Free Pearson Edexcel International A Level Physics lesson

Ideal gases and molecular motion

Connect pressure measurements to molecular kinetic energy.

What you’ll learn

  • Convert pressure, volume and absolute temperature before using the particle equation.
  • Control temperature and gas amount, and analyse a reciprocal or logarithmic graph.
  • Connect wall momentum transfer to pressure, then combine macroscopic and microscopic gas equations.

Count particles and use kelvin

The ideal-gas equation pV = NkT relates absolute pressure p, volume V, particle number N and absolute temperature T. Boltzmann’s constant k = 1.38 × 10⁻²³ J K⁻¹ connects the microscopic count to macroscopic measurements. Use pascals and cubic metres, and do not confuse N, a particle count, with n, an amount in moles.

Absolute pressure includes atmospheric pressure; a gauge may report only the excess above the surroundings. Temperature in this equation must be in kelvin, with T = θ/°C + 273.15. Ratios of Celsius temperatures are invalid because the Celsius zero is arbitrary. For fixed N, pV/T is constant between equilibrium states.

An ideal gas neglects particle volume and intermolecular interactions except elastic collisions. It works best when a real gas is dilute and sufficiently far from condensation. A fixed-mass constant-volume pressure–temperature graph extrapolates toward zero pressure near absolute zero in the ideal model, but real gases condense before the extrapolation could be followed literally.

pV = NkT fixed N: p₁V₁/T₁ = p₂V₂/T₂ T/K = θ/°C + 273.15

Symbols and units

  • p is absolute pressure in Pa, measured relative to vacuum; V is the total gas volume in m³.
  • N is a dimensionless particle count, not the amount n in moles; k = 1.38 × 10⁻²³ J K⁻¹ is Boltzmann’s constant.
  • T is absolute temperature in K; θ is the Celsius temperature, with T/K = θ/°C + 273.15.
  • Subscripts 1 and 2 identify equilibrium states of the same sealed sample when N is fixed. 1 cm³ = 10⁻⁶ m³.

How the equation explains the idea

  1. Use the stated absolute pressure, SI volume and kelvin temperature. The ideal-gas equation applies to an equilibrium gas adequately described by negligible particle volume and interactions.
  2. Isolate the particle count by dividing pV = NkT by kT: N = pV/(kT). The numerator is (1.20 × 10⁵)(2.00 × 10⁻³) = 240 Pa m³ = 240 J.
  3. The denominator is kT = (1.38 × 10⁻²³)(300) = 4.14 × 10⁻²¹ J. Thus N = 240/(4.14 × 10⁻²¹) = 5.80 × 10²² particles; the joules cancel.
  4. For two states with the same N, pV/T = Nk is unchanged, so p₂ = p₁(V₁/V₂)(T₂/T₁). Celsius temperature ratios would not preserve this proportionality.
Absolute pressure is proportional to kelvin temperature

Conditions. Computed from p = NkT/V for a sealed ideal gas: N = 5.80 × 10²² particles, V = 2.00 × 10⁻³ m³ and k = 1.38 × 10⁻²³ J K⁻¹. N and V stay fixed; every point is an equilibrium state.

Absolute temperature / K; Absolute pressure / Pa.

The slope Nk/V is 400.2 Pa K⁻¹. Doubling absolute temperature doubles pressure. Celsius temperature ratios would not give this law; a real gas can cease to be ideal outside its dilute gas range.

Open the interactive lesson to view the plotted curves and their coordinate tables.

Worked example

Count particles in a chamber

  1. Pressure is 1.20 × 10⁵ Pa, volume 2.00 × 10⁻³ m³ and temperature 300 K.
  2. N = pV/(kT) = 240/(1.38 × 10⁻²³ × 300).
  3. N = 5.80 × 10²² particles.

The numerator pV has units of energy.

Keep this idea. Convert pressure, volume and absolute temperature before using the particle equation.

Core practical 14: test an inverse relationship

To test pressure against volume, trap a fixed quantity of gas in a suitable sealed cylinder or syringe connected to a pressure sensor. Record the full gas volume, including any relevant connecting-tube dead volume, and measure absolute pressure. Compress in small stages within the apparatus rating and allow the gas to return to the same measured room temperature before each reading.

For a constant-temperature ideal gas, pV is constant. Plot p against 1/V: a straight line through the origin supports the inverse relationship and its slope equals NkT. A log p against log V plot has slope −1. Neither plot alone proves that every instrumental offset is absent; pressure zero error or an omitted dead volume can produce systematic curvature or an intercept.

Take a range large enough to test the relationship, repeat readings and compare compression with expansion after equilibration. Check for leaks by observing a settled state over time. Fast compression usually heats the gas and can give a pressure higher than the isothermal prediction. Keep the apparatus secured and within its stated pressure range rather than seeking the largest possible compression.

Fixed N,T: p ∝ 1/V ln p = constant − ln V

Symbols and units

  • p₁ and p₂ are absolute gas pressures in Pa; V₁ and V₂ are total trapped gas volumes in m³, including connecting-tube volume.
  • N is the particle count and T is absolute temperature in K; both must remain constant for pV = constant.
  • k is Boltzmann’s constant in J K⁻¹. The p against 1/V gradient is NkT in Pa m³ = J.
  • For logarithmic graphs, use dimensionless ratios p/pref and V/Vref with fixed reference pressure and volume; the log–log slope is dimensionless.

