Ideal gases

What you'll learn

How amount of substance in moles links to the number of particles.
How the kinetic theory model explains gas pressure using Newton’s laws.
How to use $p V = n R T$ , $p V = N k T$ , and $\frac{1}{3}Nm\overline{c^2}$ .
How temperature connects to molecular kinetic energy and the internal energy of an ideal gas.

Starting point: what is an ideal gas?

A gas is made from a huge number of particles — atoms or molecules — moving around randomly. In this topic, you model those particles using a simplified picture called kinetic theory.

An ideal gas is not a perfectly real gas. It is a model that works very well when a gas is at low density, not too close to liquefying, and the particles are far apart compared with their own size.

Definition

Ideal gas

An ideal gas is a gas that obeys the equation of state $p V = n R T$ and fits the kinetic theory assumptions closely.

Amount of substance: moles and particles

In A-Level Physics, you often need to move between a macroscopic amount of gas, measured in moles, and a microscopic number of particles.

Definition

Mole and Avogadro constant

One mole of any substance contains $mol−1N_A = 6.02 \times 10^{23}\ \text{mol}^{-1}$ particles. The number of particles is

N = nN_A

where $N$ is the number of particles and $n$ is the amount of substance in moles.

For a gas, the “particles” might be single atoms, such as helium atoms, or molecules, such as nitrogen molecules.

Example

Converting moles to particles

A sample contains 0.250 mol of helium. Find the number of helium atoms.

Use the relationship between amount of substance and particle number:
$N = nN_A$
Substitute the values, keeping the mole units visible:
$(0.250\ \text{mol})(6.02 \times 10^{23}\ \text{mol}^{-1})$
The mole units cancel, giving
$\times 10^{23} \approx 1.51 \times 10^{23}\ \text{atoms}$

The kinetic theory model of gases

Kinetic theory explains gas behaviour using moving particles and Newtonian mechanics.

The model assumes:

there is a large number of particles in random, rapid motion
the particles occupy negligible volume compared with the volume of the gas
collisions are perfectly elastic, so kinetic energy is conserved in collisions
the time taken for a collision is negligible compared with the time between collisions
there are negligible forces between particles except during collisions

This model is powerful because it connects things you can measure, such as pressure and temperature, to the motion of particles you cannot see directly.

Kinetic theory model of gas pressure from molecular collisions

Key Idea

Pressure comes from collisions

Gas pressure is caused by particles repeatedly colliding with the container walls and transferring momentum to them.

Explaining pressure using Newton’s laws

When a gas particle hits a wall, its velocity component perpendicular to the wall changes direction. That means its momentum changes.

By Newton’s second law, a change in momentum over time means a force has acted. By Newton’s third law, the wall exerts a force on the particle and the particle exerts an equal and opposite force on the wall.

For many particles, there are huge numbers of collisions every second. The total force from these collisions spread over the wall area gives the pressure:

\frac{F}{A}

Example

Explaining why compression increases pressure

A fixed amount of gas is compressed slowly at constant temperature. Explain why the pressure increases.

At constant temperature, the average kinetic energy of the gas particles stays the same, so their typical speeds do not increase.
Reducing the volume means the particles have less distance to travel before hitting a wall, so collisions with the walls happen more frequently.
More frequent collisions mean a greater rate of momentum transfer to the walls, so the average force on the walls increases.
Since $\frac{F}{A}$ , a larger force per unit area means the pressure increases.

The equation of state: $p V = n R T$

For an ideal gas:

p V = n R T

where:

$p$ is pressure in pascals, Pa
$V$ is volume in cubic metres, m³
$n$ is amount of substance in moles, mol
$T$ is thermodynamic temperature in kelvin, K
$R$ is the molar gas constant, $8.31\ \text{J mol}^{-1}\text{K}^{-1}$

Tip

Kelvin temperatures

Gas equations require temperature in kelvin. Convert using: temperature in K = temperature in °C + 273.15.

