Kinetic Theory

What you'll learn

Use the ideal gas equations $p V = n R T$ and $p V = N k T$ .
Convert between mass, molar mass, moles, and number of molecules.
Explain gas pressure using molecular collisions and Newton’s laws.
Show why absolute temperature is proportional to mean molecular kinetic energy.

The big idea: macroscopic and microscopic views

A gas can be described in two ways.

The macroscopic view uses measurable quantities such as pressure, volume, temperature, and amount of substance. The microscopic view describes the motion of individual molecules.

Definition

Pressure and absolute temperature

Pressure $p$ is force per unit area, measured in pascal, Pa. Absolute temperature $T$ is temperature measured in kelvin, K, and is the temperature scale used in gas equations.

Temperature must be in kelvin. To convert from degrees Celsius, add 273.15.

Common Mistake

Using degrees Celsius in gas equations

Never substitute a Celsius temperature into $p V = n R T$ or $p V = N k T$ . A gas at 20 °C has $293\ \mathrm{K}$ , not 20 K.

Moles, molecules, and molar mass

The amount of substance is measured in moles, mol. One mole contains a fixed number of specified particles: atoms, molecules, ions, or whatever particle type is being counted.

Definition

Avogadro constant and the mole

The Avogadro constant $N_A$ is the number of particles in one mole: $mol−1N_A \approx 6.02 \times 10^{23}\ \mathrm{mol^{-1}}$ . A mole is an amount of substance containing this number of specified particles.

If a sample contains $n$ moles and $N$ molecules,

N = nN_A

The relative molecular mass $M_r$ has no unit. The molar mass $M$ is the mass of one mole of the substance.

Eduqas writes the conversion as

\text{kg} = \frac{M_r}{1000}

meaning that $M$ is in kilogram per mole, kg mol^-1.

The number of moles is

\frac{\text{total mass}}{\text{molar mass}} = \frac{m_{\text{total}}}{M}

Example

Finding moles and molecules from mass

A sample of nitrogen molecules has total mass $1.40 \times 10^{-2}\ \mathrm{kg}$ . Nitrogen has $M_r = 28.0$ . Find the amount in moles and the number of molecules.

Convert relative molecular mass to molar mass:
$\frac{28.0}{1000}\ \mathrm{kg\,mol^{-1}} = 2.80 \times 10^{-2}\ \mathrm{kg\,mol^{-1}}$
Use mass divided by molar mass:
$\frac{1.40 \times 10^{-2}\ \mathrm{kg}}{2.80 \times 10^{-2}\ \mathrm{kg\,mol^{-1}}} = 0.500\ \mathrm{mol}$
Convert moles to molecules:
$nN_A = (0.500\ \mathrm{mol})(6.02 \times 10^{23}\ \mathrm{mol^{-1}}) = 3.01 \times 10^{23}$

The ideal gas equation of state

An ideal gas is a model gas that obeys the ideal gas equation exactly. Real gases are closest to ideal behaviour at low pressure and high temperature.

Definition

Equation of state

An equation of state links the pressure, volume, temperature, and amount of gas in a sample. For an ideal gas, the equation of state can be written using moles or using molecules.

Using moles:

p V = n R T

where $R$ is the molar gas constant:

8.31\ \mathrm{J\,mol^{-1}\,K^{-1}}

Using number of molecules:

p V = N k T

where $k$ is the Boltzmann constant:

\frac{R}{N_A} \approx 1.38 \times 10^{-23}\ \mathrm{J\,K^{-1}}

At constant amount of gas and constant temperature,

\frac{nRT}{V}

so pressure is inversely proportional to volume. This has the same shape as $\frac{k}{x}$ .

Graphs showing pressure against volume and pressure against reciprocal volume for an ideal gas at constant temperature

In practical or data-analysis questions, a graph of $p$ against $1V\frac{1}{V}$ should be a straight line through the origin if temperature and amount of gas are constant. The gradient is $n R T$ .

Common Mistake

Isothermal means constant temperature

If a gas is compressed quickly, its temperature may rise. Then $p V$ is not constant, because the process is not isothermal.

Example

Finding the number of molecules in a flask

A sealed flask contains gas at pressure $1.20 \times 10^5\ \mathrm{Pa}$ , volume $2.00 \times 10^{-3}\ \mathrm{m^3}$ , and temperature $K290\ \mathrm{K}$ . Find the number of molecules.

