Cambridge AS & A Level9702

Ideal gases

Physics 9702 Chapter Notes

What this chapter covers

Ideal gases - The moleIdeal gases - Equation of stateIdeal gases - Kinetic theory of gases
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In this chapter

  1. 1Counting Atoms: The Mole
  2. 2The Ideal Gas Equation
  3. 3Kinetic Theory Assumptions
  4. 4The Microscopic Origin of Pressure
  5. 5Temperature and Molecular Kinetic Energy
  6. 6Root-Mean-Square Speed

1. Counting Atoms: The Mole

In physics and chemistry, we deal with an enormous number of atoms and molecules. Counting them one by one is impossible. Instead, we use a unit called the 'mole' (mol). Think of it like a 'dozen' for eggs, but for particles. One mole of any substance contains the same number of particles: 6.02 x 10²³. This magic number is called the Avogadro constant (Nₐ). To find the total number of particles (N) in a sample, you multiply the number of moles(n) by the Avogadro constant: N = nNₐ. You can also find the number of moles by dividing the mass of the substance(m) by its molar mass (M), which is the mass of one mole of that substance.

N = n × Nₐ

n = m / M

Key term

Avogadro Constant (Nₐ): The number of constituent particles (usually atoms or molecules) that are contained in one mole of a substance, approximately 6.02 × 10²³ mol⁻¹.

Common pitfall

Confusing the number of moles (n) with the number of molecules (N). Remember N is an extremely large number representing the actual particle count, while n is usually a small number like 0.5, 1, or 10.

Fun fact

If you had one mole of standard basketballs, they would create a new planet the same size as Earth.

Worked example 13 marks

A container holds 88 g of carbon dioxide (CO₂). The molar mass of carbon dioxide is 44 g mol⁻¹. Calculate:(a) the number of moles of CO₂ in the container, and(b) the number of CO₂ molecules in the container.

  1. 1

    Step 1: Calculate the number of moles(n) using the formula n = m / M.

  2. 2

    Given mass m = 88 g and molar mass M = 44 g mol⁻¹.

  3. 3

    n = 88 g / 44 g mol⁻¹ = 2.0 mol.

  4. 4

    Step 2: Calculate the number of molecules (N) using the formula N = n × Nₐ.

  5. 5

    n = 2.0 mol and Nₐ = 6.02 × 10²³ mol⁻¹.

  6. 6

    N = 2.0 × (6.02 × 10²³) = 1.204 × 10²⁴ molecules.

  7. 7

    To 3 significant figures, N = 1.20 × 10²⁴ molecules.

Recap

  • A mole is a unit for an amount of substance.
  • One mole of any substance contains 6.02 × 10²³ particles.
  • The Avogadro constant, Nₐ, has a value of 6.02 × 10²³ mol⁻¹.
  • The number of particles N is the number of moles n multiplied by Nₐ.
  • The number of moles n can be found by dividing the mass m by the molar mass M.

Quick check

  1. How many atoms are there in 3.0 moles of helium gas?1 mark
  2. State the relationship between the number of particles N, the number of moles n, and the Avogadro constant Nₐ.1 mark

2. The Ideal Gas Equation

An 'ideal gas' is a theoretical model for a gas that simplifies its behaviour. Real gases, like air or helium, behave very much like ideal gases under normal conditions (i.e., not at very high pressures or very low temperatures). The state of an ideal gas can be described by a single equation relating its pressure (p), volume (V), temperature (T), and the amount of gas. This is called the equation of state. There are two main forms:

  1. pV = nRT, where 'n' is the number of moles of the gas and R is the molar gas constant (8.31 J mol⁻¹ K⁻¹).
  2. pV = NkT, where 'N' is the total number of molecules and 'k' is the Boltzmann constant (1.38 × 10⁻²³ J K⁻¹). The Boltzmann constant is just the molar gas constant per particle (k = R/Nₐ). Crucially, the temperature T in these equations must always be in Kelvin (K).

pV = nRT

pV = NkT

T(K) = T(°C) + 273.15

Key term

Ideal Gas: A hypothetical gas whose molecules occupy negligible volume and have no intermolecular forces, meaning it obeys the gas laws perfectly under all conditions.

