Gas Particle Speeds

\[v_\text{average} = \sqrt{\frac{3k_B T}{m}}\quad\text{Eq. (9.5)}\]

Note: If you feel uncomfortable with working with fractions, exponents, variables, or scientific notation, you should review the following tutorials before working through this tutorial:

Sections

Introduction

The equation for the average speed of gas particles plays a fundamental role in determining the pressure of a gas. This is because the speed-of-gas-particles equation tells us how hard gas particles crash into one another, and, combined with the density of particles in the gas, how often the particles collide. It is the banging of particles into one another that is the cause of gas pressure.

To understand how the gas particle speed equation behaves, first notice that two quantities inside the square root symbol are constants, 3 and \(k_B\) (Boltzmann’s constant). Only \(T\) and \(m\) can change depending on the environment and nature of the gas particles. As a result, Equation (9.5) can be rewritten as \[v_\text{average} = (\text{constant})\times \sqrt{\frac{T}{m}} = (\text{constant})\times\frac{\sqrt{T}}{\sqrt{m}}.\] The average speed of particles of mass \(m\) is proportional to the square root of the temperature \(T\) (in kelvins) divided by the square root of the particle mass (in kg). If you are wondering about how the equation ends up producing the unit of speed, \(\text{m}/\text{s}\), the constant in front contains \(\sqrt{k_B}\) which includes the missing units.

This means that for particles having the same mass \(m\), if one layer of the Sun has a temperature of \(\text{10,000 K}\) and a deeper layer has a temperature of \(\text{40,000 K}\), the deeper layer is 4 times hotter and so the average speed is \(\sqrt{4} = 2\) times faster than the more shallow layer. If an even deeper layer has a temperature of \(100,000 K\), the average speed is \(\text{100,000 K}/\text{10,000 K} = \sqrt{10} =3.16\) times greater than the \(\text{10,000 K}\) layer. For a much deeper layer with a temperature of \(\text{10,000,000 K}\) (ten million kelvins), the average speed of the particles of mass \(m\) is \(\sqrt{\text{10,000,000 K}/\text{10,000 K}} = \sqrt{\text{1000}} = \text{31.6}\).

The most common type of helium atom has approximately 4 times as much mass as a hydrogen atom. While the average speed of particles increases with increasing temperature, the average speed decreases with increasing mass (think of the particles as being more sluggish the more massive they are.) This means that a helium atom is \(\text{1}/\sqrt{\text{4}} = \text{1}/\text{2}\) the speed of a hydrogen atom at the same temperature. The most common type of an oxygen atom is \(\text{16}\) times more massive than a hydrogen atom, implying that the average speed of an oxygen atom is only \(\text{1}/\sqrt{\text{16}} = \text{1}/\text{4}\) the speed of a hydrogen atom.

(Advanced topic) Although it isn’t important to know how to use Equation (9.5) directly to find the speed of a particle, it is informative to have a feel for a typical speed of a gas particle in a star. Suppose that a proton is moving around in a gas at a temperature of \(T = \text{10,000 K}\). To carry out the calculation directly you need to know the mass of a proton (\(m_p = 1.67 \times 10^{-27}\text{ kg}\)) and the value of Boltzmann’s constant, which is one of the fundamental constants of nature: \(k_B = 1.38\times 10^{-23}\text{kg m}^2/\text{K s}^2\). Substituting those values into the equation gives \(v_\text{average} = 15.7\text{ km/s}\), which is roughly 35,000 miles per hour.

With the information from the last paragraph it is now possible to rewrite Equation (9.5) by dividing the original equation by the values we just used. This is just the same approach we have used other times in your textbook, such as for Kepler’s third law or the escape speed equation. The result is

\[v_\text{average} = (15.7\text{ km/s})\text{ }\sqrt{\frac{T/\text{10,000 K}}{m/m_p}}.\]

Examples

  1. One layer of the Sun is \(\text{10,000}\) times hotter than another layer. Are electrons moving faster or slower in the hotter level, and by how many times?

    Since we are comparing the same type of particle, the mass is the same. The square root of the temperature is in the numerator, meaning that the speed of the electrons is \(\sqrt{\text{10,000}} = \sqrt{1 \times 10^4} = 1 \times 10^2 = 100\) times greater at the hotter level.
  2. A particular hydrogen atom has a mass that is \(\text{56}\) times greater than the mass of a proton. If both types of particles are at a level in the Sun where the temperature is \(\text{100,000 K}\), is an iron atom moving faster or slower than a proton, and by how many times?

