Question

Prove 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: From kinetic Molecular theory, the increase in the temperature will increase the average kinetic energy of the molecules. The kinetic theory of gases describes the properties of gases such as pressure, temperature, volume, diffusivity, viscosity. It has to prove that The average kinetic energy of the molecule is directly proportional to the absolute temperature of the gas.

Complete answer:
From the kinetic pressure energy equation,
$P = \dfrac{1}{3}\rho {c^2}$
Where, $P$ - the pressure of the gas
$\rho $- density of the gas particles
$c$ - velocity of the gas particles
we know that the density is given by,
$\rho = \dfrac{M}{V}$
Where, M- the mass of all particles
\[V\] - the volume of the gas particles
By substituting the above density equation in the kinetic pressure equation is given by,
$P = \dfrac{1}{3} \times \dfrac{M}{V} \times {c^2}$
Total number of particles,$M = \left( {m \times N} \right)$
Where,m-mass of every individual particle
N-number of particles
Therefore, the kinetic pressure equation becomes,
$P = \dfrac{1}{3} \times \dfrac{{m \times N}}{V} \times {c^2}$
$P.V = \dfrac{1}{3} \times N \times m{c^2}$
We know that the kinetic energy equation is given by,
$K.E = \dfrac{1}{2} \times m{c^{^2}}$
Substituting the K.E equation in kinetic pressure equation,
$P.V = \dfrac{1}{3} \times N \times 2K.E$
From ideal gas law,
$P.V = nRT$
Where,n-amount of substance
$R$ - ideal gas constant
\[T\] - temperature
Substituting the ideal gas equation,
$nRT = \dfrac{2}{3} \times N \times K.E$
We know that the amount of substance,
$n = \dfrac{N}{{{N_A}}}$
Where, N- total number of particles
NA-Avagadro constant $\left( {{N_A} = 6.022 \times {{10}^{23}}mo{l^{ - 1}}} \right)$
Substituting the value of\[n\] ,
$\dfrac{N}{{{N_A}}} \times RT = \dfrac{2}{3} \times N \times K.E$
$ \Rightarrow \dfrac{{RT}}{{{N_A}}} = \dfrac{2}{3} \times K.E$
$ \Rightarrow T = \dfrac{{{N_A}}}{R} \times \dfrac{2}{3} \times K.E$
In the above equation, the value\[{N_R}\] , \[R\] are the constant terms.
Therefore $T \propto K.E$, that means the absolute temperature of the gas is directly proportional to the average kinetic energy of the molecule of an ideal gas.
Hence it is proved.

Note: The kinetic theory is based on assuming the pressure developed on the container walls due to the collision of molecules depends on the temperature of the gases. The assumptions on the kinetic theory of gases are that the gas particles are constant and in random motion, the volume of gas particles is always negligible, there are only elastic collisions of particles.