Question

At what temperature will the kinetic energy of gas molecules be double of its value at 27℃?
(A) 54℃
(B) 108℃
(C) 300℃
(D) 327℃

Answer 1

Hint
The ratio of the average kinetic energy per molecule of the gas to the temperature of the gas is $\dfrac{3}{2}{\rm{k}}$ which is constant as ‘k’ is the Boltzmann’s constant.

Complete step by step solution
Let the average kinetic energy per molecule of the gas be ${{\rm{E}}_1} = {\rm{\;}}$E.
The temperature of the gas in Kelvin will be, ${{\rm{T}}_1} = 27 + 273 = 300$ K
Let at temperature ${{\rm{T}}_2} = {\rm{\;}}$T’, the average kinetic energy per molecule of gas is ${{\rm{E}}_2} = {\rm{\;}}$2E.
We know that the average kinetic energy per molecule of gas is given by
${\rm{E}} = \dfrac{3}{2}{\rm{kT}}$
Where k is the Boltzmann constant.
∴ $\dfrac{{\rm{E}}}{{\rm{T}}} = \dfrac{3}{2}{\rm{k}}$ (constant)
According to the question,
$\dfrac{{{{\rm{E}}_1}}}{{{{\rm{T}}_1}}} = \dfrac{{{{\rm{E}}_2}}}{{{{\rm{T}}_2}}}$
$\dfrac{{\rm{E}}}{{300}} = \dfrac{{2{\rm{E}}}}{{{\rm{T'}}}}$
${\rm{T'}} = 600{\rm{\;}}$K
Therefore, (D) is the required solution.

Additional Information
The kinetic theory of gases defines the behavior of gases based on the three basic features of gases i.e. pressure, volume, and temperature. The pressure of the gas is due to the collision of atoms with the walls of the container. The volume of the gas is due to the freedom of atoms to spread throughout the container. The temperature of the gas is due to the kinetic energy of atoms.

Note
The conversion of temperature in the ‘Kelvin’ scale is required and then again for the required answer, the temperature should be converted into ‘Celsius’ scale.