A Level Physics: Ideal Gases — Revision Notes

Condensed for the final weeks. For the full explanation, use the Ideal Gases study guide.

The equation of state

pV = nRT        n = number of MOLES,      R = 8.31 J mol^-1 K^-1
pV = N k T      N = number of MOLECULES,  k = 1.38 x 10^-23 J K^-1

Related by N = n N_A and k = R / N_A, where N_A = 6.02 × 10²³ mol⁻¹ is the Avogadro constant — one mole of any substance contains this same fixed number of particles.

T must be in kelvin, always. More marks are lost to this than to any other error in the topic.

Worked example. 2.0 mol of an ideal gas at 300 K occupies 0.020 m³. Its pressure: p = nRT/V = (2.0 × 8.31 × 300) ÷ 0.020 = 2.49 × 10⁵ Pa.

The gas laws as special cases

Law	Held constant	Relationship
Boyle’s	T	p ∝ 1/V
Charles’s	p	V ∝ T
Pressure (Gay-Lussac)	V	p ∝ T

All three fall out of pV = nRT — learn the one equation, not the three laws.

Worked example (combined gas law). A fixed mass of gas at p₁ = 1.0 × 10⁵ Pa, V₁ = 2.0 × 10⁻³ m³ and T₁ = 300 K is compressed to V₂ = 1.0 × 10⁻³ m³ and heated to T₂ = 350 K. Since n is fixed, p₁V₁/T₁ = p₂V₂/T₂, so p₂ = p₁V₁T₂ / (V₂T₁) = (1.0 × 10⁵ × 2.0 × 10⁻³ × 350) ÷ (1.0 × 10⁻³ × 300) = 2.33 × 10⁵ Pa.

Kinetic theory

pV = (1/3) N m <c^2>

where <c²> is the mean square speed. Combining with pV = NkT:

(1/2) m <c^2> = (3/2) k T     (mean TRANSLATIONAL kinetic energy of a molecule)

The central result: the mean translational kinetic energy of a molecule is directly proportional to the absolute temperature. It depends on temperature only — not on pressure, volume, or the identity of the gas. Helium and xenon at the same temperature have the same mean translational molecular kinetic energy; the xenon molecules simply move more slowly because they are heavier.

Root-mean-square speed:

c_rms = sqrt(<c^2>) = sqrt(3kT/m) = sqrt(3RT/M)

Worked example. Nitrogen gas (molar mass M = 0.028 kg mol⁻¹) at 300 K: c_rms = √(3RT/M) = √((3 × 8.31 × 300) ÷ 0.028) = √(267,100) = 517 m s⁻¹. This is the r.m.s. speed of an average molecule — individual molecules have a whole distribution of speeds around this value.

Assumptions of the kinetic model

1. Large number of molecules in random motion.
2. Volume of the molecules is negligible compared with the volume of the container.
3. No intermolecular forces except during collisions.
4. Collisions are perfectly elastic.
5. Time of a collision is negligible compared with the time between collisions.

Assumptions 2 and 3 are the ones that fail in real gases, which is why real gases deviate most at high pressure and low temperature — molecules are close together, so their own volume matters and the attractions become significant.

Internal energy of an ideal gas

Because assumption 3 says there are no intermolecular forces, an ideal gas has no molecular potential energy. Its internal energy is entirely kinetic, so:

U = (3/2) n R T     for a monatomic ideal gas

Hence internal energy depends only on temperature — which is what makes isothermal processes have ΔU = 0.

Exam traps

• Using °C anywhere in pV = nRT.
• Confusing n (moles) with N (molecules), and so R with k.
• Confusing mean square speed <c²> with the square of the mean speed — they are not the same.
• Saying heavier molecules have more translational kinetic energy at the same temperature. They don’t; they are just slower.
• Forgetting that “ideal” means zero potential energy, so U is purely kinetic.

Self-test

1. State the two forms of the equation of state and say what each symbol counts.
2. What is the mean translational kinetic energy of a molecule proportional to?
3. Two gases at the same temperature — which has the greater mean translational molecular KE?
4. Under what conditions do real gases deviate most from ideal behaviour, and why?
5. Why does an ideal gas have no molecular potential energy?
6. What does the Avogadro constant represent?
7. 1.5 mol of an ideal gas at 250 K occupies 0.010 m³. Calculate its pressure.
8. Calculate the r.m.s. speed of a nitrogen molecule (M = 0.028 kg mol⁻¹) at 300 K.
9. A gas is compressed from 2.0 × 10⁻³ m³ to 1.0 × 10⁻³ m³ while its temperature rises from 300 K to 350 K, starting at 1.0 × 10⁵ Pa. Find the new pressure.

Answers: 1. pV = nRT with n in moles and R = 8.31; pV = NkT with N the number of molecules and k = 1.38 × 10⁻²³. 2. The absolute temperature, and nothing else. 3. Neither — they are equal; the heavier gas simply has a lower r.m.s. speed. 4. High pressure and low temperature, because molecular volume is no longer negligible and intermolecular attractions are no longer negligible. 5. The kinetic model assumes no intermolecular forces except during collisions, and potential energy arises from those forces. 6. It is the fixed number of particles (6.02 × 10²³) contained in one mole of any substance. 7. p = nRT/V = (1.5 × 8.31 × 250) ÷ 0.010 = 3.12 × 10⁵ Pa. 8. c_rms = √(3RT/M) = √((3 × 8.31 × 300) ÷ 0.028) = 517 m s⁻¹. 9. p₂ = p₁V₁T₂/(V₂T₁) = (1.0 × 10⁵ × 2.0 × 10⁻³ × 350) ÷ (1.0 × 10⁻³ × 300) = 2.33 × 10⁵ Pa.