Assumptions of the Kinetic Theory of Gases
- 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 gas behave as identical, 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 forces of attraction or repulsion between the molecules
- The molecules are in continuous random motion
- The number of molecules of gas in a container is very large, therefore the average behaviour (eg. speed) is usually considered
Exam Tip
Make sure to memorise all the assumptions for your exams, as it is a common exam question to be asked to recall them.
Root-Mean-Square Speed
- The pressure of an ideal gas equation includes the mean square speed of the particles:
$
\left\langle\mathrm{c}^2\right\rangle
$
- Where
- $c=$ average speed of the gas particles
- $\circ\left\langle\mathrm{c}^2\right\rangle$ has the units $\mathbf{m}^2 \mathbf{s}^{-2}$
- Since particles travel in all directions in 3D space and velocity is a vector, some particles will have a negative direction and others a positive direction
- When there are a large number of particles, the total positive and negative velocity values will cancel out, giving a net zero value overall
- In order to find the pressure of the gas, the velocities must be squared
- This is a more useful method, since a negative or positive number squared is always positive
- To calculate the average speed of the particles in a gas, take the square root of the mean square speed:
$
\sqrt{<c^2>}=c_{r . m . s}
$
- $\mathrm{c}_{\mathrm{r} \text {.m.s }}$ is known as the root-mean-square speed and still has the units of $\mathbf{m ~ s} \mathbf{s}^{-1}$
- The mean square speed is not the same as the mean speed
Worked example: Root-mean-square speed
An ideal gas has a density of $4.5 \mathrm{~kg} \mathrm{~m}^{-3}$ at a pressure of $9.3 \times 10^5 \mathrm{~Pa}$ and a temperature of $504 \mathrm{~K}$. Determine the root-mean-square (r.m.s.) speed of the gas atoms at $504 \mathrm{~K}$.
Answer/Explanation
Step 1:
Write out the equation for the pressure of an ideal gas with density
$
p=\frac{1}{3} \rho<c^2>
$
Step 2:
Rearrange for mean square speed
$
<c^2>=\frac{3 p}{\varrho}
$
Step 3:
Substitute in values
$
\left\langle c^2\right\rangle=\frac{3 \times\left(9.3 \times 10^5\right)}{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
$
\mathrm{c}_{\mathrm{r} . \mathrm{m} . \mathrm{s}}=\sqrt{\left\langle c^2\right\rangle}=\sqrt{6.2 \times 10^5}=787.4=790 \mathrm{~m} \mathrm{~s}^{-1}(2 \mathrm{~s} . \mathrm{f})
$
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
- Picture a single molecule in a cube-shaped box with sides of equal length I
- The molecule has a mass $m$ and moves with speed $c$, parallel to one side of the box
- It collides at regular intervals with the ends 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 all the molecules can be deduced
5 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 they rebound in the opposite direction to their initial velocity, their final velocity is -c
- The change in momentum is therefore:
$
\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 $2 /$ with speed $\mathrm{c}$ :
$
\text { Time between collisions }=\frac{\text { distance }}{\text { speed }}=\frac{2 l}{c}
$
- Note: $\mathrm{c}$ is not taken as the speed of light in this scenario
3. Find the change in momentum per second
- The force the molecule exerts on one wall is found using Newton’s second law of motion:
$
\text { Force }=\text { rate of change of momentum }=\frac{\Delta p}{\Delta t}=\frac{2 m c}{\frac{2 l}{c}}=\frac{m c^2}{l}
$
- The change in momentum is $+2 \mathrm{mc}$ 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 $\mathrm{N}$ molecules
- The area of one wall is $l^2$
- The pressure is defined using the force and area:
- $
\text { Pressure } \mathrm{p}=\frac{\text { Force }}{\text { Area }}=\frac{m c^2}{l}=\frac{m c^2}{l^2}
$ - This is the pressure exerted from one molecule
- To account for the large number of $N$ molecules, the pressure can now be written as:
$
\mathrm{p}=\frac{N m c^2}{l^3}
$
- Each molecule has a different velocity and they all contribute to the pressure
- The mean squared speed of $\mathrm{c}^2$ is written with left and right-angled brackets $\left\langle\mathrm{c}^2\right\rangle$
- The pressure is now defined as:
$
\mathrm{p}=\frac{N m<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
- Splitting the velocity into its components $c_x, c_y$ and $c_z$ to denote the amount in the $x, y$ and $z$ directions, $c^2$ can be defined using pythagoras’ theorem in 3D:
$
c^2=c_x^2+c_y^2+c_z^2
$
- Since there is nothing special about any particular direction, it can be determined that:
$
\left\langle\mathrm{c}_x^2\right\rangle=\left\langle\mathrm{c}_y^2\right\rangle=\left\langle\mathrm{c}_z^2\right\rangle
$
- Therefore, $\left\langle\mathrm{c}_x^2\right\rangle$ can be defined as:
$
\left\langle\mathrm{C}_x^2\right\rangle=\frac{1}{3}\left\langle\mathrm{c}^2\right\rangle