Examiner insight

Examiners award full marks for calculations only when the temperature is correctly converted to Kelvin before being used in any gas law formula.

Fun fact

The concept of an 'ideal gas' was developed by combining the experimental laws of Boyle, Charles, and Gay-Lussac, which describe how real gases behave under near-normal conditions.

Worked example 13 marks

A cylinder of volume 0.025 m³ contains 0.55 moles of an ideal gas at a temperature of 25 °C. Calculate the pressure of the gas in the cylinder.

  1. 1

    Step 1: Convert the temperature from Celsius to Kelvin.

  2. 2

    T(K) = T(°C) + 273.15 = 25 + 273.15 = 298.15 K.

  3. 3

    Step 2: State the ideal gas equation and rearrange it for pressure (p).

  4. 4

    pV = nRT => p = nRT / V.

  5. 5

    Step 3: Substitute the known values into the equation.

  6. 6

    n = 0.55 mol, R = 8.31 J mol⁻¹ K⁻¹, T = 298.15 K, V = 0.025 m³.

  7. 7

    p = (0.55 × 8.31 × 298.15) / 0.025

  8. 8

    p = 1362.9 / 0.025 = 54516 Pa.

  9. 9

    Step 4: Give the answer to an appropriate number of significant figures.

  10. 10

    p ≈ 5.5 × 10⁴ Pa (or 55 kPa).

Worked example 23 marks

A sealed container of gas has a pressure of 1.5 × 10⁵ Pa at a temperature of 20 °C. The container is heated to 100 °C. Assuming the volume is constant, what is the new pressure?

  1. 1

    Step 1: For a fixed mass and volume of gas, p/T is constant. So, p₁/T₁ = p₂/T₂.

  2. 2

    Step 2: Convert both temperatures to Kelvin.

  3. 3

    T₁ = 20 + 273.15 = 293.15 K.

  4. 4

    T₂ = 100 + 273.15 = 373.15 K.

  5. 5

    Step 3: Rearrange the formula to find the new pressure, p₂.

  6. 6

    p₂ = p₁ × (T₂ / T₁).

  7. 7

    Step 4: Substitute the values and calculate p₂.

  8. 8

    p₂ = (1.5 × 10⁵ Pa) × (373.15 K / 293.15 K)

  9. 9

    p₂ = 1.91 × 10⁵ Pa.

Recap

  • An ideal gas is a theoretical model that real gases approximate.
  • The equation of state for an ideal gas is pV = nRT or pV = NkT.
  • Temperature (T) must always be in Kelvin (K) for gas law calculations.
  • Pressure (p) must be in Pascals (Pa) and volume (V) in cubic metres (m³).
  • R is the molar gas constant (8.31 J mol⁻¹ K⁻¹) and k is the Boltzmann constant (1.38 × 10⁻²³ J K⁻¹).

Quick check

  1. State the ideal gas equation in terms of the number of moles, n.1 mark
  2. What is the value of 50 °C in Kelvin?1 mark

3. Kinetic Theory Assumptions

The kinetic theory of gases explains the macroscopic properties of a gas (like pressure and temperature) by considering the motion of its microscopic particles (atoms or molecules). To create a simple but powerful model, we make several key assumptions about the behaviour of an ideal gas:

Key term

Elastic Collision: A collision in which the total kinetic energy of the colliding objects is conserved (i.e., no energy is lost as heat or sound).

Examiner insight

Marks are often awarded for stating the assumptions precisely. For example, just saying 'molecules are small' is insufficient; you must state their volume is 'negligible compared to the container's volume'.

Fun fact

Even though we assume there are no forces between ideal gas molecules, it's the forces during collisions with the walls that create pressure!

Worked example 14 marks

The kinetic theory of gases is based on a set of simplifying assumptions. State four of these assumptions.

