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

Given that $v_{ avg }=4.6 \times 10^{2} m / s$
$ =\sqrt{\frac{8 R T}{\pi M}}$
$v_{ rms }=\sqrt{\frac{3 R T}{M}}$
$\frac{v_{ rms }}{\sqrt{\frac{3 R T}{M}}} \frac{4.6 \times 10^{2}}{\sqrt{\frac{8 R T}{\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} $
$=149.58 \simeq 150$