Internal energy as the random kinetic and potential energies of the molecules

Absolute zero, the kelvin scale and the link between temperature and mean molecular kinetic energy

The ideal gas equation pV = NkT, with N the number of molecules

Boyle's law, the pressure law and Charles's law as special cases, and their graphs

Core Practical 14: how the pressure of a fixed mass of gas depends on its volume at constant temperature

The kinetic theory assumptions and the derivation of ½m⟨c²⟩= (3/2)kT

Root-mean-square speed and how it depends on temperature and molecular mass

1. A gas obeys pV = NkT from outside and is a huge number of randomly moving molecules from inside, and putting the two views together shows that absolute temperature measures the mean kinetic energy of the molecules.

2. Internal energy is the sum of the randomly distributed kinetic and potential energies of all the molecules; in an ideal gas it is all kinetic.

3. Always use kelvin: T/K = θ/°C + 273. At absolute zero the molecules would have their minimum possible energy.

4. ½m⟨c²⟩= (3/2)kT: mean kinetic energy depends on T only, not on the type of gas.

5. Fix two of N, T, p, V and you get a gas law: p ∝1/V (Boyle), p ∝T (pressure law), V ∝T (Charles).

6. Kinetic theory gives pV = ⅓Nm⟨c²⟩; comparing with pV = NkT gives cᵣₘₛ = √(3kT/m).

The kinetic part comes from the random motion of the molecules: vibration in a solid, vibration and sliding past each other in a liquid, free flight in a gas. The potential part comes from the forces between molecules; pulling molecules further apart against attractive forces increases it.

"Randomly distributed" matters: at any instant some molecules are fast and some slow, and collisions keep swapping energy between them. Only the total and the average are fixed by the state of the body.

Heating during a change of state (melting, boiling) increases the potential energy of the molecules at constant temperature; heating without a change of state increases mainly their kinetic energy, so the temperature rises.

An ideal gas has no forces between molecules except during collisions, so no potential energy: its internal energy is the total random kinetic energy of its molecules.

The kinetic energy of the body moving as a whole, or its gravitational potential energy, is not internal energy; only random, molecular-scale energies count.

Measure the pressure of a fixed volume of gas at several temperatures and plot p against θ in °C: the graph is a straight line. Extend it backwards and it reaches p = 0 at −273 °C, for every gas and every starting pressure.

The kelvin (absolute) scale starts at absolute zero and has degrees the same size as Celsius degrees:

This is not in the formulae list (learn it). A temperature difference is the same in K and °C; a temperature ratio is only meaningful in K.

The mean kinetic energy of the molecules is proportional to the absolute temperature: double T in kelvin and you double it (the kinetic theory section shows why).

Only kelvin works: 10 °C to 20 °C is 283 K to 293 K, a 3.5% rise, not a doubling.

At the same temperature, molecules of different gases have the same mean kinetic energy; heavier ones simply move more slowly.

WORKED EXAMPLE · QUESTION

A sample of gas is heated from 27 °C to 327 °C. (a) Calculate the ratio of the mean kinetic energy of the molecules at 327 °C to that at 27 °C. (b) Calculate the Celsius temperature at which the mean kinetic energy would be half its value at 27 °C.

SOLUTION

(a) Convert: 27 + 273 = 300 K and 327 + 273 = 600 K. Mean kinetic energy ∝ T, so the ratio is 600 ÷ 300: 2.0 (not 327 ÷ 27 = 12.1).

(b) Half the mean kinetic energy needs half the kelvin temperature: T = 300 ÷2 = 150 K, so θ= 150 - 273: −123 °C.

p is pressure (Pa), V volume (m³), N the number of molecules (no unit), k the Boltzmann constant, 1.38 × 10⁻²³ J K⁻¹, and T the absolute temperature (K). The equation is in the formulae list and k is on the data sheet. It holds for an ideal gas; real gases behave like this at low pressures and at temperatures well above those at which they liquefy.

