Derivation of the Kinetic Theory of Gases Equation (Cambridge International A-Level Physics)

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:

$\Delta p=p_{f}-p_{i}$

$\Delta p = - m c - ( + m c ) = - m c - m c = - 2 m c$

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

• 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:

$F=\frac{\Delta p}{\Delta t}$

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

• 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:

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

$p=\frac{\frac{mc^{2}}{l}}{l^{2}}=\frac{mc^{2}}{l^{3}}$

• 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:

$p=\frac{Nmc^{2}}{l^{3}}$

• 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:

$p = \frac{N m ( \text{angle} \text{brackets} c^{2} ) \text{angle} \text{brackets} }{l^{3}}$

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 $\langle c_{x}^{2} \rangle$ accounts for a third of the total mean square speed < c ^2 >

$( \text{angle} \text{brackets} c_{x}^{2} ) \text{angle} \text{brackets} = 1 \frac{1}{3} ( \text{angle} \text{brackets} c^{2} ) \text{angle} \text{brackets}$

• 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 $l^{3}$

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

$p V = 1 \frac{1}{3} N m ( \text{angle} \text{brackets} c^{2} ) \text{angle} \text{brackets}$

• 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:

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

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

$p = 1 \frac{1}{3} \\rho ( \text{angle} \text{brackets} c^{2} ) \text{angle} \text{brackets}$

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

$p=\frac{1}{3}ρ<c^{2}>$

Step 2: Rearrange for mean square speed

$<c^{2}>=\frac{3p}{ρ}$

Step 3: Substitute in values

< c^{2} > = \frac{3 \times ( 9.3 \times 10^{5 ) }{4.5} = 6.2 \times 10^{5} \mathrm{m}^{2} \mathrm{s}^{-2} }

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

$c_{r. m. s} = \sqrt{< c^{2} >} = \sqrt{6.2 \times 10^{5} = 787.4 = 790 \mathrm{m} \mathrm{s}^{-1} } ( 2 \mathrm{s} . \mathrm{f} . )$

Root-Mean-Square Speed

Spec point — 15.3.1

• 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)
  • $\langle c \rangle$ = 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 $\langle c \rangle$ 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:

$c_{r. m. s} = \sqrt{( \text{angle} \text{brackets} c^{2} ) \text{angle} brackets}$

• $c_{r.m.s}$ 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 $\langle c \rangle$