$
- The box is a cube and all the sides are of length I
- This means $\beta^3$ is equal to the volume of the cube, $V$
- Substituting the new values for $\left\langle c^2\right\rangle$ and $\beta^3$ back into the pressure equation obtains the final equation:
$
\mathrm{pV}=\frac{1}{3} \mathrm{Nm}<\mathrm{c}^2>
$
- This is known as the Kinetic Theory of Gases equation
- Where:
- $p=$ pressure $(\mathrm{Pa})$
- $\mathrm{V}=$ volume $\left(\mathrm{m}^3\right)$
- $\mathrm{N}=$ number of molecules
- $\mathrm{m}=$ mass of one molecule of gas $(\mathrm{kg})$
- $\left\langle\mathrm{c}^2\right\rangle=$ mean square speed of the molecules $\left(\mathrm{m} \mathrm{s}^{-1}\right)$
- This can also be written using the density $\rho$ of the gas:
$
\rho=\frac{m a s s}{\text { volume }}=\frac{N m}{V}
$
- Rearranging the pressure equation for $p$ and substituting the density $\rho$ :
$
p=\frac{1}{3} \rho<c^2>
$
Average Kinetic Energy of a Molecule
- An important property of molecules in a gas is their average kinetic energy
- This can be deduced from the ideal gas equations relating pressure, volume, temperature and speed
- Recall the ideal gas equation:
$
\mathbf{p V}=\mathbf{N k T}
$
- Also recall the equation linking pressure and mean square speed of the molecules:
$
\mathrm{pV}=\frac{1}{3} \mathrm{Nm}<\mathrm{c}^2>
$
- The left hand side of both equations are equal (pV)
- This means the right hand sides are also equal:
$
\frac{1}{3} \mathrm{Nm}<\mathrm{c}^2>=\mathrm{NkT}
$
- $\mathrm{N}$ will cancel out on both sides and multiplying by 3 obtains the equation:
$
m\left\langle c^2\right\rangle=3 k T
$
- Recall the familiar kinetic energy equation from mechanics:
$
\text { Kinetic energy }=\frac{1}{2} m v^2
$
- Instead of $v^2$ for the velocity of one particle, $\left\langle c^2\right\rangle$ is the average speed of all molecules
Multiplying both sides of the equation by $1 / 2$ obtains the average translational kinetic energy of the molecules of an ideal gas:
$
E_K=\frac{1}{2} m<c^2>=\frac{3}{2} k T
$
- Where:
- $\mathrm{E}_{\mathrm{K}}=$ kinetic energy of a molecule (J)
- $\mathrm{m}=$ mass of one molecule $(\mathrm{kg})$
- $\left\langle\mathrm{c}^2\right\rangle=$ mean square speed of a molecule $\left(\mathrm{m}^2 \mathrm{~s}^{-2}\right)$
- $\mathrm{k}=$ Boltzmann constant
- $\mathrm{T}=$ temperature of the gas $(\mathrm{K})$
- Note: this is the average kinetic energy for only one molecule of the gas
- A key feature of this equation is that the mean kinetic energy of an ideal gas molecule is proportional to its thermodynamic temperature
$
\mathbf{E}_{\mathrm{k}} \propto \mathbf{T}
$
- Translational kinetic energy is defined as:
The energy a molecule has as it moves from one point to another
- Translational kinetic energy is defined as:
- A monatomic (one atom) molecule only has translational energy, whilst a diatomic (twoatom) molecule has both translational and rotational energy
Worked example: Average kinetic energy of a molecule
Helium can be treated as an ideal gas. Helium molecules have a root-mean-square (r.m.s.) speed of $730 \mathrm{~m} \mathrm{~s}^{-1}$ at a temperature of $45^{\circ} \mathrm{C}$. Calculate the r.m.s. speed of the molecules at a temperature of $80^{\circ} \mathrm{C}$.
Answer/Explanation
Step 1:
Write down the equation for the average translational kinetic energy:
$
\left.E_K=\frac{1}{2} m<c^2\right\rangle=\frac{3}{2} k T
$
Step 2: Find the relation between $c_{\text {r.m.s }}$ and temperature $T$
Since $\mathrm{m}$ and $\mathrm{k}$ are constant, $\left\langle\mathrm{c}^2\right\rangle$ is directly proportional to $\mathrm{T}$
$
<\mathrm{c}^2>\propto \mathbf{T}
$
Therefore, the relation between $c_{r . m . s}$ and $\mathrm{T}$ is:
$
\mathrm{c}_{\mathrm{r} . \mathrm{m} . \mathrm{s}}=\sqrt{\left\langle c^2\right\rangle} \propto \sqrt{T}
$
Step 3:
Write the equation in full
$
\mathrm{c}_{\mathrm{r} . \mathrm{m} . \mathrm{s}}=\mathrm{a} \sqrt{T}
$
where $a$ is the constant of proportionality
Step 4: Calculate the constant of proportionality from values given by rearranging for a:
$
T=45^{\circ} \mathrm{C}+273.15=318.15 \mathrm{~K}
$
$\mathrm{a}=\frac{c_{r . m . s}}{\sqrt{T}}=\frac{730}{\sqrt{318.15}}=40.92 \ldots$
Step 5:
Calculate $c_{\text {r.m.s }}$ at $80{ }^{\circ} \mathrm{C}$ by substituting the value of a and new value of $\mathrm{T}$
$
T=80^{\circ} \mathrm{C}+273.15=353.15 \mathrm{~K}
$
$
c_{r . m . s}=\frac{730}{\sqrt{318.15}} \times \sqrt{353.15}=769=770 \mathrm{~m} \mathrm{~s}^{-1}(2 \text { s.f. })
$
Exam Tip
Keep in mind this particular equation for kinetic energy is only for one molecule in the gas. If you want to find the kinetic energy for all the molecules, remember to multiply by $\boldsymbol{N}$, the total number of molecules. You can remember the equation through the rhyme ‘Average K.E is threehalves $\mathrm{kT}^{\prime}$.