Root-Mean-Square Speed
~10 min read · Tier toggle changes depth (T1 plain · T2 standard · T3 rigour)
You should be comfortable with the ideal gas law and the definition of the Boltzmann constant. Familiarity with basic vector algebra and the concept of an average over a population is also helpful.
Why this matters
The root-mean-square speed provides a single, physically meaningful measure of how fast gas molecules are moving.
In the previous lecture, we established that temperature is a direct measure of the average translational kinetic energy of gas molecules. However, because molecules in a gas move in random directions with a wide range of speeds, there is no single "average speed" that captures the energy content as effectively as the root-mean-square (rms) speed. This quantity is crucial because it links the microscopic motion of individual particles to the macroscopic variables of pressure and temperature. Understanding \(v_{\mathrm{rms}}\) allows us to predict how changes in temperature or molecular mass affect the behaviour of gases, from the diffusion of scents in a room to the escape of atmospheres from planets. It is the bridge between the statistical nature of molecular motion and the deterministic laws of thermodynamics.
What is Root-Mean-Square Speed?
Imagine a room filled with a large number of people walking around randomly. If you asked, "How fast are they moving?", you might calculate the average speed by adding up all their speeds and dividing by the number of people. However, this simple arithmetic mean can be misleading if some people are moving very fast and others very slowly, or if we are interested in the energy they carry. In physics, when we deal with random motion and energy, we often care more about the square of the speed than the speed itself. This is because kinetic energy depends on the square of the velocity (\(v^2\)).
To get a representative speed that correctly reflects the energy of the system, we use a specific statistical average called the root-mean-square speed. The name describes the three steps involved in its calculation. First, we take the square of the speed of every single molecule in the gas. This step ensures that all values are positive and emphasises the faster-moving molecules, which contribute more significantly to the total energy and pressure. Second, we calculate the mean (average) of these squared speeds. This gives us the "mean-square speed," denoted as \(\langle v^2 \rangle\). Finally, we take the square root of this average value. This last step converts the quantity back into units of speed, giving us the "root-mean-square speed," \(v_{\mathrm{rms}}\).
This method is preferred over the simple arithmetic mean because it is directly related to the physical quantities we measure, such as pressure and temperature. In the kinetic theory of gases, the pressure exerted by a gas on the walls of its container is the result of molecules colliding with those walls. The force of these collisions depends on the momentum transferred, which is related to the square of the velocity. Therefore, the average of the squared speeds is the fundamental quantity that determines the pressure. By taking the square root, we obtain a speed value that is consistent with the energy scale of the gas. It is important to note that \(v_{\mathrm{rms}}\) is not the speed of any specific molecule, nor is it the most probable speed. It is a statistical property of the entire ensemble of molecules that best represents the energy content of the gas.
Deriving the Expression for \(v_{\mathrm{rms}}\)
To find the mathematical expression for \(v_{\mathrm{rms}}\), we start with the fundamental result of kinetic theory from the previous lectures. The pressure \(P\) exerted by an ideal gas is given by the equation \(PV = \frac{1}{3}Nm\langle v^2 \rangle\), where \(N\) is the number of molecules, \(m\) is the mass of a single molecule, and \(\langle v^2 \rangle\) is the mean-square speed. This equation tells us that the product of pressure and volume is proportional to the total kinetic energy of the gas molecules.
We also know from the ideal gas law that \(PV = NkT\), where \(k\) is the Boltzmann constant and \(T\) is the absolute temperature in Kelvin. By equating the two expressions for \(PV\), we can relate the microscopic motion to the macroscopic temperature. Setting \(\frac{1}{3}Nm\langle v^2 \rangle = NkT\), we can cancel \(N\) from both sides and rearrange to solve for the mean-square speed: \(\langle v^2 \rangle = \frac{3kT}{m}\). This result shows that the mean-square speed is directly proportional to the temperature and inversely proportional to the mass of the molecule.
Since \(v_{\mathrm{rms}}\) is defined as the square root of the mean-square speed, we take the square root of both sides: \(v_{\mathrm{rms}} = \sqrt{\frac{3kT}{m}}\). While this form is useful for single molecules, it is often more convenient to work with moles of gas rather than individual molecules. We can convert this expression using the relationships \(N_A = \frac{1}{k}R\) (where \(R\) is the gas constant and \(N_A\) is Avogadro's constant) and \(M = mN_A\) (where \(M\) is the molar mass). Substituting \(k = \frac{R}{N_A}\) and \(m = \frac{M}{N_A}\) into the equation, we get \(v_{\mathrm{rms}} = \sqrt{\frac{3(R/N_A)T}{M/N_A}}\). The \(N_A\) terms cancel out, leaving us with the standard formula used in chemistry: \(v_{\mathrm{rms}} = \sqrt{\frac{3RT}{M}}\). This final expression clearly shows that the rms speed depends only on the temperature and the molar mass of the gas.