How the equation explains the idea

  1. At fixed N and T, pV = NkT gives p₁V₁ = p₂V₂. Divide by V₂ to obtain p₂ = p₁V₁/V₂.
  2. The volume units cancel in the ratio: p₂ = 100 × (60/40) = 150 kPa. Convert for a product check: (100000)(60 × 10⁻⁶) = (150000)(40 × 10⁻⁶) = 6.0 J.
  3. Rewriting p = (NkT)(1/V) predicts a straight p–1/V line. Taking logs of reference-scaled quantities gives ln(p/pref) = constant − ln(V/Vref), with gradient −1.
  4. A rapid compression can increase T, so pressure can exceed 150 kPa before equilibration. A leak or omitted dead volume also breaks the stated test conditions.
An isothermal pressure–volume curve is an inverse curve

Conditions. Same sealed ideal gas for both isotherms, with Nk = 0.0200 J K⁻¹. At T = 300 K, pV = 6.00 J; at 450 K, pV = 9.00 J. Volume includes all connecting-tube gas; each curve holds its own temperature fixed.

Total gas volume / m³; Absolute pressure / Pa.

Along either curve, halving volume doubles pressure. At the same volume, the 450 K gas has 1.5 times the pressure of the 300 K gas. A rapid compression that also heats the gas need not follow the cooler isotherm.

Open the interactive lesson to view the plotted curves and their coordinate tables.

Worked example

Check a pressure–volume pair

  1. At 60 cm³ the absolute pressure is 100 kPa.
  2. For the same equilibrated gas at 40 cm³, p₂ = 100 × 60/40 = 150 kPa.
  3. Both states have pV = 6.0 Pa m³ after SI conversion.

An inverse relationship requires unchanged gas amount and temperature.

Keep this idea. Control temperature and gas amount, and analyse a reciprocal or logarithmic graph.

Derive mean molecular kinetic energy

Gas pressure comes from molecules changing momentum as they collide with a wall. In the ideal model, collisions are elastic and random motion is isotropic: the mean squared velocity components in three perpendicular directions are equal. For a molecule of mass m hitting a wall normally with velocity component cx, the momentum change magnitude is 2mcx.

In a cube of side l, the time between successive collisions with that same wall is 2l/|cx|. Momentum change divided by time gives its average force contribution mcx²/l. Summing N particles and dividing by wall area l² gives pV = Nm〈cx²〉. Isotropy gives 〈c²〉 = 3〈cx²〉, so pV = Nm〈c²〉/3. Combining this with pV = NkT yields ½m〈c²〉 = 3kT/2.

This is mean translational kinetic energy per particle, not the energy of every particle and not necessarily every internal degree of freedom of a molecule. Root-mean-square speed is √〈c²〉 = √(3kT/m). At the same temperature particles of different mass have the same mean translational kinetic energy, while lighter particles have greater rms speed. The ideal classical model tends toward zero translational kinetic energy at absolute zero; it is not a claim that all quantum motion disappears.

pV = ⅓Nm〈c²〉 ½m〈c²〉 = ³⁄₂kT crms = √(3kT/m)

Symbols and units

  • p is pressure in Pa, V is volume in m³, N is particle count and m is the mass of one molecule in kg, not the whole sample mass.
  • c is molecular speed in m s⁻¹ in this section, not the speed of light or specific heat capacity. 〈c²〉 is mean squared speed in m² s⁻²; crms = √〈c²〉.
  • cx is the velocity component normal to a wall in m s⁻¹; l is the side length of the model cube in m; angle brackets denote an average over particles.
  • T is absolute temperature in K and k = 1.38 × 10⁻²³ J K⁻¹. The quantity ½m〈c²〉 is mean translational kinetic energy per particle in J.

How the equation explains the idea

  1. An elastic wall collision transfers normal momentum 2m|cx|. Dividing by the return time 2l/|cx| gives average force mcx²/l; summing particles and dividing by area l² gives pV = Nm〈cx²〉.
  2. Isotropic random motion gives 〈cx²〉 = 〈c²〉/3. Equate the resulting pV = Nm〈c²〉/3 to NkT, cancel N and multiply by 3/2: ½m〈c²〉 = 3kT/2.
  3. At 300 K, mean translational energy = 1.5 × 1.38 × 10⁻²³ × 300 = 6.21 × 10⁻²¹ J. This is an average, not the energy of every molecule or every molecular internal mode.
  4. Rearrange for speed: crms = √(3kT/m) = √(1.242 × 10⁻²⁰/(4.60 × 10⁻²⁶)) ≈ 520 m s⁻¹. At fixed T, a larger particle mass lowers rms speed while leaving mean translational energy unchanged.

Worked example

Compare molecules at 300 K

  1. Mean translational energy = 1.5 × 1.38 × 10⁻²³ × 300 = 6.21 × 10⁻²¹ J.
  2. For particle mass 4.60 × 10⁻²⁶ kg, crms = √(3 × 1.38 × 10⁻²³ × 300 / 4.60 × 10⁻²⁶).
  3. crms ≈ 520 m s⁻¹; another species at 300 K has the same mean energy but need not have that rms speed.

Temperature fixes mean translational energy, not a common molecular speed.

Keep this idea. Connect wall momentum transfer to pressure, then combine macroscopic and microscopic gas equations.

Open the interactive lesson for free practice and reviewed three-question samples when available. New full question sets and timed unit checks are in preparation. Existing complete Physics revision packs retain paid access; existing free teaching PDFs remain free.

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