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

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 (e.g. mass and speed) to the macroscopic properties of gases (e.g. 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 (e.g. speed) is usually considered

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 $<c>$

How to derive the kinetic theory 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

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

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

Step-by-step derivation

   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 $\rho$:

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