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Derivation of the Kinetic Theory of Gases Equation (SL IB Physics)

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Derivation of the Kinetic Theory of Gases Equation

  • Gas pressure arises due to the collisions of the gas particles with the walls of the container that holds it
    • When a gas particle collides with a wall in the container, it exerts a force on the wall
    • The random motion of a large number of molecules exerting a force on the walls creates an overall pressure
    • This is because pressure = force per unit area

Derivation

  • Picture a single molecule in a cube-shaped box with sides of equal length l
  • The molecule has a mass m and moves with speed c, parallel to one side of the box
  • It collides at regular intervals with the ends of the box, exerting a force and contributing to the pressure of the gas
  • By calculating the pressure this one molecule exerts on one end of the box, the total pressure produced by all the molecules can be deduced

2-3-6-kinetic-theory-derivation

A single molecule in a box collides with the walls and exerts a pressure

     1. Find the change in momentum as a single molecule hits a wall perpendicularly

  • One assumption of the kinetic theory is that molecules rebound elastically
    • This means there is no kinetic energy lost in the collision, so initial and final velocities are equal in magnitude
  • When a gas particle of mass m, travels at an average speed v, and hits a wall of the container, it undergoes a change in momentum due to the force exerted by the wall on the particle

initial momentum = m v

final momentum = negative m v

  • Therefore, the change in momentum increment p is:

increment p = final momentum – initial momentum

increment p space equals space minus m v space minus space open parentheses m v close parentheses space equals space minus 2 m v

  • Force is equal to the rate of change of momentum (Newton's Second Law), so the force exerted by the wall on a molecule is:

F space equals space fraction numerator increment p over denominator increment t end fraction space equals space minus fraction numerator 2 m v over denominator increment t end fraction

  • According to Newton's Third Law, the particle exerts an equal and opposite force on the wall
  • Therefore force exerted on the wall by the particle is:

F space equals space fraction numerator 2 m v over denominator increment t end fraction

2-3-6-gas-pressure

A particle hitting a wall of the container in which the gas is held experiences a force from the wall and a change in momentum. The particle exerts an equal and opposite force on the wall

     2. Calculate the number of collisions per second by the molecule on a wall

  • The time between collisions from one wall to the other, and then back again, over a distance of 2 l with speed v is:

time between collisions = fraction numerator d i s t a n c e over denominator s p e e d end fraction space equals space fraction numerator space 2 l over denominator v end fraction

     3. Find the change in momentum per second

  • The force the molecule exerts on one wall, F, from step 1 now becomes the following when substituting the time between collisions for Δt:

 F space equals space fraction numerator 2 m v over denominator increment t end fraction space equals space fraction numerator 2 m v space over denominator fraction numerator 2 l over denominator v end fraction end fraction space equals fraction numerator space m v squared over denominator l end fraction

     4. Calculate the total pressure from N molecules

  • The area of one wall is l squared
  • The pressure, P can be written as:

P space equals space fraction numerator space F over denominator A end fraction space equals space fraction numerator fraction numerator space m v squared over denominator l end fraction over denominator l squared end fraction space equals space fraction numerator space m v squared over denominator l cubed end fraction

  • This is the pressure exerted from one molecule
  • To account for the large number of N molecules, the pressure can now be written as:

P space equals fraction numerator space N m v squared over denominator l cubed end fraction

     5. Consider the effect of the molecule moving in 3D space

  • The pressure equation written above still assumes all the molecules are travelling in the same x direction and colliding with the same pair of opposite faces of the cube
    • To reflect this, it can be rewritten as:

P space equals fraction numerator space N m v subscript x squared over denominator l cubed end fraction

  • Where vx is the x component of the average velocity of all the particles
  • In reality, all molecules will be moving in three dimensions equally
  • Splitting the velocity into its components vx, vy and vz to denote the amount in the x, y and z directions, v2 (the square of the average speed) can be defined using pythagoras’ theorem in 3D:

v squared space equals space v subscript x squared space plus thin space v subscript y squared space plus space v subscript z squared

  • Since there is nothing special about any particular direction, it can be determined that the square of the average speed in each direction is equal:

v subscript x squared space equals space v subscript y squared space equals space v subscript z squared

v squared space equals space 3 v subscript x squared

  • Therefore, by combining the two equations, v can be defined as:

v subscript x squared space equals space 1 third v squared

     6. Re-write the pressure equation

  • The box is a cube with a side length l
  • Therefore, volume of the cube V space equals space l cubed

P space equals fraction numerator space N m v squared over denominator 3 V end fraction

 

  • This can also be written using the density ρ of the gas:

rho space equals space fraction numerator space m a s s over denominator v o l u m e end fraction space equals space fraction numerator N m space over denominator V end fraction

    • m is the mass of a single particle so Nm is the total mass of all the particles
  • Rearranging the pressure equation for p and substituting the density ρ:

P space equals space 1 third rho v squared

  • Where:
    • P = pressure of the gas (Pa)
    • ρ = density of the gas (kg m–3)
    • v = mean square speed (m s–1)
  • This is known as the Kinetic theory of gases equation

Worked example

An ideal gas has a density of 4.5 kg m-3 at a pressure of 9.3 × 105 Pa and a temperature of 504 K.

Determine the mean square speed of the gas molecules at 504 K.

Answer:

Step 1: List the known quantities

  • Density of the gas, ρ = 4.5 kg m-3 
  • Gas pressure, P = 9.3 × 105 Pa
  • Temperature, T = 504 K

Step 2: Rearrange the pressure equation for the mean square speed

P thin space equals space 1 third rho v squared

v space equals space square root of fraction numerator 3 P over denominator rho end fraction end root

Step 3:  Substitute in the values

v space equals space square root of 3 space cross times fraction numerator space 9.3 space cross times space 10 to the power of 5 over denominator 4.5 end fraction end root space equals space 787.4 space equals space 790 space straight m space straight s to the power of negative 1 end exponent

Examiner Tip

Although you will not be asked to recall the derivation, you need to understand the physics that underlines it and how to use the equation in the final result.

The 'mean square' speed just means an average speed, as we're assuming that all the molecules travel at the same speed.

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Ashika

Author: Ashika

Expertise: Physics Project Lead

Ashika graduated with a first-class Physics degree from Manchester University and, having worked as a software engineer, focused on Physics education, creating engaging content to help students across all levels. Now an experienced GCSE and A Level Physics and Maths tutor, Ashika helps to grow and improve our Physics resources.

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