Pressure, Temperature and RMS Speed | iCalculator™

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Thermodynamics Learning Material
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13.7Pressure, Temperature and RMS Speed

In this Physics tutorial, you will learn:

  • What is rms speed of a gas?
  • What is the relationship between pressure temperature and rms speed in a gas?
  • Why are the factors affecting the rms speed of ideal gas molecules?
  • What is the translational kinetic energy of gas particles?
  • What is mean free path of gas molecules?
  • What are the factors affecting mean free path of gas molecules?


Can you figure out how molecules of a gas move in space? What can you say if they are limited within the walls of a closed container?

What can you say about the momentum and kinetic energy of ideal gases? What about real gases?

What kind of movements can molecules of a gas do? Do you think they can move linearly? Can they spin around themselves? Can they oscillate around fixed positions just as the springs do?

All these things and many others will be explained in this tutorial, in which a quantitative analysis of gas molecules movement is made. Therefore, many concepts explained earlier, associated with an advanced mathematical apparatus is used here to describe the physical quantities involved. This means, you may need to recall your previous knowledge obtained in mathematics in order to better understand this tutorial.

Relationship between Pressure, Temperature and RMS Speed in a Gas

Let's suppose we have n moles of a gas enclosed within a cubic container of volume V and side length L, as shown in the figure below.

Physics Tutorials: This image provides visual information for the physics tutorial Pressure, Temperature and RMS Speed

The container walls are held at constant temperature T. Let's find the relationship between the gas pressure exerted at the walls and the speed of molecules.

Molecules move in every direction within the container; they hit each other and the container walls and then, they bounce back. If we consider the gas as ideal, the collisions between particles can be neglected. Thus, we consider only the elastic collisions of gas molecules with the container walls.

When a molecule of mass m collides normally with the walls of the container at velocity v, it turns back at velocity -v as shown in the figure.

Physics Tutorials: This image provides visual information for the physics tutorial Pressure, Temperature and RMS Speed

This means the change in momentum Δp of the molecule is

∆p = p1 - p2
= m × ∆v
= m × v - m × (-v)
= 2 × m × v

This process is repeated continuously between the two opposite walls. Given that the distance between the container walls is L, the time interval Δt needed for the molecule to turn again to the original position is

∆t = 2L/v

From the Newton's Second Law of Motion, we have for the force by which the molecule hits the container wall:

F = m × a
= m × ∆v/∆t
= m × ∆v/∆t
= ∆p/∆t

Substituting the values obtained earlier, we get:

F = ∆p/∆t
= 2 × m × v/2L/v
= m × v2/L

Given that Pressure = Force / Area, we obtain for the pressure exerted by the molecule on the container wall:

P = F/A
= m × v2/L/L2
= m × v2/L3

For a large number of molecules, we can write

P = m/L3 × (v21 + v22 + ... + v2N)

where N is the total number of molecules in the container.

Thus, given that N = n × NA and (v21 + v22 + ... + v2N ) = v2average, we obtain

P = m × n × Na/L3 × v2average

We have m × NA = M and L3 = V, where M is the molar mass of gas and V is the volume of container. Thus, we obtain

P = n × M/V × v2average

We have considered only one direction so far (let's say only the x-direction. If we consider all three directions available for the motion of molecules, we obtain:

v2 = v2x + v2y + v2z = 3v2x

Therefore, the equation of gas pressure for a single direction becomes

P = n × M/3V × v2average

We call the quantity v2average as root mean square speed (vrms). Hence, we can write

P = n × M/3V × v2rms

The last formula indicates how a macroscopic quantity such as pressure depends on a microscopic quantity such as the speed of molecules.

From the ideal gas law, we know that

P × V = n × R × T

Therefore, combining the last two formulae, we obtain

n × R × T = n × M/3 × v2rms
R × T = M/3 × v2rms
v2rms = 3 × R × T/M
vrms = √3 × R × T/M

Example 1

What is the rms speed of CO2 molecules at 27°C if their molar mass is 44 g/mol?

Solution 1


M = 44 g/mol = 0.44 kg/mol
T = 27°C = 300 K
R = 8.31 J/molK
vrms = ?

Using the equation

vrms = √3 × R × T/M

we obtain after substitutions

vrms = √3 × 8.31 × 300/0.044 = 412 m/s

This is a very big value; it results the CO2 molecules move at the speed of bullets when they come out of the riffle muzzle. However, the real rms speed is much smaller as besides the collision with the container walls molecules collide with each other as well - a speed which we neglected when considering the gas as ideal. Also, the value found is an average value; there are many molecules which move slower.

