Ideal gases

All three subtopics here are printed under "A Level subject content" in the 9702 syllabus, and every objective carries the tier "A Level". None of it is AS. It is examined on Paper 4. Paper 1 is the AS multiple-choice paper, and the whole 9702 bank on this site comes from Paper 1, so no practice is tagged to this topic.

Syllabus points

15.1 The mole

• Understand that amount of substance is an SI base quantity with the base unit mol.
• Use molar quantities where one mole of any substance is the amount containing a number of particles of that substance equal to the Avogadro constant N_A.

15.2 Equation of state

• Understand that a gas obeying pV ∝ T, where T is the thermodynamic temperature, is known as an ideal gas.
• Recall and use the equation of state for an ideal gas expressed as pV = nRT, where n = amount of substance (number of moles), and as pV = NkT, where N = number of molecules.
• Recall that the Boltzmann constant k is given by k = R/N_A.

15.3 Kinetic theory of gases

• State the basic assumptions of the kinetic theory of gases.
• Explain how molecular movement causes the pressure exerted by a gas and derive and use the relationship pV = ⅓Nm⟨c²⟩, where ⟨c²⟩ is the mean-square speed (a simple model considering one-dimensional collisions and then extending to three dimensions using ⟨c²⟩ = 3⟨c_x²⟩ is sufficient).
• Understand that the root-mean-square speed c_r.m.s. is given by the square root of ⟨c²⟩.
• Compare pV = ⅓Nm⟨c²⟩ with pV = NkT to deduce that the average translational kinetic energy of a molecule is (3/2)kT, and recall and use this expression.

The mole

Amount of substance is one of the seven SI base quantities, and its base unit is the mole. That is worth noticing: questions on base units regularly include the mole, and it is easy to assume, wrongly, that only the kilogram, metre, second, ampere, kelvin and candela qualify.

One mole of any substance contains a number of particles equal to the Avogadro constant, N_A = 6.02 × 10²³ mol⁻¹.

number of molecules N = nN_A

Worked example. How many molecules are there in 3.5 mol of helium?

N = 3.5 × 6.02 × 10²³ = 2.107 × 10²⁴

so about 2.1 × 10²⁴ molecules.

To find the amount of substance from a mass, divide by the molar mass.

Worked example. 8.0 g of helium has a molar mass of 4.0 g mol⁻¹.

n = 8.0 / 4.0 = 2.0

so 2.0 mol.

The equation of state

An ideal gas is one that obeys pV ∝ T at all pressures and temperatures, where T is the thermodynamic temperature in kelvin.

Two forms of the equation of state, and both are used:

pV = nRT

where n is the amount of substance in moles and R is the molar gas constant, 8.31 J K⁻¹ mol⁻¹, and

pV = NkT

where N is the number of molecules and k is the Boltzmann constant, 1.38 × 10⁻²³ J K⁻¹.

The two are the same equation. Since N = nN_A:

k = R / N_A

Check that. 8.31 / (6.02 × 10²³) = 1.380 × 10⁻²³, which is k.

Choose the form that matches the data. If the question gives you moles or a mass, use nRT; if it gives you a number of molecules or asks about one molecule, use NkT.

Worked example. A cylinder of volume 0.020 m³ holds gas at a pressure of 4.0 × 10⁵ Pa and a temperature of 27 °C. Find the amount of gas.

Convert the temperature first, because the equation needs kelvin:

T = 27 + 273 = 300

n = 4.0 × 10⁵ × 0.020 / (8.31 × 300) = 3.209

so n = 3.2 mol.

Worked example, a change of conditions. When only some quantities change, use the fact that pV/T is constant for a fixed mass of gas:

p₁V₁ / T₁ = p₂V₂ / T₂

A sealed container of gas at 300 K and 1.0 × 10⁵ Pa is heated to 450 K at constant volume. The volume cancels:

p₂ = 1.0 × 10⁵ × 450 / 300 = 1.5 × 10⁵

so the pressure rises to 1.5 × 10⁵ Pa. Had the temperatures been given in Celsius, 27 °C to 177 °C, and used without conversion, the answer would have come out as 6.6 × 10⁵ Pa, which is wrong by a factor of more than four. This is the single most common error in the topic.

Kinetic theory: the assumptions

The model treats a gas as a very large number of tiny particles in constant random motion, and it makes these assumptions:

• The gas contains a large number of molecules, so statistical averages are meaningful.
• The molecules are in continuous random motion, with a range of speeds and directions.
• The volume of the molecules is negligible compared with the volume of the container.
• Intermolecular forces are negligible, except during collisions.
• Collisions between molecules, and between molecules and the walls, are perfectly elastic, so no kinetic energy is lost.
• The time of a collision is negligible compared with the time between collisions.
• Newtonian mechanics applies to the molecules.

The third and fourth assumptions are the ones worth watching, because they are the ones a real gas breaks. At high pressure the molecules are close together, so their own volume is no longer negligible; at low temperature they move slowly enough for the attractive forces between them to matter. A real gas therefore behaves most nearly ideally at low pressure and high temperature.

