Kinetic theory of gases: Root mean square speed

In a nutshell

The root mean square speed is a quantity used to describe the motion of particles in a gas and it is the square root of the average square velocity of all the particles. It can be used to find the pressure, volume and temperature of a gas.

Equations

description	equation
Pressure and volume of an ideal gas	$pV=\dfrac{1}{3}Nm\overline{c^2}$
Mean kinetic energy	$\dfrac{1}{2}m\overline{c^2}=\dfrac{3}{2}kT$

Constants

constant	symbol	value
$Boltzmann\ constant$	$k$	$1.38\times10^{-23}\ J\ K^{-1}$

Variable definitions

quantity name	symbol	derived units	si base units
$pressure$	$p$	$Pa$	$kg \ m^{-1} \ s^{-2}$
$volume$	$V$	$m^3$	$m^3$
$temperature$	$T$	$K$	$K$
$number\ of\ atoms$	$N$
$mass$	$m$	$kg$	$kg$
$velocity$	$c$	$m\ s^{-1}$	$m\ s^{-1}$

The root mean square speed

The root mean square speed $c_{r.m.s.}$ (r.m.s. speed) is a quantity used to describe the motion of particles and it is essentially the average speed of each particle.

To calculate it you need to square the velocity $c$ of each particle in a gas and then find the average known as the mean square speed $\overline{c^2}$ . The square root of this quantity is the r.m.s. speed:

$\sqrt{\overline{c^2}}=c_{r.m.s.}$

Note: The bar ( $\overline{c}$ ) means average.

The reason that r.m.s speed is used, rather than the velocity, is because velocity is a vector. There could be both negative and positive values for velocity, but this is not the case for speed as it is a scalar.

Therefore, if you added all of the velocities together, they will cancel out to zero, because of the large number of particles you are taking an average from. So, by squaring the velocities, this removes the signs, and gives the squared speed.

Pressure and volume of an ideal gas

The mean square speed can be used to find the volume and pressure of an ideal gas with $N$ particles of mass $m$ through the equation:

$pV=\dfrac{1}{3}Nm\overline{c^2}$

This equation is significant as it describes the properties of a macroscopic object based on the properties of its constituent microscopic parts.

Mean kinetic energy of an ideal gas

By substituting the equation $pV=NkT$ into the expression above, you can find that the temperature of an ideal gas depends on the mean kinetic energy of the particles with no regard for their number.

Equate the two equations:

$\dfrac{1}{3}Nm\overline{c^2} = NkT$

Cancel $N$ on both sides, divide by $2$ and multiply by $3$ :

$\dfrac{1}{2}m\overline{c^2}=\dfrac{3}{2}kT$

Where the mean kinetic energy of the particles $E_k=\dfrac{1}{2}m\overline{c^2}$ .

Example

Calculate the temperature of helium gas when its atoms have an r.m.s. speed of $1.36\times10^3\ m\ s^{-1}$ . A helium atom has a mass of $6.64\times10^{-27}\ kg$ .

Firstly, write down what you know:

$c_{r.m.s.}=1.36\times10^{3}\ m\ s^{-1}\newline m=6.64\times10^{-27}\ kg$ 

Write down the equation that relates temperature and the mean kinetic energy and rearrange it for $T$ :

$\dfrac{1}{2}m\overline{c^2}=\dfrac{3}{2}kT$ 

$T=\dfrac{m\overline{c^2}}{3k}$ 

The mean square speed is simply:

$\overline{c^2}=c_{r.m.s}^2$ 

Substitute all the values into the equation:

$T=\dfrac{6.64\times10^{-27}\times(1.36\times10^3)^2}{3\times1.38\times10^{-23}}$ 

Calculate the temperature, and give your answer to the lowest number of significant figures given in the question:

$T=297\ K$ 

The temperature of the helium gas is $\underline{297\ K}$ .