Common Mistake

Using degrees Celsius in $p V = n R T$

Never substitute a temperature in degrees Celsius into $p V = n R T$ or $p V = N k T$ . A temperature of 0 °C is not zero thermal energy; it is 273.15 K.

Example

Using the ideal gas equation

A gas sample contains 0.0400 mol and occupies $2.50 \times 10^{-3}\ \text{m}^3$ at 300 K. Find its pressure.

Choose the ideal gas equation because $n$ , $V$ , and $T$ are given:
$p V = n R T$
Rearrange for pressure:
$\frac{nRT}{V}$
Substitute values with units:
$\frac{(0.0400\ \text{mol})(8.31\ \text{J mol}^{-1}\text{K}^{-1})(300\ \text{K})}{2.50 \times 10^{-3}\ \text{m}^3}$
Calculate:
$\times 10^{4}\ \text{Pa}$

Gas-law investigations

For a fixed amount of gas, $n$ is constant, so the ideal gas equation can be rearranged as:

pVT=constant\frac{pV}{T} = \text{constant}

This gives two important experimental relationships.

Boyle’s law: $\text{constant}$

If temperature is constant, then:

\text{constant}

So pressure is inversely proportional to volume:

\propto \frac{1}{V}

In a practical investigation, you might trap air in a sealed syringe or Boyle’s law apparatus. You change the volume, measure the pressure, and keep the temperature constant by making changes slowly and allowing the gas to return to room temperature.

A graph of $p$ against $V$ is a curve, but a graph of $p$ against $1V\frac{1}{V}$ should be a straight line through the origin.

Pressure law: $pT=constant\frac{p}{T} = \text{constant}$

If volume is constant, then:

pT=constant\frac{p}{T} = \text{constant}

So pressure is directly proportional to kelvin temperature:

\propto T

In a practical setup, a fixed volume of gas can be heated in a water bath while a pressure sensor records the pressure. If you plot pressure against temperature in degrees Celsius, the straight line can be extrapolated back to where pressure would be zero. This gives an estimate of absolute zero.

Boyle's law graphs and pressure-temperature extrapolation to absolute zero

Definition

Absolute zero

Absolute zero is 0 K, equal to approximately -273.15 °C. It is the temperature at which an ideal gas would extrapolate to zero pressure at constant volume.

Example

Estimating absolute zero

At constant volume, a gas has pressure 101 kPa at 0 °C and 138 kPa at 100 °C. Estimate absolute zero from these data.

Find the gradient of the pressure-temperature graph:
$∘C−1\text{gradient} = \frac{138\ \text{kPa} - 101\ \text{kPa}}{100\ ^\circ\text{C} - 0\ ^\circ\text{C}} = 0.370\ \text{kPa}\ ^\circ\text{C}^{-1}$
At 0 °C, the pressure is 101 kPa, so the straight-line model is:
$(0.370\ \text{kPa}\ ^\circ\text{C}^{-1})\theta + 101\ \text{kPa}$
Set $p = 0$ and solve for the Celsius temperature $θ\theta$ :
$∘C\theta = -\frac{101\ \text{kPa}}{0.370\ \text{kPa}\ ^\circ\text{C}^{-1}} = -273\ ^\circ\text{C}$

Common Mistake

Extrapolation, not direct measurement

A real gas may liquefy before reaching very low temperatures, so the absolute-zero experiment estimates by extrapolating the straight-line trend.

Kinetic theory equation for pressure

The OCR equation linking pressure to molecular motion is:

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

where:

$N$ is the number of particles
$m$ is the mass of one particle in kilograms, kg
$c2‾\overline{c^2}$ is the mean square speed, in m² s⁻²

You do not need to derive this equation, but you do need to know how to use it and what it means physically.

Definition

Mean square speed and r.m.s. speed

The mean square speed is the average value of the squared speeds of the particles. The root mean square speed is

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

Common Mistake

Squaring the mean speed

$c2‾\overline{c^2}$ is not the same as the mean speed squared. You square each particle’s speed first, then take the average.

