91Ó°ÊÓ

• Four particles have speed$200\text{″¾}/\text{s}$
• Two particles have speed$500\text{ }m/s$
• Four particles have speed$600\text{″¾}/\text{s}$

Average speed is calculated by adding the speed of each molecules and then by dividing the total number of molecules.

The RMS speed of molecules is given by,

${\text{v}}_{rms}=\sqrt{\frac{{\displaystyle {∑}_{}^{}{\text{v}}^2}}{\text{N}}}$

Average speed is given by

${\text{V}}_{avg}=\frac{1}{\text{N}}\underset{i=1}{\overset{N}{}}{\text{V}}_i$

We have total of$10$particles, so$\text{N}=10$

Four particles have speed of $200\text{″¾}/\text{s}.$

So,${\text{V}}_1=4×200=800$

Two particles have speed of .$400\text{″¾}/\text{s}$

So,${\text{V}}_2=2×500=1000$

Four particles have speed of.$600\text{″¾}/\text{s}$

So,${\text{V}}_3=4×600=2400$

$V_{avg}=\frac{1}{N}(V_1+V_2+V_3)$

$\begin{array}{c}{\text{V}}_{avg}=\frac{1}{10}(800+1000+2400)\\ =420\text{″¾}/\text{s}\end{array}$

Therefore the average speed is .$420\text{″¾}/\text{s}$

The rms speed is given by

${\text{V}}_{rms}=\sqrt{\frac{1}{\text{N}}\underset{i=1}{\overset{N}{}}{\text{V}}_i^2}$

$\begin{array}{c}{\text{V}}_{rms}=\sqrt{\frac{1}{10}[4{\left(200\frac{\text{m}}{\text{s}}\right)}^2+2{\left(500\frac{\text{m}}{\text{s}}\right)}^2+4{\left(600\frac{\text{m}}{\text{s}}\right)}^2}\\ =458\text{″¾}/\text{s}\end{array}$

Therefore the RMS speed is $458\text{″¾}/\text{s}$.

Yes; ${\text{V}}_{rms}>{\text{V}}_{avg}$