Question

Two gases X (Molecular weight ${{M}_{X}}$) and Y (Molecular weight ${{M}_{Y}},{{M}_{Y}}>{{M}_{X}}$) are at the same temperature in T in two different containers. Their root mean square velocities are ${{C}_{X}}\text{ and }{{\text{C}}_{Y}}$ respectively. If the average kinetic energies per molecule of the two gases are X and Y are ${{E}_{X}}\text{ and }{{\text{E}}_{Y}}$ respectively then which of the following relation(s) is (are) true?
[A] ${{E}_{X}}>{{E}_{Y}}$
[B] ${{C}_{X}}>{{C}_{Y}}$
[C] ${{E}_{X}}={{E}_{Y}}=\dfrac{3}{2}RT$
[D] ${{E}_{X}}={{E}_{Y}}=\dfrac{3}{2}{{K}_{B}}T$

Answer 1

Hint: To solve this question, simply use the respective formulas to find out the root mean square velocity and the average kinetic energy of both the gases. Comparing them will give us the correct options. To find the root mean square velocity, use \[C=\sqrt{\dfrac{3RT}{M}}\] and to find out the average kinetic energy use the formula of kinetic energy, $K.E=\dfrac{3}{2}nRT$.

Complete answer:
In the question, the options that are given to us are related to the average kinetic energy and their root mean square velocities. To find out which options are correct, let us calculate the average kinetic energy and the root mean square velocities of the two gases.
Firstly, we know that the formula to find out the root mean square velocity is
     \[C=\sqrt{\dfrac{3RT}{M}}\]
Where, C is the root mean square velocity, R is the universal gas constant, T is the temperature and M is the molecular mass.
Here, both the gases are at the same temperature T. Therefore, the root mean square velocity of both the gases will be-
${{C}_{X}}=\sqrt{\dfrac{3RT}{{{M}_{X}}}}\text{ and }{{C}_{Y}}=\sqrt{\dfrac{3RT}{{{M}_{Y}}}}$
As we can see that the root mean square velocity is indirectly proportional to the molecular masses of the gases, we can write that
     \[\dfrac{{{C}_{X}}}{{{C}_{Y}}}=\sqrt{\dfrac{{{M}_{Y}}}{{{M}_{X}}}}\]
As it is given in the question that ${{M}_{Y}}>{{M}_{X}}$, therefore, ${{C}_{X}}>{{C}_{Y}}$.
Therefore, option [B] is correct.
Now let us calculate the average kinetic energy of the two gases.
We know that kinetic energy can be written as $K.E=\dfrac{3}{2}nRT$ where, ‘n’ is the number of moles.
Therefore, option [C] is incorrect, as it does not consider the number of moles.
In option [D] we have ${{E}_{X}}={{E}_{Y}}=\dfrac{3}{2}{{K}_{B}}T$ where, ${{K}_{B}}$ is the Boltzmann’s constant which gives us an alternative form of the ideal gas equation as $pV=NKT$.
Similarly, we can say that here, nR is replaced by ${{K}_{B}}$ on the introduction of Boltzmann’s constant. Therefore, option [D] is correct.
As ${{E}_{X}}={{E}_{Y}}=\dfrac{3}{2}{{K}_{B}}T$ is correct, ${{E}_{X}}>{{E}_{Y}}$ cannot be true.

Therefore, the correct options are option [B] ${{C}_{X}}>{{C}_{Y}}$ and option [D] ${{E}_{X}}={{E}_{Y}}=\dfrac{3}{2}{{K}_{B}}T$.

Note: The Boltzmann’s constant is proportionality constant and it gives us a relation between the kinetic energy of the particles in a gas and the temperature of the gas. It shows that the average kinetic translational energy is $\dfrac{1}{2}m{{v}^{2}}=\dfrac{3}{2}KT$.