Use the molecular form because the question asks for molecules:
$\frac{pV}{kT}$
Substitute the values, noting that $m3\mathrm{Pa\,m^3}$ is equivalent to joule:
$\frac{(1.20 \times 10^5\ \mathrm{Pa})(2.00 \times 10^{-3}\ \mathrm{m^3})} {(1.38 \times 10^{-23}\ \mathrm{J\,K^{-1}})(290\ \mathrm{K})}$
Calculate and quote a sensible number of significant figures:
$\frac{240\ \mathrm{J}}{4.00 \times 10^{-21}\ \mathrm{J}} = 5.99 \times 10^{22} \approx 6.0 \times 10^{22}$

Assumptions of the kinetic theory of gases

The kinetic theory of gases explains gas behaviour using moving molecules and Newton’s laws.

For the ideal gas model, assume:

The gas contains a very large number of molecules.
The molecules are in continuous random motion.
The volume of the molecules is negligible compared with the volume of the container.
There are no intermolecular forces except during collisions.
Collisions between molecules and with the container walls are perfectly elastic.
The time taken for a collision is negligible compared with the time between collisions.
Energy is randomly distributed among the molecules, so molecules do not all have the same speed or kinetic energy.

Key Idea

Averages are essential

Because molecular speeds are randomly distributed, kinetic theory uses average quantities such as the mean square speed $c2‾\overline{c^2}$ rather than trying to track one molecule.

How molecular motion causes pressure

A gas molecule colliding with a wall changes momentum. The wall exerts a force on the molecule, and by Newton’s third law the molecule exerts an equal and opposite force on the wall. Many collisions per second create a steady pressure.

Kinetic theory diagram showing gas molecules colliding with container walls and producing pressure

For a gas of density $ρ\rho$ , molecule mass $m$ , volume $V$ , and number of molecules $N$ ,

\frac{1}{3}\rho \overline{c^2} = \frac{1}{3}\frac{N}{V}m\overline{c^2}

Here, $c$ is molecular speed and $c2‾\overline{c^2}$ is the mean square speed.

The factor $13\frac{1}{3}$ appears because molecular motion is random in three dimensions. On average, one third of the squared speed contributes to motion in each direction.

The root mean square speed is

crms=c2‾c_{\mathrm{rms}} = \sqrt{\overline{c^2}}

Common Mistake

Mean square is not mean speed squared

Do not replace $c2‾\overline{c^2}$ with $(c‾)2\left(\overline{c}\right)^2$ . You square each molecule’s speed first, then average those squared speeds.

Example

Calculating root mean square speed

A gas has pressure $1.00 \times 10^5\ \mathrm{Pa}$ and density $m−30.160\ \mathrm{kg\,m^{-3}}$ . Find $crmsc_{\mathrm{rms}}$ .

Start from the pressure equation and use $crms2=c2‾c_{\mathrm{rms}}^2 = \overline{c^2}$ :
$\frac{1}{3}\rho c_{\mathrm{rms}}^2$
Rearrange for $crmsc_{\mathrm{rms}}$ :
$crms=3pρc_{\mathrm{rms}} = \sqrt{\frac{3p}{\rho}}$
Substitute values and check the units become speed squared inside the square root:
$s−1c_{\mathrm{rms}} = \sqrt{\frac{3(1.00 \times 10^5\ \mathrm{Pa})}{0.160\ \mathrm{kg\,m^{-3}}}} = \sqrt{1.875 \times 10^6\ \mathrm{m^2\,s^{-2}}} = 1.37 \times 10^3\ \mathrm{m\,s^{-1}}$

Temperature and molecular kinetic energy

Multiplying the pressure equation by volume gives

\frac{1}{3}Nm\overline{c^2}

For an ideal gas,

p V = N k T

Combining them:

13Nmc2‾=NkT13mc2‾=kT12mc2‾=32kT\begin{aligned} \frac{1}{3}Nm\overline{c^2} &= NkT \\ \frac{1}{3}m\overline{c^2} &= kT \\ \frac{1}{2}m\overline{c^2} &= \frac{3}{2}kT \end{aligned}

So the mean translational kinetic energy of one molecule is

Ek,mean=32kTE_{\mathrm{k,mean}} = \frac{3}{2}kT

For one mole of a monatomic gas, there are $N_A$ molecules, so the total translational kinetic energy is

Ek,mole=NA(32kT)=32RTE_{\mathrm{k,mole}} = N_A\left(\frac{3}{2}kT\right) = \frac{3}{2}RT

Key Idea

Temperature measures mean molecular kinetic energy

On the kelvin scale, $T$ is proportional to the mean translational kinetic energy of the molecules. A hotter gas has a larger mean molecular kinetic energy, but not every molecule has the same energy.