  1. 1
    1. The gas contains a very large number of identical molecules/atoms.
  2. 2
    1. The molecules are in continuous, random motion.
  3. 3
    1. The volume of the molecules themselves is negligible compared to the volume of the container they occupy.
  4. 4
    1. There are no forces of attraction or repulsion between molecules (except during collisions).
  5. 5
    1. All collisions, both between molecules and between molecules and the container walls, are perfectly elastic.
  6. 6
    1. The time taken for a collision is negligible compared to the time spent between collisions.

Recap

  • Gas molecules are in constant, random motion.
  • Collisions with container walls and other molecules are perfectly elastic.
  • The volume of the molecules is negligible compared to the container's volume.
  • There are no intermolecular forces between molecules.
  • The duration of a collision is negligible.

Quick check

  1. What is meant by an 'elastic collision' in the context of the kinetic theory?1 mark
  2. State the assumption made about the volume of gas molecules.1 mark

4. The Microscopic Origin of Pressure

Why does a gas exert pressure? According to the kinetic theory, it's due to the constant bombardment of the container walls by gas molecules. Each time a molecule hits a wall and bounces off, it changes its momentum. This change in momentum means the wall has exerted a force on the molecule, and by Newton's third law, the molecule exerts an equal and opposite force on the wall. The pressure is the total force from all these collisions averaged over the area of the wall. This microscopic view can be mathematically linked to macroscopic properties by the kinetic theory equation: pV = (1/3)Nm<c²>. Here, N is the number of molecules, m is the mass of one molecule, and <c²> is the mean square speed of the molecules.

pV = (1/3)Nm<c²>

Key term

Mean Square Speed (<c²>): The average of the squared speeds of all the molecules in a gas, calculated by summing the square of each molecule's speed and dividing by the total number of molecules.

Examiner insight

Students who can clearly explain the link between molecular momentum change and the force exerted on the container wall demonstrate a superior understanding of the topic.

Common pitfall

Confusing the mean square speed, <c²>, with the square of the mean speed, (<c>)². For a typical gas, these two values are not the same.

Worked example 13 marks

A container of volume 2.0 × 10⁻² m³ holds 5.0 × 10²⁴ molecules of a gas. The pressure of the gas is 1.2 × 10⁵ Pa. The mass of one molecule is 4.7 × 10⁻²⁶ kg. Calculate the mean square speed of the gas molecules.

  1. 1

    Step 1: State the kinetic theory pressure equation.

  2. 2

    pV = (1/3)Nm<c²>.

  3. 3

    Step 2: Rearrange the equation to make the mean square speed, <c²>, the subject.

  4. 4

    <c²> = 3pV / (Nm).

  5. 5

    Step 3: Substitute the given values into the rearranged equation.

  6. 6

    p = 1.2 × 10⁵ Pa, V = 2.0 × 10⁻² m³, N = 5.0 × 10²⁴, m = 4.7 × 10⁻²⁶ kg.

  7. 7

    <c²> = (3 × (1.2 × 10⁵) × (2.0 × 10⁻²)) / ((5.0 × 10²⁴) × (4.7 × 10⁻²⁶)).

  8. 8

    <c²> = 7200 / 235.

  9. 9

    <c²> = 30.638... m²s⁻².

  10. 10

    Step 4: Give the final answer to an appropriate number of significant figures.

  11. 11

    <c²> ≈ 31 m²s⁻².

Recap

  • Gas pressure is caused by molecular collisions with the container walls.
  • Each collision with a wall causes a change in the molecule's momentum.
  • The force on the wall is the rate of change of momentum of the colliding molecules.
  • The pressure equation is pV = (1/3)Nm<c²>.
  • The term <c²> represents the mean of the squared speeds of the molecules.

Quick check

  1. In the equation pV = (1/3)Nm<c²>, what do the symbols N, m, and <c²> represent?3 marks

5. Temperature and Molecular Kinetic Energy

One of the most profound results of the kinetic theory is the direct link it reveals between temperature and the energy of molecules. By comparing the ideal gas equation (pV = NkT) with the kinetic theory pressure equation (pV = (1/3)Nm<c²>), we can equate the right-hand sides: NkT = (1/3)Nm<c²>. After cancelling N on both sides, we get kT = (1/3)m<c²>. We can rearrange this to find an expression for the average translational kinetic energy of a single molecule, Eₖ = (1/2)m<c²>. Multiplying our rearranged equation by 3/2 gives (3/2)kT = (1/2)m<c²>. This means the average translational kinetic energy of a gas molecule is directly proportional to the absolute temperature (in Kelvin). Temperature is a measure of the average kinetic energy of the particles.

