Question

Use kinetic theory of gases to show that the average kinetic energy of a molecule of an ideal gas is directly proportional to the absolute temperature of the gas.

Answer 1

Hint: In order to solve this question, firstly we will use the immediate deduction of kinetic energy i.e. pressure expression $p = \dfrac{{mn{v^2}}}{3}$. Then we will use the ideal gas equation i.e. $pV = \mu RT$ and kinetic energy formula i.e. $\dfrac{{m{v^2}}}{2} = E$ to get the required answer.

Formula used-
1. $p = \dfrac{{mn{v^2}}}{3}$
2. $pV = \mu RT$
3. $\dfrac{{m{v^2}}}{2} = E$

Complete step by step answer:
As we know that-
An immediate deduction of kinetic theory is equivalent to the pressure expression,
I.e. $p = \dfrac{{mn{v^2}}}{3}$
Where m is mass of a gas molecule,
n is number of molecules per unit volume
and v is the r.m.s speed.
Thus, $n = \dfrac{N}{V}$
Where N is the number of molecules.
Now substituting the value of n in pressure expression i.e $p = \dfrac{{mn{v^2}}}{3}$,
We get-
$pV = \dfrac{{mN{v^2}}}{3}$
But, we know that the $\dfrac{{m{v^2}}}{2} = E$, E is the kinetic energy of a molecule.
Therefore, $pV = \dfrac{{2NE}}{3}$
Also from ideal gas equation,
$pV = \mu RT$
Where$\mu = \dfrac{N}{{{N_A}}}$is the number of moles.
Using these results,
We get-
$\mu RT = \dfrac{{2NE}}{3}$
$ \Rightarrow \dfrac{{NRT}}{{{N_A}}} = \dfrac{{2NE}}{3}$
Which finally gives,
$E = \dfrac{{3kT}}{2}$
Where, $k = \dfrac{R}{{{N_A}}}$ is the Boltzmann constant.
Thus, we conclude that the average of the kinetic energy of a molecule of an ideal gas is directly proportional to the absolute temperature of the gas.
Therefore, this is the kinetic interpretation of temperature.

Note:
While solving this question, we should know that the kinetic molecular theory can be used to explain both Charles law and Boyle's law. Also we must know that the number of molecules per unit volume is calculated by dividing the number of molecules from r.m.s speed.