Question

Derive the relation between kinetic energy of ideal gas and temperature.

Answer 1

We know that if heat is not transferred when the two systems are in contact, then it is said to be in thermal equilibrium state and their temperature is same.
Let $1g$ molecule of gas with volume and temperature be V and T respectively and number of molecules of $1$ mol of gas $\left(N_A\right)=6.02\times {10}^{23}$ molecules. According to kinetic theory of gases, the pressure of gas for $1g$ molecules is:
$P=\frac{1}{3}m{N}_A\stackrel{-}{v^2}$ ......(14.50)
Here m is the mass of the molecule and $\stackrel{-}{v^2}$ is the root mean square speed.
Using ideal gas equation in equation (14.50),
$\frac{1}{3}m{N}_A\stackrel{-}{v^2}=RT$
or, $\stackrel{-}{v^2}=\frac{3RT}{m{N}_A}$ .....(14.51)
If $E_1,E_2,E_3,.....,E_n$ are kinetic energies of each molecule, then the average kinetic energy of each molecule is
$E=\frac{E_1+E_2+E_3+....+E_N}{N_A}$
or, $E=\frac{\frac{1}{2}m{v}_1^2+\frac{1}{2}m{v}_2^2+\frac{1}{2}m{v}_3^2+.......+\frac{1}{2}m{v}_N^2}{N_A}$
where $v_1,v_2,v_3,.....v_n$ respectively are the speeds of each molecule, then
$E=\frac{1}{2}m\left[\frac{v_1^2+v_2^2+v_3^2+....+v_N^2}{N_A}\right]$
$\stackrel{-}{E}=\frac{1}{2}m{v}^2$
Using equation (14.51),
$\stackrel{-}{E}=\frac{1}{2}m\times \frac{3RT}{m{N}_A}$
$\stackrel{-}{E}=\frac{3}{2}\left(\frac{R}{N_A}\right)T$ .........(14.52)
Since, Boltzmann constant, $k_B=\frac{R}{N_A},$ then from equation (14.52),
$\stackrel{-}{E}=\frac{3}{2}{k}_{B}T$ ..........(14.53)
Equation (14.53) is a relation between the average kinetic energy of each molecule and temperature. It is clear that E and T are directly proportional to each other, i.e.,
$\stackrel{-}{E}\propto T$
i.e., kinetic energy of each molecule of a gas is a measure of temperature and at absolute zero temperature, i.e.,
$T\to 0\Rightarrow \stackrel{-}{E}\to 0$ and $\stackrel{-}{v}\to 0$
So, at absolute zero temperature, the molecules are motionless. Thus the absolute zero temperature is the temperature at which the speed of all molecules is zero.