Question

There are $2.0\times 10^{22}$ molecules of an ideal gas,each of mass $6\times 10^{-23}g$ in a sample with average speed $4.6\times 10^{4}cm/s$. The total translational kinetic energy of gas molecules (in $J$) is :
(Given: $\sqrt{\frac{8}{3\pi }}=0.92$ )

Answer 1

Answer: 150

Given that $v_{avg}=4.6\times 10^{2}m/s$
$=\sqrt{\frac{8RT}{\pi M}}$
$v_{rms}=\sqrt{\frac{3RT}{M}}$
$\frac{v_{rms}}{\sqrt{\frac{3RT}{M}}}\frac{4.6\times 10^2}{\sqrt{\frac{8RT}{\pi M}}}$
$v_{rms}=4.6\times 10^2\times \sqrt{\frac{3\pi }{8}}$
$v_{rms}^2=24.93\times 10^4(m/s{)}^2$
Total translational KE of gas molecules
$=\left(\frac{1}{2}m{v}_{rms}^2\right)\times (\text{ no. of molecules })$
$=\frac{1}{2}\times 6\times 10^{26}\times 24.93\times 10^4\times 2\times 10^{22}$
$=149.58\simeq 150$