Question

The speed of the four particles are given as 2, 3, 4 and 5 cm/s respectively. What is their rms speed?
A. 3.5 cm/s
B. $\left( {\dfrac{{27}}{2}} \right)$ cm/s
C. $\sqrt {54} $ cm/s
D. $\left( {\dfrac{{\sqrt {54} }}{2}} \right)$ cm/s

Answer 1

Hint:First of all we should know about the rms speed in order to solve this equation. As we know that in an Ideal gas for measuring a typical speed of a molecule in terms of temperature, thus the speed which is used to determine is known as the root-mean-square or rms speed of a molecule.

Formula used:The formula used is
${v_{_{rms}}}\, = \sqrt {\dfrac{{3RT}}{M}} $ -(i)
Here, R is universal gas constant. T is known as temperature and M is known as the Molar mass of the gas.

Complete step-by-step solution:Root mean square or rms speed of a molecule is just the square root of the average speed.The root-mean-square speed generally depends upon the two factor that is molecular weight and the temperature.
The main reason for using the rms velocity instead of the average velocity is that for a sample of typical gas the net velocity is zero since the particles are moving in all directions.
From using the eq (i)
We will find the rms speed of four particles having speed as 2, 3, 4 and 5 cm/s.
${v_{rms}} = \,\sqrt {\dfrac{{{2^2}\, + \,{3^2}\, + \,{4^2}\, + \,{5^2}}}{4}\,\,} \,\,\,\,\, = \,\sqrt {\dfrac{{54}}{4}} $
Thus the rms speed is $\sqrt {\dfrac{{54}}{2}} $

Hence the correct option is D.

Note:The main reason for using the rms velocity instead of the average velocity is that for a sample of typical gas the net velocity is zero since the particles are moving in all directions. This is a key formula because the velocity of the particles determines both the diffusion and effusion rates.