Question

The average speed of molecules in a sample of ozone gas is $1.84\times {10}^{4}cm{s}^{-1}$, the average translation kinetic energy per molecules is $xJ$. The value of $\frac{x}{2}\times {10}^{22}$ is:
(Given: $N_A=6\times {10}^{23};\sqrt{\frac{8}{3\pi }}=0.92$)

Answer 1

$⟨C⟩=\sqrt{\frac{8RT}{\Pi M}}$

$\Rightarrow 1.84\times {10}^{4}cm{s}^{-1}=\sqrt{\frac{8\times (8.314\times {10}^{7}erg{K}^{-1}{mol}^{-1})}{3.14\times 48gm.{mol}^{-1}}}$

$\Rightarrow T=76.72K$

$\therefore$ The average translation kinetic energy,

$E=\frac{3}{2}{K}_T=\frac{3}{2}\frac{R}{N_A}.T$

$=\frac{3}{2}\times \frac{8.314}{6.0\times {10}^{23}}\times 76.7$

$=\frac{318.8\times {10}^{-23}}{2}J$

$\therefore$ $\frac{x}{2}\times {10}^{22}=31.88$