JEE Main & Advanced Physics Kinetic Theory of Gases Kinetic Energy of Ideal Gas

In ideal gases, the molecules are considered as point particles. For point particles, there is no internal excitation, no vibration and no rotation. The point particles can have only translational motion and thus only translational energy. For an ideal gas the internal energy can only be tranlational kinetic energy.

Hence kinetic energy (or internal energy) of 1 mole ideal gas

\[E=\frac{1}{2}Mv_{rms}^{2}=\frac{1}{2}M\times \frac{3RT}{M}=\frac{3}{2}RT\]  

Various Translational kinetic energies  

(1) Kinetic energy per molecule of gas does not depends upon the mass of the molecule but only depends upon the temperature of the gas. As  \[E=\frac{3}{2}\,kT\,\]  or   \[E\propto T\] i.e. molecules of different gases say He, \[{{H}_{2}}\] and \[{{O}_{2}}\] etc. at same temperature will have same translational kinetic energy though their r.m.s. speed are different.

(2) For two gases at the same temperature \[{{m}_{1}}({{v}_{rms}})_{1}^{2}={{m}_{2}}({{v}_{rms}})_{2}^{2}\]

(3) Kinetic energy per mole of gas depends only upon the temperature of gas.

(4) Kinetic energy per gram of gas depend upon the temperature as well as molecular weight (or mass of one molecule) of the gas. \[{{E}_{gram}}=\frac{3}{2}\frac{k}{m}T\,\] \[\Rightarrow \] \[{{E}_{gram}}\propto \frac{T}{m}\]

(5) From the above expressions it is clear that higher the temperature of the gas, more will be the average kinetic energy possessed by the gas molecules at T = 0, E = 0 i.e. at absolute zero the molecular motion stops.