to the square root of temperature
(B)- Inversely proportional to absolute temperature
(C)- Directly proportional to the square of temperature
(D)- Directly proportional to the square root of temperature

Answer 1

Hint: The velocity obtained from the kinetic energy which squared in order to nullify the directional component. Further the square root of the squared molecular velocity is taken to obtain its relation to absolute temperature.

Complete step by step answer:
In the Kinetic Molecular Theory, the gas particles move constantly and in random motion, thus colliding and have different speeds and direction of motion. Thus, the molecular velocity describes its movement. But, due to their constantly changing velocities, their average behaviour is taken into account, as the distribution of velocities remains the same.
For a collection of gas particles, the average velocity may become zero, as an equal number of particles may be moving in opposite directions. Thus, to exclude the directional component, the root-mean square speed is preferred, which has the square root of the average velocity-squared of the molecule.
\[{{v}_{rms}}=\text{ }\sqrt{\overline{{{v}^{2}}}}=\sqrt{\dfrac{{{V}_{1}}^{2}+{{V}_{2}}^{2}+{{V}_{3}}^{2}+...+{{V}_{N}}^{2}}{N}}\]
As the average kinetic energy of the gas molecules is given by $\overline{{{E}_{K}}}=\dfrac{3}{2}{{k}_{B}}T=\dfrac{3}{2}\dfrac{RT}{{{N}_{A}}}$ ,

where ${{k}_{B}}$ is the Boltzmann constant, ${{N}_{A}}$ is the Avogadro’s number and R is the gas constant.
and also, we know, $\overline{{{E}_{K}}}=\dfrac{1}{2}M\overline{{{v}^{2}}}$

Then, on equating the above two equations for kinetic energy, we get,
\[\dfrac{\text{3}}{\text{2}}\dfrac{\text{RT}}{{{\text{N}}_{\text{A}}}}\,\text{=}\,\,\dfrac{1}{2}M\overline{{{v}^{2}}}\]
\[\overline{{{v}^{2}}}=\dfrac{\text{3RT}}{\text{M}{{\text{N}}_{\text{A}}}}\,\]
Then, taking the square root of the square of average velocity, we get ${{v}_{rms}}=\sqrt{\dfrac{3RT}{M}}$

where M is the molar mass in $kg\,mo{{l}^{-1}}$, R is the gas constant, and T is the absolute temperature in kelvin.
Thus, the molecular velocity of gas is directly proportional to the square root of temperature.
So, the correct answer is “Option B”.

Note: The root-mean-square speed depends on both the molecular weight and temperature, which directly affect the kinetic energy of the molecule. As the temperature increases, the root-mean square molecular speed increases.