The K.E. of hydrogen gas per gram mole at $0^{\circ} \mathrm{C}$ is $(\mathrm{R}=8.3 \mathrm{~J} / \

Question

The K.E. of hydrogen gas per gram mole at $0^{\circ} \mathrm{C}$ is $(\mathrm{R}=8.3 \mathrm{~J} / \mathrm{mole}-\mathrm{K})$\

Professor PC Thomas Classes · Answer

KE per molecule $=\mathrm{n}\left(\frac{1}{2} k Tight)$\ where $\mathrm{n}=$ number of degree of freedom KE per $\mathrm{mol}=\mathrm{N} \quad\left(n \frac{1}{2} k Tight)$\ where $\mathrm{N}=$ Avogadro's number. But $\mathrm{Nk}=\mathrm{R}$\ $	herefore$ KE per mol $=\mathrm{n}^{\frac{1}{2}} \mathrm{kT}$\ For hydrogen $\mathrm{n}=5, \mathrm{R}=8.3, \mathrm{~T}=273$\ $	herefore \mathrm{KE}$ per $\mathrm{mol}=$\