Question

derive expression for kinetic energy of translation for one mole of gas and hence that for a molecule of gas .hence explain the kinetic interpretation of temperature

Answer 1

Dear Student,

Consider a box of length L, with 1 mole of gas insed it, where each particle moving at speed v.

$$\begin{gathered} \overline{K}=\overline{\frac{1}{2}m{v}^2}\Rightarrow \frac{1}{2}m{v_{rms}}^2 \\ {v}_{rms}=\sqrt{\frac{3RT}{m}} \\ or,\overline{K}=\frac{1}{2}m\times \frac{3RT}{m}\Rightarrow \frac{3RT}{2} \\ or,\overline{K}=\frac{3RT}{2}(for1mole) \\ for{N}_{A}molecules\overline{K}=\frac{3RT}{2} \\ for1molecule\overline{K}=\frac{3RT}{2N_A} \\ \Rightarrow \frac{3}{2}kT(kisboltzmancons\tan t) \\ as,k=\frac{R}{N_A} \end{gathered}$$

As you can see, temperature is the only variable on which the kinetic energy of an ideal gas depends. The kinetic energy is not affected by the mass of the molecules.

Regards.