20.3.2 The Root-Mean-Square Speed

We need to be careful when averaging velocities. The velocity component $v_{x}$ has a sign. At any instant of time, half the molecules in a container move to the right and have positive $v_{x}$ while the other half move to the left and have negative $v_{x}$ . Thus the average velocity is ${(v_{x})}_{avg} = 0$ . If this weren’t true, the entire container of gas would move away!

The speed of a molecule is $v = {(v_{x}^{2} + v_{y}^{2} + v_{z}^{2})}^{1 / 2}$ . Thus the average of the speed squared is

{(v^{2})}_{avg} = {(v_{x}^{2} + v_{y}^{2} + v_{z}^{2})}_{avg} = {(v_{x}^{2})}_{avg} + {(v_{y}^{2})}_{avg} + {(v_{z}^{2})}_{avg}

(20.12)

The square root of ${(v^{2})}_{avg}$ is called the root-mean-square speed $v_{rms} :$

v_{rms} = \sqrt{{(v^{2})}_{avg}} (root-mean-square speed)

(20.13)

This is usually called the rms speed. You can remember its definition by noting that its name is the opposite of the sequence of operations: First you square all the speeds, then you average the squares (find the mean), then you take the square root. Because the square root “undoes” the square, $v_{rms}$ must, in some sense, give an average speed.

NOTE We could compute a true average speed $v_{avg}$ , but that calculation is difficult. More important, the root-mean-square speed tends to arise naturally in many scientific and engineering calculations. It turns out that $v_{rms}$ differs from $v_{avg}$ by less than 10%, so for practical purposes we can interpret $v_{rms}$ as being essentially the average speed of a molecule in a gas.

There’s nothing special about the x-axis. The coordinate system is something that we impose on the problem, so on average it must be the case that

{(v_{x}^{2})}_{avg} = {(v_{y}^{2})}_{avg} = {(v_{z}^{2})}_{avg}

(20.14)

Hence we can use Equation 20.12 and the definition of $v_{rms}$ to write

v_{rms}^{2} = {(v_{x}^{2})}_{avg} + {(v_{y}^{2})}_{avg} + {(v_{z}^{2})}_{avg} = 3 {(v_{x}^{2})}_{avg}

(20.15)

Consequently, ${(v_{x}^{2})}_{avg}$ is

{(v_{x}^{2})}_{avg} = \frac{1}{3} v_{rms}^{2}

(20.16)

Using this result in Equation 20.11 gives us the net force on the wall of the container:

F_{on wall} = \frac{1}{3} \frac{N}{V} m v_{rms}^{2} A

(20.17)

Thus the pressure on the wall of the container due to all the molecular collisions is

p = \frac{F_{on wall}}{A} = \frac{1}{3} \frac{N}{V} m v_{rms}^{2}

(20.18)

We have met our goal. Equation 20.18 expresses the macroscopic pressure in terms of the microscopic physics. The pressure depends on the number density of molecules in the container and on how fast, on average, the molecules are moving.

EXAMPLE 20.2 The rms speed of helium atoms

A container holds helium at a pressure of 200 kPa and a temperature of $60 ° C$ . What is the rms speed of the helium atoms?

SOLVE The rms speed can be found from the pressure and the number density. Using the ideal-gas law gives us the number density:

\frac{N}{V} = \frac{p}{k_{B} T} = \frac{200, 000 Pa}{(1.38 \times 10^{- 23} J/K)(333 K)} = 4.35 \times 10^{25} m^{- 3}

The mass of a helium atom is $m = 4 u = 6.64 \times 10^{- 27} kg$ . Thus

v_{rms} = \sqrt{\frac{3 p}{(N / V) m}} = 1440 m / s

REVIEW We found in Chapter 16 that the speed of sound in helium is roughly 1000 m/s. Individual atoms probably move somewhat faster than a wave front, so 1440 m/s seems quite reasonable.