N is a number of molecules, typically 10²⁰ to 10²⁷, not an amount in moles. If you are given the Avogadro constant N_A (6.02 × 10²³ mol⁻¹) and an amount n in moles, use N = nN_A.

Units check: Pa m³ = N m⁻² × m³ = N m = J, and kT is J K⁻¹ × K = J. So pV is an energy.

Convert first: cm³ → m³ (× 10⁻⁶), litres → m³ (× 10⁻³), kPa → Pa, °C → K.

Use the absolute pressure of the gas, not a tyre-gauge reading, which shows pressure above atmospheric.

WORKED EXAMPLE · QUESTION

A classroom measures 8.0 m × 6.0 m × 3.0 m. The air is at 1.01 × 10⁵ Pa and 22 °C; treat it as an ideal gas. (a) Calculate the number of molecules in the room. (b) The average mass of an air molecule is 29 u. Calculate the mass of air in the room.

SOLUTION

(a) V = 8.0 ×6.0 ×3.0 = 144 m³ and T = 22 + 273 = 295 K. N = pV/kT = (1.01 ×10⁵ ×144)/(1.38 ×10⁻²³ ×295): N = 3.57 ×10²⁷.

(b) Mass = number × mass of one molecule (1 u = 1.66 × 10⁻²⁷ kg): 3.57 ×10²⁷ ×29 ×1.66 ×10⁻²⁷: 172 kg.

Given N_A = 6.02 ×10²³ mol⁻¹, the amount would be N/N_A = 5930 mol.

For a fixed mass of gas N is constant, so pV/T = Nk is constant. Hold one more quantity constant and a simpler law drops out.

A p–V curve cannot show proportionality by eye, so replot as p against 1/V. Since p = NkT ×(1/V), the graph is a straight line through the origin with gradient NkT; a hotter isotherm is a steeper line.

WORKED EXAMPLE · QUESTION

A car tyre contains 0.025 m³ of air at an absolute pressure of 3.10 × 10⁵ Pa and 7 °C. After a long drive the air is at 37 °C; the volume does not change. (a) Calculate the new pressure. (b) The driver lets air out until the pressure is back to 3.10 × 10⁵ Pa, still at 37 °C. Calculate the fraction of the molecules let out, and the number this is.

SOLUTION

(a) Fixed N and V, so p/T is constant, in kelvin: p₂ = 3.10 ×10⁵ ×(310/280): 3.43 ×10⁵ Pa.

(b) At fixed V and T, N ∝p: fraction remaining = 3.10 ÷3.432 = 0.903, so 9.7% let out. Original number: N = (3.10 ×10⁵ ×0.025)/(1.38 ×10⁻²³ ×280) = 2.01 ×10²⁴, so let out 0.0968 ×2.01 ×10²⁴: 1.9 ×10²³ molecules.

This is Core Practical 14. A fixed mass of air is trapped and compressed in steps at constant temperature, and its pressure recorded at each volume.

Measure the tube's internal diameter d (or use the value given), so V = πd²/4 l. The area is constant, so V ∝l and the length alone tests the relationship.

Record l and p at atmospheric pressure, then pump to raise the pressure in steps, closing the valve each time.

Compressing a gas warms it: wait about a minute after each change so the air returns to room temperature, then read the length (eye level with the oil, to avoid parallax) and the pressure.

Take 6–8 readings across the gauge's range; repeat while releasing the pressure and average.

Plot p against 1/l (or 1/V): a straight line through the origin shows p ∝1/V at constant temperature.

Safety: eye protection, and stay within the working pressure of gauge and tube, since a glass tube under pressure can shatter.

WORKED EXAMPLE · QUESTION

The results plotted above were taken at 20 °C in a tube of internal diameter 1.20 cm.

(a) Explain how the data support Boyle's law. (b) Use the gradient, 39.1 kPa m, to calculate the number of air molecules trapped. (c) The diameter is known to ± 0.02 cm, the gradient to ± 3.0% and the temperature to ± 1 K. Calculate the percentage uncertainty in N.