Example

Finding root mean square speed

A box contains $5.00 \times 10^{22}$ helium atoms. Each atom has mass $6.64 \times 10^{-27}\ \text{kg}$ . The gas pressure is $1.01 \times 10^{5}\ \text{Pa}$ and the volume is $2.00 \times 10^{-3}\ \text{m}^3$ . Find $crmsc_\text{rms}$ .

Rearrange the kinetic theory equation:
$\frac{1}{3}Nm\overline{c^2} \Rightarrow \overline{c^2} = \frac{3pV}{Nm}$
Substitute the values:
$m2s−2\overline{c^2} = \frac{3(1.01 \times 10^{5}\ \text{Pa})(2.00 \times 10^{-3}\ \text{m}^3)} {(5.00 \times 10^{22})(6.64 \times 10^{-27}\ \text{kg})} = 1.83 \times 10^{6}\ \text{m}^2\text{s}^{-2}$
Take the square root:
$s−1c_\text{rms} = \sqrt{1.83 \times 10^{6}\ \text{m}^2\text{s}^{-2}} = 1.35 \times 10^{3}\ \text{m s}^{-1}$

Maxwell-Boltzmann distribution

In a gas, particles do not all travel at the same speed. Random collisions produce a spread of speeds called a Maxwell-Boltzmann distribution.

The curve starts near zero at zero speed, rises to a peak, then has a long tail at high speeds. Increasing temperature makes the distribution broader and shifts it towards higher speeds. The area under the curve represents the total number of particles, so for the same amount of gas the area stays the same.

Maxwell-Boltzmann speed distribution for two temperatures

Key Idea

Temperature changes the distribution

At a higher temperature, more particles have high speeds, the peak is lower and further to the right, and $crmsc_\text{rms}$ increases.

Boltzmann constant and the particle form of the gas equation

The Boltzmann constant connects temperature to energy per particle.

Definition

Boltzmann constant

The Boltzmann constant is

\frac{R}{N_A}

with value $\times 10^{-23}\ \text{J K}^{-1}$ .

Since $\frac{N}{N_A}$ and $\frac{R}{N_A}$ , the ideal gas equation can be written as:

p V = N k T

This version is useful when you are working with individual particles rather than moles.

Example

Finding the number of gas particles

A gas at pressure $1.20 \times 10^{5}\ \text{Pa}$ fills a volume of $3.00 \times 10^{-3}\ \text{m}^3$ at 290 K. Find the number of particles.

Use the particle form because the question asks for $N$ :
$p V = N k T$
Rearrange and substitute:
$\frac{pV}{kT} = \frac{(1.20 \times 10^{5}\ \text{Pa})(3.00 \times 10^{-3}\ \text{m}^3)} {(1.38 \times 10^{-23}\ \text{J K}^{-1})(290\ \text{K})}$
Calculate, noting that $m3\text{Pa m}^3$ is equivalent to joules:
$\times 10^{22}\ \text{particles}$

Mean kinetic energy of a gas particle

You must know how to derive the key energy result by combining:

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

and

p V = N k T

Example

Deriving mean kinetic energy

Set the two expressions for $p V$ equal to each other:
$13Nmc2‾=NkT\frac{1}{3}Nm\overline{c^2} = NkT$
Cancel $N$ from both sides:
$13mc2‾=kT\frac{1}{3}m\overline{c^2} = kT$
Multiply both sides by $32\frac{3}{2}$ :
$12mc2‾=32kT\frac{1}{2}m\overline{c^2} = \frac{3}{2}kT$
Recognise the left-hand side as the mean translational kinetic energy of one particle:
$Ek=32kTE_k = \frac{3}{2}kT$

Key Idea

Temperature measures average kinetic energy

For an ideal gas, the mean translational kinetic energy of a particle depends only on the absolute temperature, not on the type of gas.

Internal energy of an ideal gas

The internal energy of a system is the total energy stored microscopically in its particles. In general, this includes random kinetic energy and potential energy.

For the ideal gas model, intermolecular forces are negligible except during collisions, so there is no stored intermolecular potential energy. The internal energy is therefore the total random kinetic energy of the particles.

