Question

The average velocity of molecules in a gas in equilibrium is:
(A) proportional to $\sqrt{T}$
(B) proportional to T
(C) proportional to $T^2$
(D) equal to zero

Answer 1

Hint
The property on which kinetic theory of a gas's velocity depends upon the temperature. The average velocity vector of random motion of molecules is always zero. The speed of the molecules in a gas is proportional to the temperature and is inversely proportional to molar mass of the gas. So, if the temperature of a sample of gas is increased, the molecules speed up and the root mean square molecular speed increases as a result.

Complete step by step answer
Average velocity of gas is given by the formula:
$\Rightarrow v_{average}=\sqrt{{\displaystyle \frac{8RT}{\pi M}}}$
Here R is the gas constant, T is absolute temperature, and M${}_m$ is the molar mass of the gas particles in kg/mol.
At equilibrium the temperature is constant so average velocity will be zero.
$\Rightarrow v_{average}=0$
Hence, Correct option is (D).

Note
According to the kinetic molecular theory, the typical K.E. of gas particles is proportional to absolutely the temperature of the gas. This will be expressed with the subsequent equation where k represents the Boltzmann constant. The Boltzmann constant is just the universal gas constant R divided by the Avogadro’s constant (NA). The bar above certain terms indicates they're average values.
$\Rightarrow \overline{E_k}={\displaystyle \frac{3}{2}}kT$
This distribution of speeds arises from the collisions that occur between molecules within the gas phase. Although these collisions are elastic (there is not any net loss of energy), the individual speeds of every molecule involved within the collision may change. For instance, within the collision of two molecules, one molecule could also be deflected at a rather higher speed and therefore the other at a rather lower speed, but the typical K.E. doesn't change.