Ideal gas law worksheet answers for AP Chemistry (free PDF download)

Problem 1 Pressure units

1. 0.750 × 101.3 = 76.0 kPa; 0.750 × 760 = 570 mmHg.
2. 95.0 / 101.3 = 0.938 atm.

Problem 2 Kelvin conversions

1. 298 K, 233 K, 373 K (add 273.15).
2. 350 − 273.15 = 77 °C.

Problem 3 Compressing a syringe

1. Boyle’s law: P₁V₁ = P₂V₂ at constant n and T.
2. P₂ = 100 × 6.0 / 2.0 = 300 kPa.

Problem 4 Heating a balloon

1. 300.15 K and 360.15 K.
2. V₂ = 2.40 × 360.15 / 300.15 = 2.88 L.

Problem 5 Reading a Boyle graph

1. 200 kPa at 2 L, 50 kPa at 8 L.
2. 200 × 2 = 400 kPa·L; 50 × 8 = 400 kPa·L. PV is constant, so P ∝ 1/V.
3. P = 400 / 10 = 40 kPa.

Problem 6 Kinetic theory statements

1. False: collisions are elastic, so no kinetic energy is lost.
2. True.
3. False: the volume of the particles is negligible compared with the space between them.

Problem 7 Pressure from the ideal gas law

T = 293.15 K.
P = nRT/V = 0.500 × 8.314 × 293.15 / 12.0 = 102 kPa.

Problem 8 Identify a gas from its molar mass

1. n = PV/RT = 100.0 × 0.500 / (8.314 × 300.15) = 0.0200 mol.
2. M = 1.29 / 0.02004 = 64.4 g/mol.
3. The gas is C₂H₅Cl.

Problem 9 Density of carbon dioxide

ρ = PM/RT = 101.3 × 44.01 / (8.314 × 298.15) = 1.80 g/L.
Air (about 29 g/mol) has a density near 1.18 g/L, so CO₂ is denser and sinks.

Problem 10 Partial pressures in a tank

1. n_total = 1.50 mol; P = 1.50 × 8.314 × 298.15 / 20.0 = 186 kPa.
2. x(O₂) = 0.300/1.50 = 0.200; P(O₂) = 0.200 × 186 = 37.2 kPa.

Problem 11 Gas over water

1. P(H₂) = 101.3 − 3.17 = 98.1 kPa.
2. n = 98.13 × 0.125 / (8.314 × 298.15) = 4.95 × 10⁻³ mol.

Problem 12 Two speed distributions

1. B is hotter: its peak is at higher speed and the curve is wider.
2. A peaks at about 420 m/s and B at about 600 m/s. T_B = 300 × (600/420)² = about 600 K.
3. The areas are equal, because both represent the same number of molecules.

Problem 13 Pressure and temperature

Higher temperature gives a higher average kinetic energy, so particles hit the walls harder and more often.
P₂ = 100.0 × 353.15 / 293.15 = 120 kPa.

Problem 14 Effusion rates

Rate(H₂)/Rate(O₂) = √(32.00/2.016) = √15.87 = 3.98.

Problem 15 Gas from a reaction

n(CaCO₃) = 2.50 / 100.09 = 0.02498 mol = n(CO₂).
V = nRT/P = 0.02498 × 8.314 × 298.15 / 101.3 = 0.611 L.

Problem 16 Speeds of gas molecules

N₂: √(3 × 8.314 × 300 / 0.028014) = √267 106 = 517 m/s.
He: √(7482.6 / 0.004003) = 1.37 × 10³ m/s.
At the same temperature He atoms have the same average kinetic energy but a much smaller mass, so more of them reach escape speed.

Problem 17 A real gas calculation

1. P = 0.08314 × 300 / 1.00 = 24.9 bar.
2. P = nRT/(V − nb) − an²/V² = 24.94/0.9573 − 3.64 = 26.05 − 3.64 = 22.4 bar.
3. The real pressure is lower because attractions between CO₂ molecules reduce the force of wall collisions; the correction for particle volume is smaller than the attraction term here.

Problem 18 Mixture from masses

1. n(O₂) = 0.250 mol; n(N₂) = 14.0/28.01 = 0.500 mol.
2. P = 0.750 × 8.314 × 300 / 10.0 = 187 kPa.
3. x(O₂) = 0.250/0.750 = 0.333; P(O₂) = 0.333 × 187 = 62.4 kPa.

Problem 19 Interpreting compressibility

1. Z = 1.
2. X is hydrogen: weak attractions, so particle volume dominates and Z > 1 at all pressures. Y is methane: stronger attractions give Z < 1.
3. At moderate pressure attractions reduce the pressure below the ideal value (Z < 1). At high pressure particles are crowded, particle volume matters more, and the volume is larger than ideal (Z > 1).

Problem 20 Molar mass of a volatile liquid

1. n = PV/RT = 101.3 × 0.255 / (8.314 × 373.15) = 0.008326 mol; M = 0.617 / 0.008326 = 74.1 g/mol.
2. The gas temperature must equal the known T in PV = nRT.
3. If the liquid was not fully vaporised or some vapour leaked before sealing, the measured mass of vapour is too low.