Example

Calculating mean kinetic energy at room temperature

Find the mean translational kinetic energy of one molecule, and of one mole, for a monatomic ideal gas at $K300\ \mathrm{K}$ .

For one molecule, use the Boltzmann constant form:
$Ek,mean=32kTE_{\mathrm{k,mean}} = \frac{3}{2}kT$
Substitute the temperature:
$JE_{\mathrm{k,mean}} = \frac{3}{2}(1.38 \times 10^{-23}\ \mathrm{J\,K^{-1}})(300\ \mathrm{K}) = 6.21 \times 10^{-21}\ \mathrm{J}$
For one mole, use the molar gas constant form:
$mol−1E_{\mathrm{k,mole}} = \frac{3}{2}RT = \frac{3}{2}(8.31\ \mathrm{J\,mol^{-1}\,K^{-1}})(300\ \mathrm{K}) = 3.74 \times 10^3\ \mathrm{J\,mol^{-1}}$

Exam technique

In the exam

Convert first: use kelvin, cubic metre, kilogram, pascal, and mole before substituting.
Choose the gas equation by what is given: use $p V = n R T$ for moles and $p V = N k T$ for molecules.
For kinetic theory, use averages carefully: $c2‾\overline{c^2}$ and $crms=c2‾c_{\mathrm{rms}} = \sqrt{\overline{c^2}}$ .
In graph questions, constant temperature and fixed amount give $\propto \frac{1}{V}$ , so plot $p$ against $1V\frac{1}{V}$ for a straight line.

Self review

Check yourself

Why must temperature be measured in kelvin in the ideal gas equations?
How would you find the number of molecules from the mass of a gas sample?
Starting from $\frac{1}{3}Nm\overline{c^2}$ and $p V = N k T$ , can you show that $Ek,mean=32kTE_{\mathrm{k,mean}} = \frac{3}{2}kT$ ?

What you'll learn

Use the ideal gas equations $p V = n R T$ and $p V = N k T$ .
Convert between mass, molar mass, moles, and number of molecules.
Explain gas pressure using molecular collisions and Newton’s laws.
Show why absolute temperature is proportional to mean molecular kinetic energy.

The big idea: macroscopic and microscopic views

A gas can be described in two ways.

The macroscopic view uses measurable quantities such as pressure, volume, temperature, and amount of substance. The microscopic view describes the motion of individual molecules.

Definition

Pressure and absolute temperature

Pressure $p$ is force per unit area, measured in pascal, Pa. Absolute temperature $T$ is temperature measured in kelvin, K, and is the temperature scale used in gas equations.

Temperature must be in kelvin. To convert from degrees Celsius, add 273.15.

Common Mistake

Using degrees Celsius in gas equations

Never substitute a Celsius temperature into $p V = n R T$ or $p V = N k T$ . A gas at 20 °C has $293\ \mathrm{K}$ , not 20 K.

Moles, molecules, and molar mass

The amount of substance is measured in moles, mol. One mole contains a fixed number of specified particles: atoms, molecules, ions, or whatever particle type is being counted.

Definition

Avogadro constant and the mole

If a sample contains $n$ moles and $N$ molecules,

N = nN_A

The relative molecular mass $M_r$ has no unit. The molar mass $M$ is the mass of one mole of the substance.

Eduqas writes the conversion as

\text{kg} = \frac{M_r}{1000}

meaning that $M$ is in kilogram per mole, kg mol^-1.

The number of moles is

\frac{\text{total mass}}{\text{molar mass}} = \frac{m_{\text{total}}}{M}

Example

Finding moles and molecules from mass

A sample of nitrogen molecules has total mass $1.40 \times 10^{-2}\ \mathrm{kg}$ . Nitrogen has $M_r = 28.0$ . Find the amount in moles and the number of molecules.

Convert relative molecular mass to molar mass:
$\frac{28.0}{1000}\ \mathrm{kg\,mol^{-1}} = 2.80 \times 10^{-2}\ \mathrm{kg\,mol^{-1}}$
Use mass divided by molar mass:
$\frac{1.40 \times 10^{-2}\ \mathrm{kg}}{2.80 \times 10^{-2}\ \mathrm{kg\,mol^{-1}}} = 0.500\ \mathrm{mol}$
Convert moles to molecules:
$nN_A = (0.500\ \mathrm{mol})(6.02 \times 10^{23}\ \mathrm{mol^{-1}}) = 3.01 \times 10^{23}$

The ideal gas equation of state

An ideal gas is a model gas that obeys the ideal gas equation exactly. Real gases are closest to ideal behaviour at low pressure and high temperature.