How molecular motion causes pressure

Consider one molecule of mass m in a cubical box of side L, moving with velocity c_x along the x-axis.

Step 1, the change of momentum. The molecule hits the wall and rebounds elastically, so its velocity changes from +c_x to -c_x.

Δp = 2mc_x

Step 2, the time between collisions. It must travel to the opposite wall and back before hitting this wall again, a distance 2L.

t = 2L / c_x

Step 3, the force. By Newton's second law, force is the rate of change of momentum, so divide the change of momentum by the time.

F = 2mc_x / (2L / c_x), which simplifies to F = mc_x² / L

Step 4, the pressure. Pressure is force per unit area, and the wall has area L².

p = mc_x² / L³

Since L³ is the volume V, this is p = mc_x² / V.

Step 5, many molecules. For N molecules with a range of speeds, use the mean square speed ⟨c_x²⟩.

pV = Nm⟨c_x²⟩

Step 6, three dimensions. Motion is random, so on average the three components are equal, and since c² = c_x² + c_y² + c_z²:

⟨c²⟩ = 3⟨c_x²⟩

Step 7, the result. Substituting the last line into the one before it gives

pV = ⅓Nm⟨c²⟩

Read the finished equation as a physical statement. Pressure comes from the rate of change of momentum of molecules bouncing off the walls. Heating a gas raises the mean speed, so each collision transfers more momentum and collisions happen more frequently, which is why pressure rises with temperature at constant volume. Reducing the volume at constant temperature raises the pressure because the molecules hit the walls more often, not because they hit harder.

Root-mean-square speed

⟨c²⟩ is the mean of the squares of the molecular speeds. The root-mean-square speed is its square root:

c_r.m.s. = √⟨c²⟩

Note the order of operations, because reversing it is a genuine trap: square the speeds, take the mean, then take the root. The mean speed and the r.m.s. speed are not the same number.

Worked example. Four molecules have speeds 300, 400, 500 and 600 m s⁻¹.

mean square speed = (300² + 400² + 500² + 600²) / 4 = 215000

Taking the square root, c_r.m.s. = 464 m s⁻¹.

The ordinary mean speed is

(300 + 400 + 500 + 600) / 4 = 450

which is smaller. Squaring gives the faster molecules more weight, so the r.m.s. value always exceeds the mean.

Mean kinetic energy of a molecule

Set the two expressions for pV equal:

⅓Nm⟨c²⟩ = NkT

Multiply both sides by 3/2 and divide by N:

½m⟨c²⟩ = (3/2)kT

The left side is the average translational kinetic energy of one molecule. So:

E_k = (3/2)kT

This is one of the most quotable results in the syllabus, and everything it says is examinable:

• The mean kinetic energy of a molecule depends only on the thermodynamic temperature. Not on the pressure, not on the volume, and not on the type of gas.
• Two different gases at the same temperature have the same mean molecular kinetic energy. Since E_k = ½m⟨c²⟩, the gas with the smaller molecular mass must have the greater r.m.s. speed. Helium molecules move much faster than oxygen molecules at the same temperature, and that is why helium escapes from the atmosphere.
• E_k is proportional to T in kelvin. Doubling the thermodynamic temperature doubles the mean kinetic energy, so it multiplies the r.m.s. speed by √2, not by 2.
• At absolute zero, E_k would be zero, which is another way of stating what absolute zero means.

Worked example. Find the r.m.s. speed of oxygen molecules at 27 °C. The molar mass of oxygen is 0.032 kg mol⁻¹.

Mass of one molecule:

m = 0.032 / (6.02 × 10²³) = 5.316 × 10⁻²⁶

T = 27 + 273 = 300

Using ½m⟨c²⟩ = (3/2)kT:

⟨c²⟩ = 3 × 1.38 × 10⁻²³ × 300 / (5.316 × 10⁻²⁶) = 2.336 × 10⁵

Taking the square root, c_r.m.s. = 483 m s⁻¹.

A number of that order, a few hundred metres per second, is a useful sanity check: molecular speeds in a gas at room temperature are comparable with the speed of sound, which is no coincidence.

Common mistakes

• Using degrees Celsius in pV = nRT or in E_k = (3/2)kT. Every T in this topic is in kelvin.
• Forgetting that the mole is an SI base unit.
• Mixing up R and k. R is per mole and goes with n; k is per molecule and goes with N.
• Taking the mean of the speeds and then squaring, instead of taking the mean of the squares and then rooting.
• Saying that raising the temperature doubles the r.m.s. speed when it doubles the temperature. Speed goes as the square root.
• Saying that at the same temperature a heavier gas has more kinetic energy per molecule. It has the same, and therefore a lower r.m.s. speed.
• Explaining pressure as molecules "pushing" the walls rather than as the rate of change of momentum in collisions.
• Leaving out the word "random" from the assumptions, or forgetting that collisions must be elastic.
• Saying a real gas is closest to ideal at high pressure. It is closest at low pressure and high temperature.
• Dropping the factor of ⅓ in pV = ⅓Nm⟨c²⟩ after the three-dimensional step.