For a monatomic ideal gas:

\frac{3}{2}NkT = \frac{3}{2}nRT

Example

Calculating internal energy

A sample of helium contains 0.500 mol at 300 K. Estimate its internal energy.

Helium is monatomic, so use the ideal-gas internal energy equation:
$\frac{3}{2}nRT$
Substitute the values:
$\frac{3}{2}(0.500\ \text{mol})(8.31\ \text{J mol}^{-1}\text{K}^{-1})(300\ \text{K})$
Calculate:
$\times 10^{3}\ \text{J}$

Exam technique

In the exam

Check whether the question gives moles or particles: use $p V = n R T$ for moles and $p V = N k T$ for particles.
Always convert temperature to kelvin before substituting into gas equations.
For explanation questions, link particle collisions to momentum change, rate of momentum transfer, force, then pressure.
For experiments, state what is kept constant: constant temperature for Boyle’s law, constant volume for $\propto T$ .
Quote graph conclusions carefully: $p$ against $1V\frac{1}{V}$ is linear for Boyle’s law; $p$ against Celsius temperature can be extrapolated to estimate absolute zero.

Self review

Check yourself

Why does compressing a gas at constant temperature increase its pressure?
What is the difference between mean square speed and root mean square speed?
How do you derive $12mc2‾=32kT\frac{1}{2}m\overline{c^2} = \frac{3}{2}kT$ from the two pressure equations?

What you'll learn

How amount of substance in moles links to the number of particles.
How the kinetic theory model explains gas pressure using Newton’s laws.
How to use $p V = n R T$ , $p V = N k T$ , and $\frac{1}{3}Nm\overline{c^2}$ .
How temperature connects to molecular kinetic energy and the internal energy of an ideal gas.

Starting point: what is an ideal gas?

A gas is made from a huge number of particles — atoms or molecules — moving around randomly. In this topic, you model those particles using a simplified picture called kinetic theory.

Definition

Ideal gas

An ideal gas is a gas that obeys the equation of state $p V = n R T$ and fits the kinetic theory assumptions closely.

Amount of substance: moles and particles

In A-Level Physics, you often need to move between a macroscopic amount of gas, measured in moles, and a microscopic number of particles.

Definition

Mole and Avogadro constant

One mole of any substance contains $mol−1N_A = 6.02 \times 10^{23}\ \text{mol}^{-1}$ particles. The number of particles is

N = nN_A

where $N$ is the number of particles and $n$ is the amount of substance in moles.

For a gas, the “particles” might be single atoms, such as helium atoms, or molecules, such as nitrogen molecules.

Example

Converting moles to particles

A sample contains 0.250 mol of helium. Find the number of helium atoms.

Use the relationship between amount of substance and particle number:
$N = nN_A$
Substitute the values, keeping the mole units visible:
$(0.250\ \text{mol})(6.02 \times 10^{23}\ \text{mol}^{-1})$
The mole units cancel, giving
$\times 10^{23} \approx 1.51 \times 10^{23}\ \text{atoms}$

The kinetic theory model of gases

Kinetic theory explains gas behaviour using moving particles and Newtonian mechanics.

The model assumes:

there is a large number of particles in random, rapid motion
the particles occupy negligible volume compared with the volume of the gas
collisions are perfectly elastic, so kinetic energy is conserved in collisions
the time taken for a collision is negligible compared with the time between collisions
there are negligible forces between particles except during collisions

This model is powerful because it connects things you can measure, such as pressure and temperature, to the motion of particles you cannot see directly.

Kinetic theory model of gas pressure from molecular collisions

Key Idea

Pressure comes from collisions

Gas pressure is caused by particles repeatedly colliding with the container walls and transferring momentum to them.

Explaining pressure using Newton’s laws

When a gas particle hits a wall, its velocity component perpendicular to the wall changes direction. That means its momentum changes.

For many particles, there are huge numbers of collisions every second. The total force from these collisions spread over the wall area gives the pressure:

\frac{F}{A}

Example

Explaining why compression increases pressure

A fixed amount of gas is compressed slowly at constant temperature. Explain why the pressure increases.

