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Derivation of the Kinetic Theory of Gases Equation (CIE A Level Physics)

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Derivation of the kinetic theory of gases equation

  • The equation for the kinetic theory of ideal gases contains the mean square speed of the particles:

less than c squared greater than

  • Where
    • c = average speed of the gas particles
    • <c2> has the units m2 s-2

  • Since particles travel in all directions in 3D space and velocity is a vector, some particles will have a negative direction and others a positive direction
  • When there are a large number of particles, the total positive and negative velocity values will cancel out, giving a net zero value overall
  • To calculate the average speed of the particles in a gas, take the square root of the mean square speed:

square root of less than c squared greater than end root space equals space c subscript r. m. s end subscript

  • cr.m.s is known as the root-mean-square speed and still has the units of m s-1
    • The mean square speed is not the same as the mean speed

Root-mean-square speed

  • Molecular movement causes the pressure exerted by a gas
    • When molecules rebound from their container wall, the change in momentum gives rise to a force exerted by the particles on the wall
    • Many molecules moving in random motion exert forces on the walls, which creates an average overall pressure (since pressure is the force per unit area)

Derivation of the kinetic theory of gases equation

Consider the model of a gas molecule in a container

  • Take a single molecule in a cube-shaped box with sides of equal length L
  • The molecule has a mass m and moves with speed c1, parallel to one side of the box
  • It collides at regular intervals with the sides 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 a total of N molecules can be deduced

Modelling a gas molecule in a container

Single molecule in box, downloadable AS & A Level Physics revision notes

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

Derivation of the equation for the pressure exerted by a gas

     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
  • If they rebound in the opposite direction to their initial velocity, their final velocity is -c
  • The change in momentum is therefore:

increment p space equals space minus m c space minus space open parentheses plus m c close parentheses space equals space minus m c space minus space m c space equals space minus 2 m c

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

  • The time between collisions of the molecule travelling to one wall and back is calculated by travelling a distance of 2l with speed c:

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

  • Note: c is not taken as the speed of light in this scenario

     3. Find the change in momentum per second

  • The force the molecule exerts on one wall is found using Newton’s second law of motion:

Force = rate of change of momentum = fraction numerator increment p over denominator increment t end fraction space equals space fraction numerator 2 m c space over denominator fraction numerator 2 l over denominator c end fraction end fraction space equals space fraction numerator m c squared space over denominator l end fraction

  • The change in momentum is +2mc since the force on the molecule from the wall is in the opposite direction to its change in momentum

     4. Calculate the total pressure from N molecules

  • The area of one wall is l2
  • The pressure, p is defined using the force and area:

p space equals fraction numerator space f o r c e over denominator a r e a end fraction space equals space fraction numerator fraction numerator m c squared space over denominator l end fraction over denominator l squared end fraction space equals space fraction numerator m c squared space over denominator l cubed end fraction

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

p space equals space fraction numerator N m c squared space over denominator l cubed end fraction

  • Each molecule has a different velocity and they all contribute to the pressure
  • The mean squared speed of c2 is written with left and right-angled brackets <c2>
  • The pressure is now defined as:

p space equals space fraction numerator N m less than c squared greater than space over denominator l cubed end fraction

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

  • The pressure equation still assumes all the molecules are travelling in the same direction and colliding with the same pair of opposite faces of the cube
  • In reality, all molecules will be moving in three dimensions equally
  • Splitting the velocity into its components cx, cy and cz to denote the amount in the x, y and z directions, c2 can be defined using pythagoras’ theorem in 3D:

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

  • Since there is nothing special about any particular direction, it can be determined that:

less than c subscript x greater than squared space equals space less than c subscript y greater than squared space space equals space less than c subscript z greater than squared space

  • Therefore, <cx2> can be defined as:

less than c subscript x greater than squared space equals 1 third less than c squared greater than

     6. Re-write the pressure equation

  • The box is a cube and all the sides are of length l
    • This means l3 is equal to the volume of the cube, V

  • Substituting the new values for <c2> and l3 back into the pressure equation obtains the final equation:

p V space space equals space 1 third N m less than c squared greater than

  • Where:
    • p = pressure (Pa)
    • V = volume (m3)
    • N = number of molecules 
    • m = mass of one molecule (kg)
    • <c2> = mean square speed of the molecules (m s–1)
  • This can also be written using the density ρ of the gas:

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

  • Rearranging the pressure equation for p and substituting the density ρ:

p space equals space 1 third rho less than c squared greater than

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 root-mean-square (r.m.s.) speed of the gas atoms at 504 K.

Answer:

Step 1: Write out the equation for the pressure of an ideal gas with density

p space equals space 1 third rho less than c squared greater than

Step 2: Rearrange for mean square speed

less than c squared greater than space equals fraction numerator space 3 p over denominator rho end fraction

Step 3: Substitute in values

less than c squared greater than space equals space fraction numerator 3 space cross times open parentheses 9.3 space cross times space 10 to the power of 5 close parentheses space over denominator 4.5 end fraction space equals space 6.2 space cross times space 10 to the power of 5 space straight m squared space straight s to the power of negative 2 end exponent

Step 4: To find the r.m.s value, take the square root of the mean square speed

c subscript r. m. s end subscript space equals space square root of less than c squared greater than end root space equals space square root of 6.2 space cross times space 10 to the power of 5 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 space open parentheses 2 space straight s. straight f. close parentheses

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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.