SOLUTION

(a) The products pl all lie between 39.6 and 40.8 kPa m, constant within about 2%, so p ∝1/l ∝1/V (and the graph is a straight line through the origin).

(b) Gradient = pl = pV/A = NkT/A, so N = gradient ×A/kT. A = π(0.60 ×10⁻²)² = 1.131 ×10⁻⁴ m², T = 293 K: N = (39.1 ×10³ ×1.131 ×10⁻⁴)/(1.38 ×10⁻²³ ×293): 1.09 ×10²¹.

(c) The diameter is squared, so its percentage counts twice: 0.02 ÷ 1.20 = 1.7%, so 3.3% for A; T: 1 ÷ 293 = 0.34%. Total 3.3 + 3.0 + 0.34: 6.7%, N = (1.09 ±0.07) ×10²¹.

The assumptions of the kinetic model

A gas is a very large number of identical molecules in random motion.

The volume of the molecules is negligible compared with the volume of the container.

There are no forces between molecules except during collisions.

Collisions with each other and with the walls are elastic (no loss of kinetic energy).

The duration of a collision is negligible compared with the time between collisions.

The derivation, step by step

1. Change of momentum of one molecule hitting the wall: mu - (-mu) = 2mu, given to the wall.

2. It crosses the box and back, distance 2L, before hitting the same wall again: time between hits = 2L/u.

3. Average force on the wall = rate of change of momentum: 2mu ÷(2L/u) = mu²/L.

4. For N molecules use the mean of u²: force = Nm⟨u²⟩/L, and pressure = force ÷ area L² = Nm⟨u²⟩/V.

5. The motion is random, so ⟨u²⟩= ⟨v²⟩= ⟨w²⟩, and since c² = u² + v² + w², ⟨u²⟩= ⅓⟨c²⟩.

6. So pV = ⅓Nm⟨c²⟩, or p = ⅓ρ⟨c²⟩ since the density ρ= Nm/V.

7. Compare with pV = NkT: ⅓Nm⟨c²⟩= NkT, so m⟨c²⟩= 3kT.

m is the mass of one molecule (kg), ⟨c²⟩ the mean square speed (m² s⁻²). The equation is in the formulae list, but the derivation is not supplied: learn its steps. Consequences: mean kinetic energy ∝ T; the internal energy of an ideal gas is N ×(3/2)kT; and cᵣₘₛ = √(⟨c²⟩) = √(3kT/m).

Root-mean-square speed

⟨c²⟩ is the mean of the squares of the speeds; its square root is the r.m.s. speed. Order matters: square each speed, take the mean, then the square root. The r.m.s. speed is always at least as big as the ordinary mean speed. cᵣₘₛ ∝√T for a given gas and ∝1/√m at a given temperature, so hydrogen and helium move much faster than nitrogen at the same temperature.

WORKED EXAMPLE · QUESTION

Five molecules have speeds of 200, 400, 500, 600 and 800 m s⁻¹. Calculate their mean speed and their r.m.s. speed.

SOLUTION

Mean: (200 + 400 + 500 + 600 + 800) ÷5: 500 m s⁻¹.

⟨c²⟩= (200² + 400² + 500² + 600² + 800²) ÷5 = 290 000 m² s⁻², so cᵣₘₛ = √(290 000): 539 m s⁻¹, larger than the mean because squaring weights the fast molecules more.

WORKED EXAMPLE · QUESTION

A nitrogen molecule has a mass of 28 u (1 u = 1.66 × 10⁻²⁷ kg). The air is at 20 °C. (a) Calculate the mean kinetic energy of a nitrogen molecule. (b) Calculate the r.m.s. speed. (c) Calculate the Celsius temperature at which the r.m.s. speed would be twice as large.

SOLUTION

(a) (3/2)kT = 1.5 ×1.38 ×10⁻²³ ×293: 6.07 ×10⁻²¹ J.

