Question

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

Answer 1

Hint: Let us get some idea about the ideal gas. An ideal gas is a theoretical gas made up of a large number of uniformly moving point particles with no interparticle interactions. Since it obeys the ideal gas law, a simplified equation of state, and is amenable to statistical mechanics research, the ideal gas principle is useful.


Complete step-by-step solution:
We know that when two systems are in contact and no heat is exchanged, they are said to be in thermal equilibrium and their temperatures are the same.
Let the volume and temperature of a $$1g$$ molecule of gas be$$V$$ and $$T$$, respectively, and the number of molecules in $$1mol$$of gas $$(N_A)=6.02\times {10}^{23}$$molecules. The pressure of gas for $$1g$$ molecules, according to the kinetic theory of gases, is:
$P={\displaystyle \frac{1}{3}}\times {\displaystyle \frac{m\times N_A}{V}}\times \overline{v^2}$
$or,PV={\displaystyle \frac{1}{3}}\times m\times N_A\times \overline{v^2}...(1)$
$$m$$ Is the molecule's mass, and $$\overline{v^2}$$is the molecule's velocity is the square root of the root mean square speed.
Using ideal gas equation in equation $(1)$
${\displaystyle \frac{1}{3}}\times m\times N_A\times \overline{v^2}=RT$
$or,\overline{v^2}={\displaystyle \frac{3\times R\times T}{m\times N_A}}...(2)$
If each molecule's kinetic energies are $E_1,E_2,E_3,....,E_n$ the average kinetic energy of each molecule is
$E={\displaystyle \frac{E_1+E_2+E_3+....+E_N}{N_A}}$
$E={\displaystyle \frac{{\displaystyle \frac{1}{2}}\times m\times v_1^2+{\displaystyle \frac{1}{2}}\times m\times v_2^2+{\displaystyle \frac{1}{2}}\times m\times v_3^2+.......+{\displaystyle \frac{1}{2}}\times m\times v_N^2}{N_A}}$
where $$v_1,v_2,v_3,.....,v_n$$are the speeds of each molecule,
$E={\displaystyle \frac{1}{2}}\times m\times \left[{\displaystyle \frac{v_1^2+v_2^2+v_3^2+....+v_N^2}{N_A}}\right]$
$\bar{E}={\displaystyle \frac{1}{2}}\times m\times {\displaystyle \frac{3\times R\times T}{m\times N_A}}$
$\bar{E}={\displaystyle \frac{3}{2}}\times \left({\displaystyle \frac{R}{N_A}}\right)\times T......(3)$
As we that, Boltzmann constant, $k_B={\displaystyle \frac{R}{N_A}},thenfromequation(3)$
$\bar{E}={\displaystyle \frac{3}{2}}\times k_B\times T....(4)$
The average kinetic energy of each molecule and temperature are related in $$Equation\left(4\right).$$ It is obvious that E and T are proportional to one another, i.e.
$\bar{E}\propto T$
At absolute zero temperature, kinetic energy of each molecule of a gas is a measure of temperature.
$T\to 0\Rightarrow \bar{E}\to 0and\bar{v}\to 0$
As a result, the molecules are motionless at absolute zero temperature. As a result, absolute zero is the temperature at which all molecules have the same speed.

 Note:The kinetic molecular theory of gases explains the physical behaviour of gases. The magnitude of the gas pressure is determined by the number of collisions that gas particles have with the walls of their container and the force at which they collide. Average kinetic energy is proportional to temperature.