Dependence on Temperature and Molar Mass
The equation \(v_{\mathrm{rms}} = \sqrt{\frac{3RT}{M}}\) reveals two critical dependencies. First, the rms speed is proportional to the square root of the absolute temperature (\(T^{1/2}\)). This means that if you double the absolute temperature of a gas, the rms speed does not double; instead, it increases by a factor of \(\sqrt{2}\), which is approximately 1.414. For example, if a gas at 300 K has a certain rms speed, heating it to 600 K will increase that speed by about 41%. This sub-linear relationship is a direct consequence of the fact that temperature is a measure of average kinetic energy, and energy scales with the square of speed. Therefore, to double the speed, you would need to quadruple the absolute temperature.
Second, the rms speed is inversely proportional to the square root of the molar mass (\(M^{-1/2}\)). This implies that lighter molecules move faster than heavier molecules at the same temperature. Consider two gases at the same temperature, such as hydrogen (\(\mathrm{H_2}\), \(M \approx 2\ \mathrm{g\,mol^{-1}}\)) and oxygen (\(\mathrm{O_2}\), \(M \approx 32\ \mathrm{g\,mol^{-1}}\)). Since oxygen is 16 times heavier than hydrogen, its rms speed will be \(\sqrt{16} = 4\) times slower than that of hydrogen. This difference in speeds has significant practical implications. For instance, in a mixture of gases, lighter molecules will diffuse and effuse faster than heavier ones, a phenomenon described by Graham's Law. It also explains why small molecules like helium and hydrogen can escape the Earth's atmosphere over geological time scales, while heavier molecules like nitrogen and oxygen are retained more effectively.
It is important to remember that these relationships hold for ideal gases. At very high pressures or low temperatures, intermolecular forces become significant, and the ideal gas approximation may break down. However, for most standard laboratory conditions, the dependence of \(v_{\mathrm{rms}}\) on temperature and molar mass provides an excellent description of gas behaviour. The proportionality to \(\sqrt{T}\) and \(1/\sqrt{M}\) allows us to make quick comparisons between different gases and conditions without needing to know the exact number of molecules or the volume of the container.
Comparison with Other Average Speeds
While the root-mean-square speed is the most commonly used measure in thermodynamics, it is not the only way to define an "average" speed for a gas. The Maxwell–Boltzmann distribution describes the probability of finding a molecule with a specific speed. From this distribution, we can derive two other important average speeds: the most probable speed (\(v_p\)) and the mean speed (\(v_{\mathrm{avg}}\)). The most probable speed is the speed at which the distribution curve reaches its maximum. It represents the speed that the largest number of molecules possess. Mathematically, \(v_p = \sqrt{\frac{2RT}{M}}\).
The mean speed, \(v_{\mathrm{avg}}\), is the arithmetic average of all molecular speeds, calculated by integrating the speed multiplied by the probability distribution over all possible speeds. This value is given by \(v_{\mathrm{avg}} = \sqrt{\frac{8RT}{\pi M}}\). Comparing the three expressions, we can see that they all share the same dependence on \(\sqrt{T/M}\), but they differ by numerical constants. Specifically, \(v_p = \sqrt{2} \sqrt{\frac{RT}{M}}\), \(v_{\mathrm{avg}} = \sqrt{\frac{8}{\pi}} \sqrt{\frac{RT}{M}} \approx 1.596 \sqrt{\frac{RT}{M}}\), and \(v_{\mathrm{rms}} = \sqrt{3} \sqrt{\frac{RT}{M}} \approx 1.732 \sqrt{\frac{RT}{M}}\). Therefore, for any gas at a given temperature, the relationship between these speeds is \(v_p < v_{\mathrm{avg}} < v_{\mathrm{rms}}\).
The reason \(v_{\mathrm{rms}}\) is larger than the other averages is that it weights higher speeds more heavily due to the squaring process. Since the distribution of speeds is skewed towards higher values (there is no upper limit to speed, though the probability becomes very small), the average of the squares is pulled up more than the average of the speeds themselves. In practical applications, such as calculating the rate of effusion or the mean free path, the choice of average speed can matter. However, for calculating pressure and energy, \(v_{\mathrm{rms}}\) is the correct quantity to use because it is directly derived from the second moment of the velocity distribution, which corresponds to the kinetic energy.
