The problems are on the ideal gas equation worksheet. Try them on your own first, then compare with the answers below.
Quick answer: An ideal gas obeys pV = nRT, where pressure, volume and kelvin temperature are linked through the amount of gas. These worked answers correct the 20 problems of the Physics 11-12 worksheet on ideal gas equation, from Kelvin conversions to why hydrogen escapes: each solution shows the formula, the substitution and the final answer with its unit.
Each answer shows the formula, the substitution and the result with its unit. Check that you converted to kelvin, cubic metres and kilograms before you compare your numbers, and note that small rounding differences in the last digit are acceptable.
Level: Physics grades 11-12 (A Level · IB · Class 11-12) · Curriculum: IB B.1-B.3
Problem 1 Kelvin conversions
T = θ + 273.1. 37 + 273 = 310 K.
2. 85 − 273 = −188 °C.
3. A change of 1 °C equals a change of 1 K, so the rise is 25 K.
Problem 2 Moles and molecules
1. n = m / M = 3.2 / 32 = 0.10 mol.2. N = nNA = 0.10 × 6.02 × 10²³ = 6.0 × 10²² molecules.
3. m = nM = 0.75 × 44 = 33 g.
Problem 3 Pressure from the equation
V = 2.0 × 10⁻³ m³ and T = 300 K.p = nRT / V = (0.080 × 8.31 × 300) / (2.0 × 10⁻³) = 99720 Pa, so p ≈ 1.0 × 10⁵ Pa.
Problem 4 Boiling water for tea
Q = mcΔT = 2.0 × 4180 × (100 − 20) = 668800 J, so Q ≈ 6.7 × 10⁵ J.Problem 5 Cool box ice
Q = mL = 0.60 × 3.34 × 10⁵ = 200400 J, so Q ≈ 2.0 × 10⁵ J. The ice stays at 0 °C throughout.Problem 6 Ideas behind the model
1. Any three of: the molecules have negligible volume; there are no forces between them except in collisions; collisions are elastic; their motion is random; the number of molecules is very large.2. With no forces between molecules there is no potential energy. The internal energy is therefore just the sum of the random kinetic energies, which is set by the temperature. U depends on T only.
Problem 7 Diver’s cylinder
T = 293 K, V = 0.010 m³.1. n = pV / RT = (2.0 × 10⁶ × 0.010) / (8.31 × 293) = 8.21 mol, so n ≈ 8.2 mol.
2. m = nM = 8.214 × 28 = 230 g, so m ≈ 2.3 × 10² g.
Problem 8 Helium at the fair
T = 300 K.1. E = (3/2)kT = 1.5 × 1.38 × 10⁻²³ × 300 = 6.21 × 10⁻²¹ J.
2. U = (3/2)nRT = 1.5 × 0.50 × 8.31 × 300 = 1870 J, so U ≈ 1.9 × 10³ J.
Problem 9 Light atoms move fast
1. c = √(3RT / M) = √(3 × 8.31 × 300 / 0.0040) = √(1869750) = 1367 m/s.2. At equal temperature c ∝ 1/√M, so the ratio is √(0.040 / 0.0040) = √10 = 3.2.
Problem 10 Motorbike tyre
p/T is constant, with T₁ = 298 K and T₂ = 328 K.p₂ = p₁T₂ / T₁ = 2.20 × 10⁵ × 328 / 298 = 242148 Pa, so p₂ ≈ 2.42 × 10⁵ Pa.
Problem 11 Metal block heater
E = Pt = 60 × 300 = 18000 J.c = E / (mΔT) = 18000 / (0.50 × 40) = 900 J/(kg K), which matches aluminium.
Problem 12 Tea and milk
Energy lost by tea = energy gained by milk: 0.20c(90 − T) = 0.050c(T − 5).18 − 0.20T = 0.050T − 0.25, so 0.25T = 18.25 and T = 73 °C.
Problem 13 Reading a pressure-volume set
1. pV = 100 kPa × 6.0 L = 600; 150 × 4.0 = 600; 200 × 3.0 = 600; 300 × 2.0 = 600 (kPa L, i.e. J). The product is constant, so T is constant.2. n = pV / RT = 600 / (8.31 × 290) = 0.249 mol, so n ≈ 0.25 mol.
3. p = 600 / 2.4 = 250 kPa.
Problem 14 Spot the error
1. The temperature was left in °C (it must be 300 K) and the volume was left in litres (it must be 4.0 × 10⁻³ m³).2. p = (0.20 × 8.31 × 300) / (4.0 × 10⁻³) = 124650 Pa, so p ≈ 1.2 × 10⁵ Pa.
Problem 15 Air in a classroom
1. ρ = m/V = pM / RT = (1.01 × 10⁵ × 0.0290) / (8.31 × 293) = 1.203 kg/m³, so ρ ≈ 1.20 kg/m³.2. V = 8.0 × 6.0 × 3.0 = 144 m³, so m = ρV = 1.2030 × 144 = 173.2 kg, so m ≈ 1.7 × 10² kg.
Problem 16 Two flasks, two temperatures
nA = pV / RT = (2.0 × 10⁵ × 3.0 × 10⁻³) / (8.31 × 300) = 0.2407 mol.nB = (1.0 × 10⁵ × 5.0 × 10⁻³) / (8.31 × 400) = 0.1504 mol.
Total n = 0.3911 mol. p = nRT / V = (0.3911 × 8.31 × 350) / (8.0 × 10⁻³) = 142188 Pa, so p ≈ 1.4 × 10⁵ Pa.
Problem 17 Piston lifting a load
1. p = p₀ + mg/A = 1.01 × 10⁵ + (4.0 × 9.81) / (5.0 × 10⁻³) = 108848 Pa ≈ 1.09 × 10⁵ Pa.2. ΔV = nRΔT / p = (0.020 × 8.31 × 60) / 108848 = 9.16 × 10⁻⁵ m³. Rise = ΔV / A = 1.83 cm, so ≈ 1.8 cm.
3. W = pΔV = nRΔT = 0.020 × 8.31 × 60 = 9.97 J, so W ≈ 10 J.
Problem 18 Heating gas against water
1. ΔU = (3/2)nRΔT = 1.5 × 1.0 × 8.31 × 100 = 1246.5 J (≈ 1.25 × 10³ J).2. t = E / P = 1246.5 / 50 = 25 s.
3. Q = mcΔT = 0.018 × 4180 × 100 = 7524 J. Ratio = 7524 / 1246.5 = 6.0. Water stores far more energy per kelvin because its molecules have many more ways to store energy.
Problem 19 Ice for a cold drink
Energy released by the water cooling to 0 °C: 0.30 × 4180 × 25 = 31350 J.Energy absorbed per kg of ice: warming 2100 × 5 = 10 500 J plus melting 3.34 × 10⁵ J = 344 500 J.
m = 31350 / 344 500 = 0.0910 kg, so m ≈ 0.091 kg (91 g).
Problem 20 Why hydrogen escapes
T = M c² / (3R) with c = 1.12 × 10⁴ m/s, so c² = 1.254 × 10⁸ m²/s².1. T = 0.0020 × 1.254 × 10⁸ / (3 × 8.31) = 10063 K ≈ 1.0 × 10⁴ K.
2. T = 0.028 × 1.254 × 10⁸ / (3 × 8.31) = 140887 K ≈ 1.4 × 10⁵ K.
3. In a speed distribution a few hydrogen molecules in the fast tail exceed the escape speed even at atmospheric temperatures, so hydrogen leaks away over geological time. Nitrogen needs about fourteen times more energy, so almost none reaches escape speed.
