Derivation of the Kinetic Theory of Gases Equation (Cambridge (CIE) A Level Physics): Revision Note

Exam code: 9702

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Derivation of the kinetic theory of gases equation

  • 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)

Setting up the model

  • Take a single molecule in a cube-shaped box with sides of equal length L

  • The molecule has mass m and moves with speed c 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

Deriving the equation step-by-step

     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 a molecule rebounds in the opposite direction to its initial velocity c, its final velocity will be negative c

  • 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 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 2 l with speed c:

Time between collisions = distance over speed space equals space fraction numerator 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 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 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 plus 2 m c 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 A of one wall is l squared

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

space p space equals space F over A

space p italic 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:

space 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 square speed c squared is written with left and right-angled brackets open angle brackets c squared close angle brackets

  • The pressure is now defined as:

space p space equals space fraction numerator N m open angle brackets c squared close angle brackets 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 and randomly, with different velocities c:

c squared space equals space c subscript x superscript 2 space plus space c subscript y superscript 2 space plus space c subscript z superscript 2

  • Where c subscript x, c subscript y, and c subscript z are the x-, y-, and z- components of the velocity

    • This equation is a result of Pythagoras' theorem in 3D

  • Since pressure is due to the force exerted on one face of the cube, only the component of velocity perpendicular to that face (e.g. the x-direction) matters

  • For a gas with molecules moving randomly in all directions, the mean square speed is evenly distributed among all three directions:

open angle brackets c subscript x superscript 2 close angle brackets space equals space open angle brackets c subscript y superscript 2 close angle brackets space equals space open angle brackets c subscript z superscript 2 close angle brackets

  • Therefore, the mean square speed in the x-direction open angle brackets c subscript x superscript 2 close angle brackets accounts for a third of the total mean square speed open angle brackets c squared close angle brackets

open angle brackets c subscript x superscript 2 close angle brackets space equals 1 third open angle brackets c squared close angle brackets

  • This is why only one-third of the total mean square speed contributes to the pressure on one wall

     6. Re-write the pressure equation

  • The box is a cube, and all the sides are of length l

    • This means the volume V of the cube is equal to l cubed

  • Substituting the new values for open angle brackets c squared close angle brackets and l cubed back into the pressure equation obtains the final equation:

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

  • Where:

    • space p = pressure (Pa)

    • V = volume (m3)

    • N = number of molecules 

    • m = mass of one molecule (kg)

    • open angle brackets c squared close angle brackets = mean square speed of the molecules (m s–1)

  • This can also be written using the density ρ 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 pressure equation for space p and substituting the density space rho:

space p space equals space 1 third rho open angle brackets c squared close angle brackets

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

Root-mean-square speed

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

open angle brackets c squared close angle brackets

  • Where

    • c = velocity of one gas molecule (m s-1)

    • open angle brackets c close angle brackets = average velocity of the molecules in a gas (m s-1)

    • open angle brackets c squared close angle brackets = average of the square of the speeds of the molecules in a gas (m2 s-2)

  • Since there are a large number of molecules in a gas, travelling in all directions in 3D space, this results in a large range of values of c

  • The average velocity open angle brackets c close angle brackets of all the molecules in a gas gives a net zero value overall

    • This is because velocity is a vector, so some molecules will have a negative direction and others a positive direction

  • Squaring the velocities gives positive values only

  • Averaging the values of c squared gives open angle brackets c squared close angle brackets, the mean square speed

  • Then, taking the square root of the mean square speed gives:

c subscript r. m. s end subscript space equals space square root of open angle brackets c squared close angle brackets end root

  • c subscript r. m. s end subscript is known as the root-mean-square speed and has units of m s-1

    • The root-mean-square speed c subscript r. m. s end subscript is not the same as the mean speed open angle brackets c close angle brackets

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Ashika

Author: Ashika

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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.