Definition

Equation of state

An equation of state links the pressure, volume, temperature, and amount of gas in a sample. For an ideal gas, the equation of state can be written using moles or using molecules.

Using moles:

p V = n R T

where $R$ is the molar gas constant:

8.31\ \mathrm{J\,mol^{-1}\,K^{-1}}

Using number of molecules:

p V = N k T

where $k$ is the Boltzmann constant:

\frac{R}{N_A} \approx 1.38 \times 10^{-23}\ \mathrm{J\,K^{-1}}

At constant amount of gas and constant temperature,

\frac{nRT}{V}

so pressure is inversely proportional to volume. This has the same shape as $\frac{k}{x}$ .

Graphs showing pressure against volume and pressure against reciprocal volume for an ideal gas at constant temperature

In practical or data-analysis questions, a graph of $p$ against $1V\frac{1}{V}$ should be a straight line through the origin if temperature and amount of gas are constant. The gradient is $n R T$ .

Common Mistake

Isothermal means constant temperature

If a gas is compressed quickly, its temperature may rise. Then $p V$ is not constant, because the process is not isothermal.

Example

Finding the number of molecules in a flask

A sealed flask contains gas at pressure $1.20 \times 10^5\ \mathrm{Pa}$ , volume $2.00 \times 10^{-3}\ \mathrm{m^3}$ , and temperature $K290\ \mathrm{K}$ . Find the number of molecules.

Use the molecular form because the question asks for molecules:
$\frac{pV}{kT}$
Substitute the values, noting that $m3\mathrm{Pa\,m^3}$ is equivalent to joule:
$\frac{(1.20 \times 10^5\ \mathrm{Pa})(2.00 \times 10^{-3}\ \mathrm{m^3})} {(1.38 \times 10^{-23}\ \mathrm{J\,K^{-1}})(290\ \mathrm{K})}$
Calculate and quote a sensible number of significant figures:
$\frac{240\ \mathrm{J}}{4.00 \times 10^{-21}\ \mathrm{J}} = 5.99 \times 10^{22} \approx 6.0 \times 10^{22}$

Assumptions of the kinetic theory of gases

The kinetic theory of gases explains gas behaviour using moving molecules and Newton’s laws.

For the ideal gas model, assume:

The gas contains a very large number of molecules.
The molecules are in continuous random motion.
The volume of the molecules is negligible compared with the volume of the container.
There are no intermolecular forces except during collisions.
Collisions between molecules and with the container walls are perfectly elastic.
The time taken for a collision is negligible compared with the time between collisions.
Energy is randomly distributed among the molecules, so molecules do not all have the same speed or kinetic energy.

Key Idea

Averages are essential

Because molecular speeds are randomly distributed, kinetic theory uses average quantities such as the mean square speed $c2‾\overline{c^2}$ rather than trying to track one molecule.

How molecular motion causes pressure

Kinetic theory diagram showing gas molecules colliding with container walls and producing pressure

For a gas of density $ρ\rho$ , molecule mass $m$ , volume $V$ , and number of molecules $N$ ,

\frac{1}{3}\rho \overline{c^2} = \frac{1}{3}\frac{N}{V}m\overline{c^2}

Here, $c$ is molecular speed and $c2‾\overline{c^2}$ is the mean square speed.

The factor $13\frac{1}{3}$ appears because molecular motion is random in three dimensions. On average, one third of the squared speed contributes to motion in each direction.

The root mean square speed is

crms=c2‾c_{\mathrm{rms}} = \sqrt{\overline{c^2}}

Common Mistake

Mean square is not mean speed squared

Do not replace $c2‾\overline{c^2}$ with $(c‾)2\left(\overline{c}\right)^2$ . You square each molecule’s speed first, then average those squared speeds.

Example

Calculating root mean square speed

A gas has pressure $1.00 \times 10^5\ \mathrm{Pa}$ and density $m−30.160\ \mathrm{kg\,m^{-3}}$ . Find $crmsc_{\mathrm{rms}}$ .