Translational Kinetic Energy of Gas

Let's consider again the molecule in the cubic container of the previous paragraph. The kinetic energy of this molecule is

KE = m × v2rms/2

We substitute in this equation the value of rms speed found earlier:

KE = m × 3 × R × T/M/2


KE = 3/2 × m × R × T/M

Given that NA = M/m, where m is the mass of a single molecule, we obtain:

KE = 3/2 × R × T/Na

Having k = R/NA, where k is the Boltzmann constant, we obtain

KE = 3/2 × k × T

Therefore, we obtain a very important conclusion based on the last formula:

"All ideal gas molecules, no matter what kind of gas they belong, have the same translational kinetic energy at a given temperature."

This means that when we measure the temperature of an ideal gas, we are actually measuring the average translational kinetic energy of its particles. We have sustained earlier this theory of correspondence between average kinetic energy of particles and temperature when explaining the meaning of temperature but now, we are also proving it mathematically.

Example 2

What is the translational kinetic energy of 0.02 moles of an ideal gas at 227°C?

Solution 2

The formula

KE = 3/2 × k × T

gives the translational kinetic energy of a single molecule of an ideal gas, not of the entire gas.

Given this, we must multiply the kinetic energy of a single particle to the number of particles in 0.02 moles of ideal gas, considering also the fact that 227°C correspond to 500 K (227 + 273 = 500). Thus, we have:

Total KEtrans = (3/2 × k × T) × N
= (3/2 × k × T) × n × Na
= 3/2 × 1.38 × 10-23 × 5 × 102 × 2 × 10-2 × 6.02 × 1023
= 124.6 J

Mean Free Path

If you throw a few potato starch on water surface, you will notice that starch grains move in random direction. Furthermore, grain particles change direction continuously as shown in the figure.

Physics Tutorials: This image provides visual information for the physics tutorial Pressure, Temperature and RMS Speed

This phenomenon occurs because starch particles are hit by water molecules, which we cannot see as they are very small.

The same phenomenon occurs in gas particles as well. They collide continuously with each other; however, they move linearly at constant speed during the time interval between two consecutive collisions.

We use a parameter known as "mean free path" (in symbols, λ) to describe this random motion of gas particles. It represents the average distance travelled by a gas particle between two collisions. Mean free path depends on the following factors:

1- Density or concentration of gas. In this case, we express density in terms of concentration of molecules, i.e. as number of molecules per unit volume (in short, N/V) instead of mass per unit volume, as here we are interested on microscopic parameters such as the number of molecules.

Based on this factor, we can say that higher the concentration of particles, smaller their mean free path. Therefore, the molecular mean free path is inversely proportional to gas concentration.

2- Size of molecules. Obviously, larger the gas molecules, shorter the mean free path between them. molecules are, the smaller the mean free path. Mean free path λ, should vary (inversely) as the square of the molecular diameter because the cross section of a molecule - not its diameter - determines its effective target area.

Let's find the correct expression for the mean free path. Consider two spherical molecules of diameter d. If we insert them in the opposite ends of a cylindrical tube, the molecules will collide with each other only if the diameter of cylinder is smaller than or equal to 2d as shown in the figure below.

Physics Tutorials: This image provides visual information for the physics tutorial Pressure, Temperature and RMS Speed

A single molecule moving at speed v travels the distance v × Δt between two collisions. This corresponds to the height of the cylinder discussed earlier. Therefore the volume of cylinder is

V = π × d2 × v × ∆t

Therefore, mean free path is

Mean free path = Path length during ∆t/Number of collisions in ∆t
= v × ∆t/N
= v × ∆t/V × N/V
= v × ∆t/π × d2 × v × ∆t × N/V
= 1/π × d2 × N/V

The above equation does assumes all the other molecules at rest except one. This is obviously not true. If we want to take a more accurate result, we must consider a correction factor 1/2. Therefore, we obtain for the mean free path of gas molecules:

λ = 1/2 × π × d2 × N/V

Example 3

: 2 moles of a gas are enclosed inside a cubic container of side length equal to 4 m. The molecules of gas have a diameter of 10-10 m each. What is the mean free path of gas molecules?

Solution 3

We have a = 4 m, so V = a3 = (4 m)3 = 64 m3.

Also, N = n × NA = 2 moles × 6.02 × 1023 molecules/mole = 12.04 × 1023 molecules.

Therefore, we obtain for the mean free path λ:

λ = 1/2 × π × d2 × N/V
= 1/2 × 3.14 × (10-10)2 × 12.04 × 1023/64
= N/0.835 × 103
= 1.198 × 10-3 m
= 1.198 mm

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