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ScienceAges 13-15

Boyle’s law & gas compression

Apply Boyle’s law (P₁V₁ = P₂V₂) to calculate the pressure of an ideal gas after isothermal compression. Review suitability for your learners before teaching.

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Worksheet StudioPractice pack · Ages 13-15

Boyle’s law & gas compression

Take your time. Show your thinking.

  1. 1. A sealed cylinder contains 84 dm³ of ideal gas at an initial pressure of 245 kPa. The gas is compressed at constant temperature to a volume of 12 dm³. Using Boyle’s law (P₁V₁ = P₂V₂), calculate the final pressure.

    P₁ = 245 kPa, V₁ = 84 dm³ · Compressed to V₂ = 12 dm³ → P₂ = ?

  2. 2. A sealed cylinder contains 4 dm³ of ideal gas at an initial pressure of 280 kPa. The gas is compressed at constant temperature to a volume of 2 dm³. Using Boyle’s law (P₁V₁ = P₂V₂), calculate the final pressure.

    P₁ = 280 kPa, V₁ = 4 dm³ · Compressed to V₂ = 2 dm³ → P₂ = ?

  3. 3. A sealed cylinder contains 12 dm³ of ideal gas at an initial pressure of 140 kPa. The gas is compressed at constant temperature to a volume of 4 dm³. Using Boyle’s law (P₁V₁ = P₂V₂), calculate the final pressure.

    P₁ = 140 kPa, V₁ = 12 dm³ · Compressed to V₂ = 4 dm³ → P₂ = ?

  4. 4. A sealed cylinder contains 52 dm³ of ideal gas at an initial pressure of 205 kPa. The gas is compressed at constant temperature to a volume of 13 dm³. Using Boyle’s law (P₁V₁ = P₂V₂), calculate the final pressure.

    P₁ = 205 kPa, V₁ = 52 dm³ · Compressed to V₂ = 13 dm³ → P₂ = ?

  5. 5. A sealed cylinder contains 15 dm³ of ideal gas at an initial pressure of 180 kPa. The gas is compressed at constant temperature to a volume of 5 dm³. Using Boyle’s law (P₁V₁ = P₂V₂), calculate the final pressure.

    P₁ = 180 kPa, V₁ = 15 dm³ · Compressed to V₂ = 5 dm³ → P₂ = ?

  6. 6. A sealed cylinder contains 30 dm³ of ideal gas at an initial pressure of 220 kPa. The gas is compressed at constant temperature to a volume of 10 dm³. Using Boyle’s law (P₁V₁ = P₂V₂), calculate the final pressure.

    P₁ = 220 kPa, V₁ = 30 dm³ · Compressed to V₂ = 10 dm³ → P₂ = ?

  7. 7. A sealed cylinder contains 40 dm³ of ideal gas at an initial pressure of 210 kPa. The gas is compressed at constant temperature to a volume of 8 dm³. Using Boyle’s law (P₁V₁ = P₂V₂), calculate the final pressure.

    P₁ = 210 kPa, V₁ = 40 dm³ · Compressed to V₂ = 8 dm³ → P₂ = ?

  8. 8. A sealed cylinder contains 64 dm³ of ideal gas at an initial pressure of 175 kPa. The gas is compressed at constant temperature to a volume of 8 dm³. Using Boyle’s law (P₁V₁ = P₂V₂), calculate the final pressure.

    P₁ = 175 kPa, V₁ = 64 dm³ · Compressed to V₂ = 8 dm³ → P₂ = ?

  9. 9. A sealed cylinder contains 18 dm³ of ideal gas at an initial pressure of 285 kPa. The gas is compressed at constant temperature to a volume of 3 dm³. Using Boyle’s law (P₁V₁ = P₂V₂), calculate the final pressure.

    P₁ = 285 kPa, V₁ = 18 dm³ · Compressed to V₂ = 3 dm³ → P₂ = ?

  10. 10. A sealed cylinder contains 30 dm³ of ideal gas at an initial pressure of 60 kPa. The gas is compressed at constant temperature to a volume of 10 dm³. Using Boyle’s law (P₁V₁ = P₂V₂), calculate the final pressure.

    P₁ = 60 kPa, V₁ = 30 dm³ · Compressed to V₂ = 10 dm³ → P₂ = ?

  11. 11. A sealed cylinder contains 8 dm³ of ideal gas at an initial pressure of 220 kPa. The gas is compressed at constant temperature to a volume of 4 dm³. Using Boyle’s law (P₁V₁ = P₂V₂), calculate the final pressure.

    P₁ = 220 kPa, V₁ = 8 dm³ · Compressed to V₂ = 4 dm³ → P₂ = ?

  12. 12. A sealed cylinder contains 56 dm³ of ideal gas at an initial pressure of 170 kPa. The gas is compressed at constant temperature to a volume of 7 dm³. Using Boyle’s law (P₁V₁ = P₂V₂), calculate the final pressure.

