A2 · Practice questions · Kinetic theory of gases

Pressure from collisions, temperature from kinetic energy.

Six original Cambridge-style questions on the assumptions of kinetic theory, pV = ⅓Nm⟨c²⟩, root-mean-square speed and the mean translational kinetic energy (3/2)kT of a molecule.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

Keep these straight

Mean square speed is not the square of the mean.

Kinetic theory result: pV = ⅓Nm⟨c²⟩, where m is the mass of one molecule and ⟨c²⟩ is the mean of the squares of the speeds; with density ρ = Nm / V this is p = ⅓ρ⟨c²⟩.
Link to temperature: comparing with pV = NkT gives the mean translational kinetic energy per molecule ½m⟨c²⟩ = (3/2)kT, proportional to the thermodynamic temperature and the same for every gas at that temperature.
R.m.s. speed: c_rms = √⟨c²⟩ = √(3kT / m), proportional to √T and larger for lighter molecules.

Analysis

[3 marks]

Explain, in terms of the motion of its molecules, how a gas exerts a pressure on the walls of its container. Hence explain why, at constant volume and temperature, doubling the number of molecules doubles the pressure.

Molecules move randomly and collide with the walls; each collision changes the momentum of the molecule, so by Newton's second law the wall exerts a force on it and, by the third law, the molecule exerts a force on the wall. ✓
The very large number of collisions per second gives a steady average force on the wall, and this force per unit area is the pressure. ✓
At the same temperature the molecules have the same distribution of speeds, so doubling N doubles the number of wall collisions per second and the momentum transferred per second, doubling the pressure (as pV = ⅓Nm⟨c²⟩ shows). ✓

Calculation

[3 marks]

A helium atom has mass 6.64 × 10⁻²⁷ kg. Calculate the root-mean-square speed of helium atoms at 300 K, and determine the ratio of this speed to the r.m.s. speed of oxygen molecules (mass 5.3 × 10⁻²⁶ kg) at the same temperature.

½m⟨c²⟩ = (3/2)kT gives ⟨c²⟩ = 3kT / m = (3 × 1.38 × 10⁻²³ × 300) / 6.64 × 10⁻²⁷ = 1.87 × 10⁶ m² s⁻². ✓
c_rms = √(1.87 × 10⁶) = 1.4 × 10³ m s⁻¹. ✓
At the same temperature both gases have the same mean kinetic energy, so c_rms ∝ 1 / √m and the ratio is √(5.3 × 10⁻²⁶ / 6.64 × 10⁻²⁷) = 2.8; helium atoms move about 2.8 times faster. ✓

Calculation

[3 marks]

Air at a pressure of 1.0 × 10⁵ Pa has a density of 1.2 kg m⁻³. Use the kinetic theory equation to calculate the root-mean-square speed of the air molecules, and state why this value is not the same as the mean speed of the molecules.

pV = ⅓Nm⟨c²⟩ with Nm / V = ρ gives p = ⅓ρ⟨c²⟩, so ⟨c²⟩ = 3p / ρ = (3 × 1.0 × 10⁵) / 1.2 = 2.5 × 10⁵ m² s⁻². ✓
c_rms = √(2.5 × 10⁵) = 5.0 × 10² m s⁻¹. ✓
The r.m.s. speed is the square root of the mean of the squared speeds; because the molecules have a spread of speeds, squaring weights the faster ones more, so c_rms is slightly larger than the mean speed. ✓

Analysis

[3 marks]

A container holds a mixture of hydrogen (molar mass 2.0 g mol⁻¹) and oxygen (molar mass 32 g mol⁻¹) at a single temperature. Compare the mean translational kinetic energies of the two kinds of molecule, and compare their root-mean-square speeds, giving the numerical ratio.

The mean translational kinetic energy of a molecule is (3/2)kT, which depends only on the temperature, so the hydrogen and oxygen molecules have the same mean kinetic energy. ✓
Since ½m⟨c²⟩ is the same for both, the lighter hydrogen molecules must have the larger mean square speed: ⟨c²⟩ ∝ 1 / m. ✓
c_rms ∝ 1 / √m, so c_rms(H₂) / c_rms(O₂) = √(32 / 2.0) = 4.0; hydrogen molecules move four times as fast on average. ✓

Calculation

[3 marks]

Calculate the thermodynamic temperature at which the mean translational kinetic energy of a gas molecule is 1.0 × 10⁻²⁰ J. Hence calculate the total translational kinetic energy of one mole of an ideal gas at this temperature.

(3/2)kT = 1.0 × 10⁻²⁰, so T = (2 × 1.0 × 10⁻²⁰) / (3 × 1.38 × 10⁻²³) = 4.8 × 10² K (483 K). ✓
One mole contains N_A molecules, each with mean energy 1.0 × 10⁻²⁰ J, so the total is 6.02 × 10²³ × 1.0 × 10⁻²⁰ = 6.0 × 10³ J. ✓
Check: (3/2)RT = 1.5 × 8.31 × 483 = 6.0 × 10³ J, the same result because N_Ak = R. ✓

Calculation

[4 marks]

A rigid container of volume 0.030 m³ holds argon (molar mass 40 g mol⁻¹) at a pressure of 2.0 × 10⁵ Pa and a temperature of 350 K.

(a) Calculate the number of argon atoms in the container. (b) Calculate the mass of one argon atom. (c) Calculate the root-mean-square speed of the atoms. (d) Calculate the total translational kinetic energy of the argon.

(a) N = pV / kT = (2.0 × 10⁵ × 0.030) / (1.38 × 10⁻²³ × 350) = 1.2 × 10²⁴ atoms. ✓
(b) m = M / N_A = 0.040 / 6.02 × 10²³ = 6.6 × 10⁻²⁶ kg. ✓
(c) ⟨c²⟩ = 3kT / m = (3 × 1.38 × 10⁻²³ × 350) / 6.64 × 10⁻²⁶ = 2.18 × 10⁵ m² s⁻², so c_rms = 4.7 × 10² m s⁻¹. ✓
(d) Total KE = N × (3/2)kT = 1.24 × 10²⁴ × 1.5 × 1.38 × 10⁻²³ × 350 = 9.0 × 10³ J, which equals (3/2)pV = 1.5 × 2.0 × 10⁵ × 0.030. ✓

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