Start from the pressure equation and use $crms2=c2‾c_{\mathrm{rms}}^2 = \overline{c^2}$ :
$\frac{1}{3}\rho c_{\mathrm{rms}}^2$
Rearrange for $crmsc_{\mathrm{rms}}$ :
$crms=3pρc_{\mathrm{rms}} = \sqrt{\frac{3p}{\rho}}$
Substitute values and check the units become speed squared inside the square root:
$s−1c_{\mathrm{rms}} = \sqrt{\frac{3(1.00 \times 10^5\ \mathrm{Pa})}{0.160\ \mathrm{kg\,m^{-3}}}} = \sqrt{1.875 \times 10^6\ \mathrm{m^2\,s^{-2}}} = 1.37 \times 10^3\ \mathrm{m\,s^{-1}}$

Temperature and molecular kinetic energy

Multiplying the pressure equation by volume gives

\frac{1}{3}Nm\overline{c^2}

For an ideal gas,

p V = N k T

Combining them:

13Nmc2‾=NkT13mc2‾=kT12mc2‾=32kT\begin{aligned} \frac{1}{3}Nm\overline{c^2} &= NkT \\ \frac{1}{3}m\overline{c^2} &= kT \\ \frac{1}{2}m\overline{c^2} &= \frac{3}{2}kT \end{aligned}

So the mean translational kinetic energy of one molecule is

Ek,mean=32kTE_{\mathrm{k,mean}} = \frac{3}{2}kT

For one mole of a monatomic gas, there are $N_A$ molecules, so the total translational kinetic energy is

Ek,mole=NA(32kT)=32RTE_{\mathrm{k,mole}} = N_A\left(\frac{3}{2}kT\right) = \frac{3}{2}RT

Key Idea

Temperature measures mean molecular kinetic energy

Example

Calculating mean kinetic energy at room temperature

Find the mean translational kinetic energy of one molecule, and of one mole, for a monatomic ideal gas at $K300\ \mathrm{K}$ .

For one molecule, use the Boltzmann constant form:
$Ek,mean=32kTE_{\mathrm{k,mean}} = \frac{3}{2}kT$
Substitute the temperature:
$JE_{\mathrm{k,mean}} = \frac{3}{2}(1.38 \times 10^{-23}\ \mathrm{J\,K^{-1}})(300\ \mathrm{K}) = 6.21 \times 10^{-21}\ \mathrm{J}$
For one mole, use the molar gas constant form:
$mol−1E_{\mathrm{k,mole}} = \frac{3}{2}RT = \frac{3}{2}(8.31\ \mathrm{J\,mol^{-1}\,K^{-1}})(300\ \mathrm{K}) = 3.74 \times 10^3\ \mathrm{J\,mol^{-1}}$

Exam technique

In the exam

Convert first: use kelvin, cubic metre, kilogram, pascal, and mole before substituting.
Choose the gas equation by what is given: use $p V = n R T$ for moles and $p V = N k T$ for molecules.
For kinetic theory, use averages carefully: $c2‾\overline{c^2}$ and $crms=c2‾c_{\mathrm{rms}} = \sqrt{\overline{c^2}}$ .
In graph questions, constant temperature and fixed amount give $\propto \frac{1}{V}$ , so plot $p$ against $1V\frac{1}{V}$ for a straight line.

Self review

Check yourself

Why must temperature be measured in kelvin in the ideal gas equations?
How would you find the number of molecules from the mass of a gas sample?
Starting from $\frac{1}{3}Nm\overline{c^2}$ and $p V = N k T$ , can you show that $Ek,mean=32kTE_{\mathrm{k,mean}} = \frac{3}{2}kT$ ?

What you'll learn

The big idea: macroscopic and microscopic views

Moles, molecules, and molar mass

The ideal gas equation of state

Assumptions of the kinetic theory of gases

How molecular motion causes pressure

Temperature and molecular kinetic energy

Review Kinetic Theory by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Kinetic Theory

What you'll learn

The big idea: macroscopic and microscopic views

Moles, molecules, and molar mass

The ideal gas equation of state

Assumptions of the kinetic theory of gases

How molecular motion causes pressure

Temperature and molecular kinetic energy

Review Kinetic Theory by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Kinetic Theory

What you'll learn

The big idea: macroscopic and microscopic views

Moles, molecules, and molar mass

The ideal gas equation of state

Assumptions of the kinetic theory of gases

How molecular motion causes pressure

Temperature and molecular kinetic energy

Review Kinetic Theory by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Kinetic Theory Revision Guide

Kinetic Theory

What you'll learn

The big idea: macroscopic and microscopic views

Moles, molecules, and molar mass

The ideal gas equation of state

Assumptions of the kinetic theory of gases

How molecular motion causes pressure

Temperature and molecular kinetic energy

Review Kinetic Theory by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Kinetic Theory Revision Guide