At constant temperature, the average kinetic energy of the gas particles stays the same, so their typical speeds do not increase.
Reducing the volume means the particles have less distance to travel before hitting a wall, so collisions with the walls happen more frequently.
More frequent collisions mean a greater rate of momentum transfer to the walls, so the average force on the walls increases.
Since $\frac{F}{A}$ , a larger force per unit area means the pressure increases.

The equation of state: $p V = n R T$

For an ideal gas:

p V = n R T

where:

$p$ is pressure in pascals, Pa
$V$ is volume in cubic metres, m³
$n$ is amount of substance in moles, mol
$T$ is thermodynamic temperature in kelvin, K
$R$ is the molar gas constant, $8.31\ \text{J mol}^{-1}\text{K}^{-1}$

Tip

Kelvin temperatures

Gas equations require temperature in kelvin. Convert using: temperature in K = temperature in °C + 273.15.

Common Mistake

Using degrees Celsius in $p V = n R T$

Never substitute a temperature in degrees Celsius into $p V = n R T$ or $p V = N k T$ . A temperature of 0 °C is not zero thermal energy; it is 273.15 K.

Example

Using the ideal gas equation

A gas sample contains 0.0400 mol and occupies $2.50 \times 10^{-3}\ \text{m}^3$ at 300 K. Find its pressure.

Choose the ideal gas equation because $n$ , $V$ , and $T$ are given:
$p V = n R T$
Rearrange for pressure:
$\frac{nRT}{V}$
Substitute values with units:
$\frac{(0.0400\ \text{mol})(8.31\ \text{J mol}^{-1}\text{K}^{-1})(300\ \text{K})}{2.50 \times 10^{-3}\ \text{m}^3}$
Calculate:
$\times 10^{4}\ \text{Pa}$

Gas-law investigations

For a fixed amount of gas, $n$ is constant, so the ideal gas equation can be rearranged as:

pVT=constant\frac{pV}{T} = \text{constant}

This gives two important experimental relationships.

Boyle’s law: $\text{constant}$

If temperature is constant, then:

\text{constant}

So pressure is inversely proportional to volume:

\propto \frac{1}{V}

A graph of $p$ against $V$ is a curve, but a graph of $p$ against $1V\frac{1}{V}$ should be a straight line through the origin.

Pressure law: $pT=constant\frac{p}{T} = \text{constant}$

If volume is constant, then:

pT=constant\frac{p}{T} = \text{constant}

So pressure is directly proportional to kelvin temperature:

\propto T

Boyle's law graphs and pressure-temperature extrapolation to absolute zero

Definition

Absolute zero

Absolute zero is 0 K, equal to approximately -273.15 °C. It is the temperature at which an ideal gas would extrapolate to zero pressure at constant volume.

Example

Estimating absolute zero

At constant volume, a gas has pressure 101 kPa at 0 °C and 138 kPa at 100 °C. Estimate absolute zero from these data.

Find the gradient of the pressure-temperature graph:
$∘C−1\text{gradient} = \frac{138\ \text{kPa} - 101\ \text{kPa}}{100\ ^\circ\text{C} - 0\ ^\circ\text{C}} = 0.370\ \text{kPa}\ ^\circ\text{C}^{-1}$
At 0 °C, the pressure is 101 kPa, so the straight-line model is:
$(0.370\ \text{kPa}\ ^\circ\text{C}^{-1})\theta + 101\ \text{kPa}$
Set $p = 0$ and solve for the Celsius temperature $θ\theta$ :
$∘C\theta = -\frac{101\ \text{kPa}}{0.370\ \text{kPa}\ ^\circ\text{C}^{-1}} = -273\ ^\circ\text{C}$

Common Mistake

Extrapolation, not direct measurement

A real gas may liquefy before reaching very low temperatures, so the absolute-zero experiment estimates by extrapolating the straight-line trend.

