Question

If the kinetic energy in $J$ , of ${\text{C}}{{\text{H}}_4}$ (${\text{Molar mass = 16gmo}}{{\text{l}}^{ - 1}}$ ) at ${\text{T(K)}}$ is ${\text{X}}$ , the kinetic energy in $J$, ${{\text{O}}_2}$ (${\text{Molar mass = 32gmo}}{{\text{l}}^{ - 1}}$) at the same temperature is:
A.${\text{X}}$
B.${\text{2X}}$
C.${{\text{X}}^2}$
D.$\dfrac{{\text{X}}}{2}$

Answer 1

Hint: The average speed of particles is temperature dependent as well as molecular weight also but the average kinetic energy is independent of the molecular weight of gases so at the same temperature the gases will have the same average kinetic energy.

Complete step by step answer:
As we all know that relation between the average kinetic energy and temperature is \[{\text{K}}{\text{.E}}{\text{. = }}\dfrac{3}{2}kT\]. From this relation it is clear that kinetic energy is dependent on the temperature and it is also clear that the average kinetic energy is not dependent on the molecular mass of the gas. So the average kinetic energy of all the gases at same temperature is considered to be equal independent of their different molar masses.
The temperature basically affects the average speed of particles. We can easily determine that heavy particles move slowly while lighter particles move faster on average at the same temperature.
We can take an example of ${H_2}$ and ${O_2}$ gases. The molar mass of oxygen gas is $32g/mol$ and the molar mass of hydrogen gas is $2gm/mol$, so according to their weights it is clear that the average speed of hydrogen particles will be more than that of oxygen particles but the average kinetic energy of both the gases will be same at a common temperature because the average kinetic energy is independent of molar mass and dependent only on temperature.
Similarly the kinetic energy of ${\text{C}}{{\text{H}}_4}$ is equal to the kinetic energy ${{\text{O}}_2}$ of at ${\text{T(K)}}$ .
Therefore,
${\text{kinetic energy of }}{{\text{O}}_2}{\text{ = kinetic energy of C}}{{\text{H}}_4}$
${\text{kinetic energy of }}{{\text{O}}_2}{\text{ = X}}$
Hence option (A) is correct.


Note:
In reality it is very difficult to find out the speed of each and every molecule of gas therefore the term average is used, which tells us the common speed of all the gaseous particles and the same case is with kinetic energy, it is also calculated as average kinetic energy for gaseous particles because kinetic energy of each and every particle can’t be calculated.