(b) m = 28 ×1.66 ×10⁻²⁷ = 4.65 ×10⁻²⁶ kg; cᵣₘₛ = √((3 ×1.38 ×10⁻²³ ×293)/(4.65 ×10⁻²⁶)): 511 m s⁻¹.

(c) cᵣₘₛ ∝√T, so doubling it needs four times the kelvin temperature: T = 4 ×293 = 1172 K: 899 °C.

WORKED EXAMPLE · QUESTION

A cylinder contains helium at 1.2 × 10⁵ Pa and a density of 0.20 kg m⁻³. The mass of a helium atom is 4.0 u. (a) Show that the r.m.s. speed is about 1.3 × 10³ m s⁻¹. (b) Calculate the temperature of the helium. (c) Nitrogen (28 u) at the same temperature is in a second cylinder. Calculate its r.m.s. speed.

SOLUTION

(a) p = ⅓ρ⟨c²⟩, so cᵣₘₛ = √(3p/ρ) = √(3 ×1.2 ×10⁵ ÷0.20) = √(1.80 ×10⁶): 1342 m s⁻¹ ≈ 1.3 × 10³ m s⁻¹.

(b) T = m⟨c²⟩/3k = (4.0 ×1.66 ×10⁻²⁷ ×1.80 ×10⁶)/(3 ×1.38 ×10⁻²³): 289 K.

(c) Same T, same mean kinetic energy, so cᵣₘₛ ∝1/√m: 1342 ×√(4.0/28) = 1342 ×0.378: 507 m s⁻¹.

Check: p = ρkT/m = 0.20 ×1.38 ×10⁻²³ ×289 ÷6.64 ×10⁻²⁷ = 1.20 ×10⁵ Pa, consistent with the data.

1 Specific heat capacity and specific latent heat

Energy supplied to a body raises either the molecular kinetic energy (temperature rises, ΔE = mcΔθ) or the molecular potential energy (change of state, ΔE = LΔm): the two parts of internal energy.

Momentum (Unit 1) and 5.3 Binding energy, fission and fusion

The derivation is Newton's second law as rate of change of momentum, applied to molecules bouncing elastically off a wall. Nuclei fuse only if they approach very closely, which needs a large kinetic energy; since mean kinetic energy ∝ T, fusion needs temperatures of millions of kelvin.

Practical skills (Unit 6)

Core Practical 14 is a common context for planning, graph plotting and combining percentage uncertainties in the practical skills paper.

1. Water at 100 °C turns to steam at 100 °C. State what happens to the mean kinetic energy and the potential energy of its molecules, and to its internal energy.

2. Calculate the ratio of the mean kinetic energy of air molecules at 20 °C to that at −40 °C.

3. A fixed mass of gas is compressed slowly to half its volume at constant temperature. State and explain what happens to its pressure.

4. A balloon holds 2.0 litres of gas at 300 K. It is heated to 450 K at constant pressure. Calculate the new volume.

5. By what factor does the r.m.s. speed of the molecules of a gas change when its temperature rises from 300 K to 1200 K?

6. Calculate the total internal energy of the 1.09 × 10²¹ air molecules in Worked example 4, treating the air as an ideal gas at 293 K.

1. Mean kinetic energy unchanged (same temperature). Potential energy increases, because the molecules are separated against the attractive forces between them. So the internal energy increases.

2. 20 °C = 293 K and −40 °C = 233 K, so the ratio is 293 ÷ 233 = 1.26.

3. The pressure doubles (Boyle's law). The molecules have the same speeds, but in half the volume they hit each unit area of the walls twice as often, so the rate of change of momentum per unit area doubles.

4. V ∝T, so V = 2.0 ×450 ÷300 = 3.0 litres.

5. cᵣₘₛ ∝√T and T is multiplied by 4, so the r.m.s. speed doubles (factor 2).

6. U = N ×(3/2)kT = 1.09 ×10²¹ ×1.5 ×1.38 ×10⁻²³ ×293 = 6.61 J.