Kinetic theory equation for pressure

The OCR equation linking pressure to molecular motion is:

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

where:

$N$ is the number of particles
$m$ is the mass of one particle in kilograms, kg
$c2‾\overline{c^2}$ is the mean square speed, in m² s⁻²

You do not need to derive this equation, but you do need to know how to use it and what it means physically.

Definition

Mean square speed and r.m.s. speed

The mean square speed is the average value of the squared speeds of the particles. The root mean square speed is

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

Common Mistake

Squaring the mean speed

$c2‾\overline{c^2}$ is not the same as the mean speed squared. You square each particle’s speed first, then take the average.

Example

Finding root mean square speed

Rearrange the kinetic theory equation:
$\frac{1}{3}Nm\overline{c^2} \Rightarrow \overline{c^2} = \frac{3pV}{Nm}$
Substitute the values:
$m2s−2\overline{c^2} = \frac{3(1.01 \times 10^{5}\ \text{Pa})(2.00 \times 10^{-3}\ \text{m}^3)} {(5.00 \times 10^{22})(6.64 \times 10^{-27}\ \text{kg})} = 1.83 \times 10^{6}\ \text{m}^2\text{s}^{-2}$
Take the square root:
$s−1c_\text{rms} = \sqrt{1.83 \times 10^{6}\ \text{m}^2\text{s}^{-2}} = 1.35 \times 10^{3}\ \text{m s}^{-1}$

Maxwell-Boltzmann distribution

In a gas, particles do not all travel at the same speed. Random collisions produce a spread of speeds called a Maxwell-Boltzmann distribution.

Maxwell-Boltzmann speed distribution for two temperatures

Key Idea

Temperature changes the distribution

At a higher temperature, more particles have high speeds, the peak is lower and further to the right, and $crmsc_\text{rms}$ increases.

Boltzmann constant and the particle form of the gas equation

The Boltzmann constant connects temperature to energy per particle.

Definition

Boltzmann constant

The Boltzmann constant is

\frac{R}{N_A}

with value $\times 10^{-23}\ \text{J K}^{-1}$ .

Since $\frac{N}{N_A}$ and $\frac{R}{N_A}$ , the ideal gas equation can be written as:

p V = N k T

This version is useful when you are working with individual particles rather than moles.

Example

Finding the number of gas particles

A gas at pressure $1.20 \times 10^{5}\ \text{Pa}$ fills a volume of $3.00 \times 10^{-3}\ \text{m}^3$ at 290 K. Find the number of particles.

Use the particle form because the question asks for $N$ :
$p V = N k T$
Rearrange and substitute:
$\frac{pV}{kT} = \frac{(1.20 \times 10^{5}\ \text{Pa})(3.00 \times 10^{-3}\ \text{m}^3)} {(1.38 \times 10^{-23}\ \text{J K}^{-1})(290\ \text{K})}$
Calculate, noting that $m3\text{Pa m}^3$ is equivalent to joules:
$\times 10^{22}\ \text{particles}$

Mean kinetic energy of a gas particle

You must know how to derive the key energy result by combining:

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

and

p V = N k T

Example

Deriving mean kinetic energy

Set the two expressions for $p V$ equal to each other:
$13Nmc2‾=NkT\frac{1}{3}Nm\overline{c^2} = NkT$
Cancel $N$ from both sides:
$13mc2‾=kT\frac{1}{3}m\overline{c^2} = kT$
Multiply both sides by $32\frac{3}{2}$ :
$12mc2‾=32kT\frac{1}{2}m\overline{c^2} = \frac{3}{2}kT$
Recognise the left-hand side as the mean translational kinetic energy of one particle:
$Ek=32kTE_k = \frac{3}{2}kT$

Key Idea

Temperature measures average kinetic energy

For an ideal gas, the mean translational kinetic energy of a particle depends only on the absolute temperature, not on the type of gas.

Internal energy of an ideal gas

The internal energy of a system is the total energy stored microscopically in its particles. In general, this includes random kinetic energy and potential energy.

For a monatomic ideal gas:

\frac{3}{2}NkT = \frac{3}{2}nRT

Example

Calculating internal energy

A sample of helium contains 0.500 mol at 300 K. Estimate its internal energy.