The result
Kinetic theory of gases
Modelling a gas as a large number of point-like molecules undergoing elastic collisions with a container's walls, the rate of momentum transfer per unit wall area gives \(PV=\tfrac13Nm\langle v^2\rangle\), where \(N\) is the number of molecules, \(m\) their mass, and \(\langle v^2\rangle\) the mean-square speed. Comparing this to the empirically established ideal-gas-law (\(PV=NkT\), with Boltzmann constant \(k=R/N_A\)) identifies temperature directly with average molecular translational kinetic energy: \(\tfrac12m\langle v^2\rangle=\tfrac32kT\) per molecule.
A common misconception is that the rms speed is the speed of the "typical" molecule. The most probable speed \(v_p\) is lower, and the mean speed lies between the two: \(v_p : \langle v \rangle : v_{\mathrm{rms}} = 1 : 1.128 : 1.225\). Each has its own use. Pressure and kinetic energy involve \(\langle v^2 \rangle\), so they use \(v_{\mathrm{rms}}\). The rate at which molecules hit a wall or pass through a small hole, \(Z_W = \tfrac14 n \langle v \rangle\), involves the mean speed. Using \(v_{\mathrm{rms}}\) in a collision-rate formula overestimates it by about 8.5%. C
Worked example
Problem. In the upper atmosphere the temperature can reach about 1000 K. Calculate \(v_{\mathrm{rms}}\) for \(\mathrm{H_2}\) (2.016 g mol-1) and \(\mathrm{N_2}\) (28.02 g mol-1) there, and compare each with Earth's escape speed, 11.2 km s-1.
- Use \(v_{\mathrm{rms}} = \sqrt{3RT/M}\) with \(M\) in kg mol-1.
- \(\mathrm{H_2}\): \(v_{\mathrm{rms}} = \sqrt{\dfrac{3 \times 8.314 \times 1000}{2.016 \times 10^{-3}}} = \sqrt{1.237 \times 10^{7}} = 3517\ \mathrm{m\,s^{-1}}\).
- \(\mathrm{N_2}\): \(v_{\mathrm{rms}} = \sqrt{\dfrac{3 \times 8.314 \times 1000}{28.02 \times 10^{-3}}} = \sqrt{8.902 \times 10^{5}} = 943\ \mathrm{m\,s^{-1}}\).
- Ratios to the escape speed: \(11\,200/3517 = 3.2\) for \(\mathrm{H_2}\), but \(11\,200/943 = 11.9\) for \(\mathrm{N_2}\).
- Escape is governed by the far tail of the Maxwell–Boltzmann distribution, which falls off roughly as \(e^{-v^2}\). A threshold 3.2 times \(v_{\mathrm{rms}}\) is reached by a small but steady trickle of \(\mathrm{H_2}\) molecules over geological time; a threshold 12 times \(v_{\mathrm{rms}}\) is effectively never reached by \(\mathrm{N_2}\). This is why Earth has lost its hydrogen but kept its nitrogen.
Units check. \(\mathrm{J\,mol^{-1}} / \mathrm{kg\,mol^{-1}} = \mathrm{J\,kg^{-1}} = \mathrm{m^2\,s^{-2}}\), whose square root is in m s-1.
Three quick tests
- If the temperature of a gas is increased from 300 K to 1200 K, by what factor does the rms speed increase?
Answer
The rms speed is proportional to \(\sqrt{T}\). The ratio of temperatures is \(1200/300 = 4\). The factor increase in speed is \(\sqrt{4} = 2\). - Which gas has a higher rms speed at the same temperature: helium (\(M = 4\ \mathrm{g\,mol^{-1}}\)) or argon (\(M = 40\ \mathrm{g\,mol^{-1}}\))?
Answer
Helium. Since \(v_{\mathrm{rms}} \propto 1/\sqrt{M}\), the lighter gas (helium) has the higher speed. Specifically, it is \(\sqrt{40/4} = \sqrt{10} \approx 3.16\) times faster. - True or False: The rms speed is the same as the most probable speed.
Answer
False. The rms speed is higher than the most probable speed because it weights higher velocities more heavily due to the squaring operation in the average.
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Lecture notes checked line by line by the editor; every number in the worked example was recomputed.