Gas Laws Calculations

IGCSE Edexcel Physics
5.20–5.22 Boyle's law, pressure-temperature law and combined gas law
Key Concepts: Boyle's law (constant temperature): $p_1V_1 = p_2V_2$. Pressure–temperature law (constant volume): $p_1/T_1 = p_2/T_2$. Combined: $p_1V_1/T_1 = p_2V_2/T_2$. Always use temperature in kelvin ($T = \theta + 273$).

Section A — Boyle's Law (Constant Temperature)

1. State Boyle's law in words. [2]
2. A gas has volume $1.5\,\text{dm}^3$ at a pressure of 120 kPa. Calculate its volume when the pressure increases to 200 kPa at constant temperature. [2]
3. A gas occupies $4.0\,\text{m}^3$ at 100 kPa. Calculate the pressure when the gas is compressed to $0.8\,\text{m}^3$ at constant temperature. [2]
4. A syringe contains $60\,\text{cm}^3$ of gas at atmospheric pressure (101 kPa). The plunger is pushed in until the pressure is 150 kPa. Calculate the new volume. [2]

Section B — Pressure–Temperature Law (Constant Volume)

5. State the pressure–temperature law in words. [2]
6. A gas at 300 K has a pressure of 90 kPa. Calculate the pressure when the temperature increases to 360 K at constant volume. [2]
7. A sealed container of gas has pressure 120 kPa at 27 °C. Calculate the pressure at 127 °C. [3]
8. A gas has pressure 200 kPa at 400 K. At what temperature will the pressure be 150 kPa? [3]

Section C — Combined Gas Law

9. Write the combined gas law equation. [1]
10. A gas occupies $2.0\,\text{m}^3$ at 200 kPa and 300 K. Calculate the volume at 400 kPa and 600 K. [3]

Total marks: 22

Mark Scheme

1. At constant temperature [1], the pressure of a fixed mass of gas is inversely proportional to its volume [1] [2]
2. $p_1V_1 = p_2V_2$; $120 \times 1.5 = 200 \times V_2$ [1]; $V_2 = 180/200 = 0.9\,\text{dm}^3$ [1] [2]
3. $p_2 = p_1V_1/V_2 = 100 \times 4.0/0.8 = 500\,\text{kPa}$ [2]
4. $V_2 = p_1V_1/p_2 = 101 \times 60/150 = 40.4\,\text{cm}^3$ [2]
5. At constant volume [1], the pressure of a fixed mass of gas is directly proportional to its absolute (kelvin) temperature [1] [2]
6. $p_2 = p_1T_2/T_1 = 90 \times 360/300 = 108\,\text{kPa}$ [2]
7. Convert: 27 °C = 300 K; 127 °C = 400 K [1]; $p_2 = 120 \times 400/300 = 160\,\text{kPa}$ [2] [3]
8. $T_2 = T_1 \times p_2/p_1 = 400 \times 150/200 = 300\,\text{K}$ [2]; convert if required: 27 °C [1] [3]
9. $p_1V_1/T_1 = p_2V_2/T_2$ [1]
10. $V_2 = p_1V_1T_2/(T_1p_2) = (200 \times 2.0 \times 600)/(300 \times 400)$ [1] $= 240\,000/120\,000 = 2.0\,\text{m}^3$ [2] [3]