(1/2)m<c²> = (3/2)kT

Key term

Boltzmann Constant (k): A fundamental physical constant (k ≈ 1.38 × 10⁻²³ J K⁻¹) that relates the average kinetic energy of particles in a gas to the gas's thermodynamic temperature.

Examiner insight

Examiners look for the explicit statement that the average translational kinetic energy of a molecule is directly proportional to the absolute temperature.

Fun fact

Absolute zero (0 K) is the theoretical temperature at which particles have the minimum possible kinetic energy allowed by quantum mechanics. It's not quite zero motion, but it's the coldest anything can get.

Worked example 14 marks

The air in a room is heated from 15 °C to 25 °C.(a) Calculate the average translational kinetic energy of an air molecule at 15 °C.(b) By what factor does the average kinetic energy of the molecules increase?

  1. 1

    Part (a): Calculate the kinetic energy at 15 °C.

  2. 2

    Step 1: Convert temperature to Kelvin: T = 15 + 273.15 = 288.15 K.

  3. 3

    Step 2: Use the formula Eₖ = (3/2)kT.

  4. 4

    Eₖ = (3/2) × (1.38 × 10⁻²³) × 288.15

  5. 5

    Eₖ = 5.96 × 10⁻²¹ J.

  6. 6

    Part (b): Find the factor of increase.

  7. 7

    Step 1: Kinetic energy is directly proportional to absolute temperature (Eₖ ∝ T).

  8. 8

    Step 2: The ratio of the new kinetic energy to the old is the same as the ratio of the new absolute temperature to the old.

  9. 9

    T_new = 25 + 273.15 = 298.15 K. T_old = 288.15 K.

  10. 10

    Factor = T_new / T_old = 298.15 / 288.15 = 1.035.

  11. 11

    The average kinetic energy increases by a factor of 1.035.

Recap

  • Absolute temperature is a measure of the average translational kinetic energy of gas molecules.
  • The average kinetic energy of a molecule is directly proportional to the absolute temperature (T in Kelvin).
  • The formula linking them is Eₖ = (1/2)m<c²> = (3/2)kT.
  • If you double the absolute temperature of a gas, you double the average kinetic energy of its molecules.

Quick check

  1. What happens to the average kinetic energy of gas molecules if the temperature of the gas is increased from 100 K to 400 K?1 mark

6. Root-Mean-Square Speed

The molecules in a gas don't all travel at the same speed; there is a distribution of speeds. To find a typical or representative speed, we use the root-mean-square (r.m.s.) speed, cᵣₘₛ. It's a type of statistical average. To calculate it, you square all the molecular speeds, find the mean (average) of these squares (<c²>), and then take the square root of the result. So, cᵣₘₛ = √<c²>. Using the energy relationship from the previous section, (1/2)m<c²> = (3/2)kT, we can derive a direct formula for cᵣₘₛ. Rearranging for <c²> gives <c²> = 3kT/m. Therefore, the r.m.s. speed is cᵣₘₛ = √(3kT/m). This shows that hotter, lighter molecules move faster.

cᵣₘₛ = √<c²>

cᵣₘₛ = √(3kT/m)

cᵣₘₛ = √(3RT/M)

Key term

Root-Mean-Square (r.m.s.) Speed: The square root of the mean of the squares of the speeds of the molecules in a gas, which gives a statistically meaningful measure of the typical speed of a molecule.

Common pitfall

Using the molar mass (M, in kg/mol or g/mol) in the formula cᵣₘₛ = √(3kT/m). The 'm' in this equation is the mass of a single molecule in kg. You must divide the molar mass by the Avogadro constant to find 'm'.

Fun fact

The r.m.s. speed of hydrogen molecules at room temperature is about 1900 m/s, which is so fast that Earth's gravity is too weak to hold onto it over geological time. This is why there is very little hydrogen in our atmosphere.

