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Kinetic Theory of Gases Equation (AQA A Level Physics): Revision Note

Exam code: 7408

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Kinetic Theory of Gases Equation

Assumptions in Kinetic Theory

  • Gases consist of atoms or molecules randomly moving around at high speeds

  • The kinetic theory of gases models the thermodynamic behaviour of gases by linking the microscopic properties of particles (mass and speed) to macroscopic properties of particles (pressure and volume)

  • The theory is based on a set of the following assumptions:

    • Molecules of a gas behave as identical (or all have the same mass)

    • Molecules of gas are hard, perfectly elastic spheres

    • The volume of the molecules is negligible compared to the volume of the container

    • The time of a collision is negligible compared to the time between collisions

    • There are no intermolecular forces between the molecules (except during impact)

    • External forces (e.g. gravity) are ignored

    • The molecules move in continuous random motion

    • Newton's laws apply

    • There are a very large number of molecules

  • The number of molecules of gas in a container is very large, therefore the average behaviour (eg. speed) is usually considered

Derivation of the Kinetic Theory of Gases Equation

  • When molecules rebound from a wall in a container, the change in momentum gives rise to a force exerted by the particle on the wall

  • Many molecules moving in random motion exert forces on the walls, which create an average overall pressure (since pressure is the force per unit area)

Consider the Model

  • 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

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

1. Determine 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 the particle moving in a single direction with velocity c subscript 1 collides elastically with the wall, it will rebound with velocity negative c subscript 1

  • The change in momentum is therefore:

increment p space equals space p subscript f space minus space p subscript i

increment p space equals space minus m c subscript 1 space minus space m c subscript 1 space equals space minus 2 m c subscript 1

  • Where:

    • increment p = change in momentum of the molecule (kg m s-1)

    • m = mass of the molecule (kg)

    • c subscript 1 = speed of the molecule in a particular direction (m s-1)

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

  • The time between collisions of the molecule travelling to the opposite facing wall and back is calculated by travelling a distance 2 L with speed c subscript 1:

time space between space collisions space equals space distance over speed space equals space fraction numerator 2 L over denominator c subscript 1 end fraction

  • Note: c is not the speed of light in this model

3. Calculate the force exerted by the molecule on the wall

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

F space equals space fraction numerator increment p over denominator increment t end fraction

F space equals space fraction numerator 2 m c subscript 1 over denominator open parentheses fraction numerator 2 L over denominator c subscript 1 end fraction close parentheses end fraction space equals space fraction numerator m c subscript 1 superscript 2 over denominator L end fraction

  • The change in momentum is plus 2 m c subscript 1 since the force on the molecule from the wall is in the opposite direction to its change in momentum

4. Calculate the total pressure for one molecule

  • The area A of one wall is L squared

  • The pressure is defined as the force per unit area:

space p space equals space F over A

space p space equals space fraction numerator fraction numerator m c subscript 1 squared over denominator L end fraction over denominator L squared end fraction space equals space fraction numerator m c subscript 1 squared over denominator L cubed end fraction space equals space fraction numerator m c subscript 1 squared over denominator V end fraction

  • This is the pressure exerted from one molecule in a particular direction

5. Consider the effect of N molecules moving randomly in 3D space

  • For N particles travelling in the same direction as the single particle, pressure on a single face is now:

space p space equals space m over V open parentheses c subscript 1 squared space plus space c subscript 2 squared space plus space... space plus space c subscript N squared close parentheses

  • This pressure equation still assumes that all the molecules are travelling in the same direction

  • In reality, all molecules will be moving in three dimensions equally and randomly, with average velocity c:

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

  • Where c subscript x, c subscript y, and c subscript z are the components of the average velocity - this equation is a result of Pythagoras' theorem in 3D

  • The particles are moving randomly, so we can assume the magnitude of each component is equal to the other

c subscript x squared space equals space c subscript y squared space equals space c subscript z squared space equals space 1 third c squared

  • Each squared x-component of velocity accounts for a third of the actual velocity squared, allowing the equation for pressure on one wall to be amended:

space p space equals space 1 third m over V open parentheses c subscript 1 squared space plus space c subscript 2 squared space plus space... space plus space c subscript N squared close parentheses

6. Consider the speed of the molecules as an average speed

  • Each molecule has a different speed and they all contribute to the pressure

  • Therefore, the square root of the average of the square velocities is taken as the speed instead

  • This is called the root-mean-square speed c subscript r m s end subscript, which is defined as:

c subscript r m s end subscript space equals space square root of fraction numerator c subscript 1 superscript 2 space plus space c subscript 2 superscript 2 space plus space c subscript 3 superscript 2 space plus space... space plus space c subscript N superscript 2 over denominator N end fraction end root

  • Therefore:

N open parentheses c subscript r m s end subscript close parentheses squared space equals space c subscript 1 superscript 2 space plus space c subscript 2 superscript 2 space plus space c subscript 3 superscript 2 space plus space... space plus space c subscript N superscript 2

  • Finally, the pressure equation can be written as:

space p space equals space 1 third m over V N open parentheses c subscript r m s end subscript close parentheses squared

  • Multiplying both sides by the volume V gives the final Kinetic Theory of Gases Equation:

space p V space equals space 1 third N m open parentheses c subscript r m s end subscript close parentheses squared

  • Where:

    • space p = pressure (Pa)

    • V = volume (m3)

    • N = number of molecules

    • m = mass of one molecule of gas (kg)

    • c subscript r m s end subscript = root mean square speed of the molecules (m s-1)

  • The equation can also be written using the density space rho of the gas:

space rho space equals space mass over volume space equals space fraction numerator N m over denominator V end fraction

  • Rearranging the equation for pressure space p and substituting the density space rho gives the equation:

space p space equals space 1 third rho open parentheses c subscript r m s end subscript close parentheses squared

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 c subscript r m s end subscript of the gas atoms at 504 K.

Answer:

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

space p space equals space 1 third rho open parentheses c subscript r m s end subscript close parentheses squared

Step 2: Rearrange for mean square speed

open parentheses c subscript r m s end subscript close parentheses squared space equals fraction numerator space 3 p over denominator rho end fraction

Step 3: Substitute in values

open parentheses c subscript r m s end subscript close parentheses squared space equals space fraction numerator 3 cross times open parentheses 9.3 cross times 10 to the power of 5 close parentheses space over denominator 4.5 end fraction

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 fraction numerator 3 cross times open parentheses 9.3 cross times 10 to the power of 5 close parentheses space over denominator 4.5 end fraction end root space 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

Examiner Tips and Tricks

Make sure to revise and understand each step for the whole of the derivation, as you may be asked to derive all, or part, of the equation in an exam question. Ensure you also write the appropriate commentary instead of simply stating equations in your answers to get full marks.

Also, make sure to memorise all the assumptions for your exams, as it is a common exam question to be asked to recall them.

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    Katie M

    Author: Katie M

    Expertise: Physics Content Creator

    Katie has always been passionate about the sciences, and completed a degree in Astrophysics at Sheffield University. She decided that she wanted to inspire other young people, so moved to Bristol to complete a PGCE in Secondary Science. She particularly loves creating fun and absorbing materials to help students achieve their exam potential.