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

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

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 $-c$

• The change in momentum is therefore:

$∆p=p_f-p_i$

$∆p=-mc-\left(+mc\right)=-mc-mc=-2mc$

     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 = $\frac{\mathrm{distance}}{\mathrm{speed}}=\frac{2l}{c}$

• 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:

$F=\frac{∆p}{∆t}$

$F=\frac{2mc}{{\displaystyle \frac{2l}{c}}}=\frac{m{c}^2}{l}$

• 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 $A$ of one wall is $l^2$

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

$p=\frac{F}{A}$

$p=\frac{{\displaystyle \frac{m{c}^2}{l}}}{l^2}=\frac{m{c}^2}{l^3}$

• 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:

$p=\frac{Nm{c}^2}{l^3}$

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

• The mean square speed $c^2$ is written with left and right-angled brackets $<c^2>$

• The pressure is now defined as:

$p=\frac{Nm<c^2>}{l^3}$

     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 $c$:

$c^2=c_x^2+c_y^2+c_z^2$

• Where $c_x$, $c_y$, and $c_z$ 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:

$<c_x^2>=<c_y^2>=<c_z^2>$

• Therefore, the mean square speed in the $x$-direction $<c_x^2>$ accounts for a third of the total mean square speed $<c^2>$

$<c_x^2>=\frac{1}{3}<c^2>$

• 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 $l^3$

• Substituting the new values for $<c^2>$ and $l^3$ back into the pressure equation obtains the final equation:

$pV=\frac{1}{3}Nm<c^2>$

• Where:

  • $p$ = pressure (Pa)

  • $V$ = volume (m³)

  • $N$ = number of molecules 

  • $m$ = mass of one molecule (kg)

  • $<c^2>$ = mean square speed of the molecules (m² s^–2)

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

$\rho =\frac{\mathrm{mass}}{\mathrm{volume}}=\frac{Nm}{V}$

• Rearranging the pressure equation for $p$ and substituting the density $\rho$:

$p=\frac{1}{3}\rho <c^2>$

Root-mean-square speed

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

$<c^2>$

• Where

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

  • $<c>$ = average velocity of the molecules in a gas (m s^-1)

  • $<c^2>$ = 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 $<c>$ 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 $c^2$ gives $<c^2>$, the mean square speed

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

$c_{r.m.s}=\sqrt{<c^2>}$

• $c_{r.m.s}$ is known as the root-mean-square speed and has units of m s^-1

  • The root-mean-square speed $c_{r.m.s}$ is not the same as the mean speed $<c>$