Helium is monatomic, so use the ideal-gas internal energy equation:
$\frac{3}{2}nRT$
Substitute the values:
$\frac{3}{2}(0.500\ \text{mol})(8.31\ \text{J mol}^{-1}\text{K}^{-1})(300\ \text{K})$
Calculate:
$\times 10^{3}\ \text{J}$

Exam technique

In the exam

Check whether the question gives moles or particles: use $p V = n R T$ for moles and $p V = N k T$ for particles.
Always convert temperature to kelvin before substituting into gas equations.
For explanation questions, link particle collisions to momentum change, rate of momentum transfer, force, then pressure.
For experiments, state what is kept constant: constant temperature for Boyle’s law, constant volume for $\propto T$ .
Quote graph conclusions carefully: $p$ against $1V\frac{1}{V}$ is linear for Boyle’s law; $p$ against Celsius temperature can be extrapolated to estimate absolute zero.

Self review

Check yourself

Why does compressing a gas at constant temperature increase its pressure?
What is the difference between mean square speed and root mean square speed?
How do you derive $12mc2‾=32kT\frac{1}{2}m\overline{c^2} = \frac{3}{2}kT$ from the two pressure equations?

What you'll learn

Starting point: what is an ideal gas?

Amount of substance: moles and particles

The kinetic theory model of gases

Explaining pressure using Newton’s laws

The equation of state: $p V = n R T$

Gas-law investigations

Boyle’s law: $\text{constant}$

Pressure law: $pT=constant\frac{p}{T} = \text{constant}$

Kinetic theory equation for pressure

Maxwell-Boltzmann distribution

Boltzmann constant and the particle form of the gas equation

Mean kinetic energy of a gas particle

Internal energy of an ideal gas

Review Ideal gases by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Ideal gases

What you'll learn

Starting point: what is an ideal gas?

Amount of substance: moles and particles

The kinetic theory model of gases

Explaining pressure using Newton’s laws

The equation of state: $p V = n R T$

Gas-law investigations

Boyle’s law: $\text{constant}$

Pressure law: $pT=constant\frac{p}{T} = \text{constant}$

Kinetic theory equation for pressure

Maxwell-Boltzmann distribution

Boltzmann constant and the particle form of the gas equation

Mean kinetic energy of a gas particle

Internal energy of an ideal gas

Review Ideal gases by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Ideal gases

What you'll learn

Starting point: what is an ideal gas?

Amount of substance: moles and particles

The kinetic theory model of gases

Explaining pressure using Newton’s laws

The equation of state: pV=nRTpV = nRTpV=nRT

Gas-law investigations

Boyle’s law: pV=constantpV = \text{constant}pV=constant

Pressure law: pT=constant\frac{p}{T} = \text{constant}Tp​=constant

Kinetic theory equation for pressure

Maxwell-Boltzmann distribution

Boltzmann constant and the particle form of the gas equation

Mean kinetic energy of a gas particle

Internal energy of an ideal gas

Review Ideal gases by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Ideal gases Revision Guide

Ideal gases

What you'll learn

Starting point: what is an ideal gas?

Amount of substance: moles and particles

The kinetic theory model of gases

Explaining pressure using Newton’s laws

The equation of state: pV=nRTpV = nRTpV=nRT

Gas-law investigations

Boyle’s law: pV=constantpV = \text{constant}pV=constant

Pressure law: pT=constant\frac{p}{T} = \text{constant}Tp​=constant

Kinetic theory equation for pressure

Maxwell-Boltzmann distribution

Boltzmann constant and the particle form of the gas equation

Mean kinetic energy of a gas particle

Internal energy of an ideal gas

Review Ideal gases by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Ideal gases Revision Guide

The equation of state: $p V = n R T$

Boyle’s law: $\text{constant}$

Pressure law: $pT=constant\frac{p}{T} = \text{constant}$

The equation of state: $p V = n R T$

Boyle’s law: $\text{constant}$

Pressure law: $pT=constant\frac{p}{T} = \text{constant}$