    Since the temperature doesn’t change, neither does the numerator. In the denominator \(\sqrt{\text{56}} = \text{7.48}\), so an iron atom moves more slowly than a hydrogen atom at an average speed that is \(\text{1}/\text{7.48}\) times the speed of a proton.
  3. Since the average speed of a proton in a \(\text{10,000 K}\) gas is \(15.7\text{ km/s}\), what would the average speed of a proton be in a \(\text{20,000 K}\) gas?

    Since the mass of the particle is unchanged (all protons have exactly the same mass), \(m/m_p = 1\). However, the temperature has doubled: \(T/\text{10,000 K} = \text{20,000 K}/\text{10,000 K} = 2\). Given that the temperature ratio is inside the square root sign, the equation for the average speed is \(v_\text{average} = 15.7\text{ km/s} \times \sqrt{2} = 15.7\text{ km/s} \times 1.41 = 22.2\text{ km/s}\).
  4. Electrons are moving around in the \(\text{10,000 K}\) along with the protons. Electrons are only about \(\text{1}/{1800}\), however. How fast are the electrons moving on average?

    Since the temperature of the gas is \(\text{10,000 K}\), the numerator under the square root sign equals 1. The denominator is \((\text{1 }m_p/{1800})/m_p = 1/1800\) (note that the mass of the proton cancels). Now the equation becomes \[v_\text{average} = 15.7\text{ km/s}\times \sqrt{\frac{1}{1/1800}} = 15.7\text{ km/s}\times \sqrt{1800} = 15.7\text{ km/s}\times 42.4.\] Because it is so much lighter, an electron moves \(42.2\text{ km/s}\) faster than a proton, or \(666\text{ km/s}\).
  5. The temperature at the center of the Sun is about \text{15,000,000 K\), or 15 million kelvins. How fast are protons moving in that environment?

    First, \(\text{15,000,000 K}/\text{10,000 K} = 1.5\times 10^7\text{ K}/1\times 10^4\text{ K} = 1.5 \times 10^{7-4} = 1.5\times 10^3\) (the temperature unit, K, cancels). Since we are only discussing protons, the ratio in the denominator is 1. Now, \(\sqrt{1500} = 38.7\). As a result, the speed of protons in the center of the Sun on average is about \(608\text{ km/s}\), or about 1.4 million miles per hour.

Quiz Yourself

Try the following exercises to make sure that you really do understand the material just presented. By the way, computers and many calculators write scientific notation in “computer-speak,” where \(10\) raised to an integer (either positive or negative) replaces \(\times 10^\text{exponent}\) with \({\rm E(exponent)}\) or \({\rm e(exponent)}\); in other words, \(3.14159 \times 10^9\) is represented by \(3.14159\text{E}9\). If your calculator has the \(\Large{\hat{ }}\) symbol, then you would enter the number as \(3.14159 \times 10{\Large{\hat{ }}}9\). If your calculator has a \(10^x\) button then you would enter the number as \(3.14159\times9\,\,10^x\). If you have having problems with entering numbers in scientific notation, read the manual or help screens, or ask your instructor.

(Answers to the problems are given below)

  1. Identical particles in a gas are moving with an average speed of \(50\text{ km/s}\). If the temperature of the gas increases by a factor of \(25\), what would the average speed of particles in the gas become?
  2. Identical particles in a gas are moving with an average speed of \(100\text{ km/s}\). Another gas with the same temperature is composed of identical particles that are \(16\) times more massive than the first gas. What is the average speed of particles in the second gas?
  3. Based on the information in the examples above, what is the average speed of electrons in the center of the Sun?
  4. The mass of a helium atom is about four times more massive than a proton. How much faster does a helium atom move on average compared to a proton?
  5. What is the average speed of a helium atom at the center of the Sun?
  6. The average surface temperature on the surface of Earth is about \(288\text{ K}\).
    • (a) What is the average speed of a free proton near the surface of Earth?
    • (b) A hydrogen atom has essentially the same mass as a proton, while a hydrogen molecule has twice the mass of a hydrogen atom. What is the average speed of a hydrogen molecule (H2) near the surface of Earth?
    • (c) An oxygen molecule (O2) has roughly 16 times the mass of a hydrogen molecule. What is the average speed of an oxygen molecule near the surface of Earth?
  7. The solar wind that produces aurora on Earth, are particles that “blow” past Earth with a characteristic temperature of one million kelvins (\(\text{1,000,000 K}\)). What would the speed of the protons be based on the characteristic temperature?
Answers
  1. \(250\text{ km/s}\)
  2. \(25\text{ km/s}\)
  3. \(25,800\text{ km/s}\)
  4. \(1/2\)
  5. \(304\text{ km/s}\)
    • (a) \(2.66\text{ km/s}\)
    • (b) \(1.88\text{ km/s}\)
    • (c) \(0.47\text{ km/s}\)
  7. \(157\text{ km/s}\)
Scroll to Top