Home › Subjects › A-Level H2 Physics › Empirical gas laws and the kinetic theory of gases
A-Level Empirical gas laws and the kinetic theory of gases
What the A-Level syllabus expects for Empirical gas laws and the kinetic theory of gases, and how to practise it.
What the syllabus expects
- Understand that the thermodynamic temperature scale rests on an absolute zero and owes nothing to the properties of any single substance
- Turn Celsius readings into kelvin
Scope: T / K = T / °C + 273.15 - Apply the ideal-gas equation of state pV = NkT
Scope: N counts the particles present - State that a single mole holds 6.02 × 1023 particles, work with the Avogadro constant NA (its value 6.02 × 1023 per mole), and apply the relation Nk = nR that links the molar gas constant to the Boltzmann constant
Scope: n counts the moles, with N = nNA - List the founding assumptions behind the kinetic theory of gases
- Account for gas pressure as a product of the particles' random motion, then, taking pressure to be force over area, derive pV = ⅓Nm⟨c²⟩
Scope: a one-dimensional collision model later widened to three dimensions via ⟨cx²⟩ = ⅓⟨c²⟩ is enough - Use the result that an ideal-gas particle's mean translational kinetic energy rises in step with thermodynamic temperature
Scope: ½m⟨c²⟩ = 3/2 kT
How it's examined
About 4% of the past-paper style questions in Rae's bank for this subject sit in this topic.
Worked examples
Example 1 (3 marks)
Hydrogen behaves as an ideal gas and each molecule has mass 3.34×10⁻²⁷ kg. Compute a hydrogen molecule's r.m.s. speed at 400 K.
Show the worked answer
Principle: kinetic theory links the average translational kinetic energy of one molecule to the absolute temperature of the gas: ½ m ⟨c²⟩ = (3/2) k T Here m is the mass of one molecule, ⟨c²⟩ is the mean square speed, k = 1.38 × 10⁻²³ J K⁻¹ is the Boltzmann constant and T is the thermodynamic (kelvin) temperature. Step 1 - rearrange for the mean square speed. Multiply both sides by 2: m ⟨c²⟩ = 3 k T Then divide both sides by m: ⟨c²⟩ = 3 k T / m Step 2 - the root-mean-square speed is by definition the square root of that: c_rms = √(3 k T / m) Step 3 - list the quantities in SI units before substituting. k = 1.38 × 10⁻²³ J K⁻¹ T = 400 K (already absolute, so no conversion is needed) m = 3.34 × 10⁻²⁷ kg (mass of one hydrogen molecule) Step 4 - substitute and work out the top line first. 3 k T = 3 × 1.38 × 10⁻²³ × 400 = 1.656 × 10⁻²⁰ J Step 5 - divide by the molecular mass. ⟨c²⟩ = 1.656 × 10⁻²⁰ J / 3.34 × 10⁻²⁷ kg ⟨c²⟩ = 4.96 × 10⁶ m² s⁻² (the units work out: J/kg = kg m² s⁻² / kg = m² s⁻², a speed squared) Step 6 - take the square root. c_rms = √(4.96 × 10⁶) m s⁻¹ c_rms = 2.23 × 10³ m s⁻¹ The r.m.s. speed is about 2200 m s⁻¹.
More A-Level H2 Physics topics
Physical quantities, units, measurement uncertainty and vector basics · Types of force, turning effects and conditions for equilibrium · Kinematics, uniformly accelerated motion, momentum and Newton's laws · Energy stores and transfers, work, kinetic and potential energy, fields, power and efficiency · Falling freely, the gravitational potential energy of a uniform field, and how air resistance changes the motion · Impulse and the conservation of momentum and energy · all of A-Level H2 Physics