Worked example 13 marks

Helium gas is approximately an ideal gas. A molecule of helium has a mass of 6.64 × 10⁻²⁷ kg. Calculate the r.m.s. speed of a helium atom at a temperature of 27 °C.

  1. 1

    Step 1: Convert the temperature to Kelvin.

  2. 2

    T = 27 + 273.15 = 300.15 K.

  3. 3

    Step 2: State the formula for r.m.s. speed.

  4. 4

    cᵣₘₛ = √(3kT/m).

  5. 5

    Step 3: Substitute the known values into the formula.

  6. 6

    k = 1.38 × 10⁻²³ J K⁻¹, T = 300.15 K, m = 6.64 × 10⁻²⁷ kg.

  7. 7

    cᵣₘₛ = √((3 × (1.38 × 10⁻²³) × 300.15) / (6.64 × 10⁻²⁷)).

  8. 8

    cᵣₘₛ = √(1.2426 × 10⁻²⁰ / 6.64 × 10⁻²⁷).

  9. 9

    cᵣₘₛ = √(1.871 × 10⁶).

  10. 10

    cᵣₘₛ = 1368 m s⁻¹.

  11. 11

    Step 4: Give the answer to an appropriate number of significant figures.

  12. 12

    cᵣₘₛ ≈ 1370 m s⁻¹.

Recap

  • The r.m.s. speed, cᵣₘₛ, is a representative speed for molecules in a gas.
  • It is calculated as the square root of the mean square speed: cᵣₘₛ = √<c²>.
  • The formula for r.m.s. speed is cᵣₘₛ = √(3kT/m).
  • R.m.s. speed increases with temperature and decreases with molecular mass.

Quick check

  1. Two gases, hydrogen (light) and oxygen (heavy), are at the same temperature. Which gas has the higher r.m.s. speed?1 mark

End-of-chapter exercise

Test yourself on the whole chapter. Work through these before moving on.

  1. A cylinder contains 500 g of helium-4 gas (molar mass 4.0 g mol⁻¹). Calculate the number of helium atoms in the cylinder.2 marks
  2. State three assumptions of the kinetic theory of ideal gases.3 marks
  3. A weather balloon is filled with 45 moles of helium. At an altitude where the temperature is -23 °C and the pressure is 6.0 × 10⁴ Pa, what is the volume of the balloon?3 marks
  4. A car tyre has a pressure of 2.5 × 10⁵ Pa when the air inside is at 17 °C. After a long journey, the temperature of the air in the tyre rises to 57 °C. Assuming the volume of the tyre does not change, calculate the new pressure.3 marks
  5. Explain, in terms of molecular motion, why the pressure of a gas in a container of fixed volume increases when its temperature is increased.3 marks
  6. A cubical box of side length 0.20 m contains 3.0 × 10²² molecules of an ideal gas. The pressure exerted by the gas is 1.1 × 10⁵ Pa. Calculate the mean square speed, <c²>, of the gas molecules if the mass of one molecule is 5.3 × 10⁻²⁶ kg.4 marks
  7. The Sun's corona is a plasma with a temperature of about 1 × 10⁶ K. Calculate the r.m.s. speed of a proton (mass 1.67 × 10⁻²⁷ kg) in the corona, assuming it behaves as an ideal gas.3 marks
  8. A cylinder of volume 5.0 × 10⁻² m³ contains oxygen gas at a pressure of 2.0 × 10⁵ Pa and a temperature of 300 K. Assuming oxygen behaves as an ideal gas, calculate: (a) the amount, in moles, of oxygen in the cylinder, (b) the number of oxygen molecules in the cylinder, and (c) the total kinetic energy of all the gas molecules.5 marks
  9. By comparing the ideal gas equation pV = NkT with the kinetic theory equation pV = (1/3)Nm<c²>, show that the average translational kinetic energy of a molecule is given by (3/2)kT.3 marks
  10. The density of an ideal gas is 1.25 kg m⁻³ at a pressure of 1.01 × 10⁵ Pa and a temperature of 293 K. Calculate the r.m.s. speed of the gas molecules under these conditions.3 marks

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