    P₁ = 170 kPa, V₁ = 56 dm³ · Compressed to V₂ = 7 dm³ → P₂ = ?
12 questions · Seed 104729 · Age guidance, not curriculum certification
View the answer key
Worksheet StudioAnswer key · Ages 13-15

Boyle’s law & gas compression

Answers and reasoning. Review before teaching.

  1. 1. A sealed cylinder contains 84 dm³ of ideal gas at an initial pressure of 245 kPa. The gas is compressed at constant temperature to a volume of 12 dm³. Using Boyle’s law (P₁V₁ = P₂V₂), calculate the final pressure.

    1715 kPa

    By Boyle’s law at constant temperature, P₂ = (P₁ × V₁) ÷ V₂ = (245 kPa × 84 dm³) ÷ 12 dm³ = 1715 kPa.

  2. 2. A sealed cylinder contains 4 dm³ of ideal gas at an initial pressure of 280 kPa. The gas is compressed at constant temperature to a volume of 2 dm³. Using Boyle’s law (P₁V₁ = P₂V₂), calculate the final pressure.

    560 kPa

    By Boyle’s law at constant temperature, P₂ = (P₁ × V₁) ÷ V₂ = (280 kPa × 4 dm³) ÷ 2 dm³ = 560 kPa.

  3. 3. A sealed cylinder contains 12 dm³ of ideal gas at an initial pressure of 140 kPa. The gas is compressed at constant temperature to a volume of 4 dm³. Using Boyle’s law (P₁V₁ = P₂V₂), calculate the final pressure.

    420 kPa

    By Boyle’s law at constant temperature, P₂ = (P₁ × V₁) ÷ V₂ = (140 kPa × 12 dm³) ÷ 4 dm³ = 420 kPa.

  4. 4. A sealed cylinder contains 52 dm³ of ideal gas at an initial pressure of 205 kPa. The gas is compressed at constant temperature to a volume of 13 dm³. Using Boyle’s law (P₁V₁ = P₂V₂), calculate the final pressure.

    820 kPa

    By Boyle’s law at constant temperature, P₂ = (P₁ × V₁) ÷ V₂ = (205 kPa × 52 dm³) ÷ 13 dm³ = 820 kPa.

  5. 5. A sealed cylinder contains 15 dm³ of ideal gas at an initial pressure of 180 kPa. The gas is compressed at constant temperature to a volume of 5 dm³. Using Boyle’s law (P₁V₁ = P₂V₂), calculate the final pressure.

    540 kPa

    By Boyle’s law at constant temperature, P₂ = (P₁ × V₁) ÷ V₂ = (180 kPa × 15 dm³) ÷ 5 dm³ = 540 kPa.

  6. 6. A sealed cylinder contains 30 dm³ of ideal gas at an initial pressure of 220 kPa. The gas is compressed at constant temperature to a volume of 10 dm³. Using Boyle’s law (P₁V₁ = P₂V₂), calculate the final pressure.

    660 kPa

    By Boyle’s law at constant temperature, P₂ = (P₁ × V₁) ÷ V₂ = (220 kPa × 30 dm³) ÷ 10 dm³ = 660 kPa.

  7. 7. A sealed cylinder contains 40 dm³ of ideal gas at an initial pressure of 210 kPa. The gas is compressed at constant temperature to a volume of 8 dm³. Using Boyle’s law (P₁V₁ = P₂V₂), calculate the final pressure.

    1050 kPa

    By Boyle’s law at constant temperature, P₂ = (P₁ × V₁) ÷ V₂ = (210 kPa × 40 dm³) ÷ 8 dm³ = 1050 kPa.

  8. 8. A sealed cylinder contains 64 dm³ of ideal gas at an initial pressure of 175 kPa. The gas is compressed at constant temperature to a volume of 8 dm³. Using Boyle’s law (P₁V₁ = P₂V₂), calculate the final pressure.

    1400 kPa

    By Boyle’s law at constant temperature, P₂ = (P₁ × V₁) ÷ V₂ = (175 kPa × 64 dm³) ÷ 8 dm³ = 1400 kPa.

  9. 9. A sealed cylinder contains 18 dm³ of ideal gas at an initial pressure of 285 kPa. The gas is compressed at constant temperature to a volume of 3 dm³. Using Boyle’s law (P₁V₁ = P₂V₂), calculate the final pressure.

    1710 kPa

    By Boyle’s law at constant temperature, P₂ = (P₁ × V₁) ÷ V₂ = (285 kPa × 18 dm³) ÷ 3 dm³ = 1710 kPa.

  10. 10. A sealed cylinder contains 30 dm³ of ideal gas at an initial pressure of 60 kPa. The gas is compressed at constant temperature to a volume of 10 dm³. Using Boyle’s law (P₁V₁ = P₂V₂), calculate the final pressure.

    180 kPa

    By Boyle’s law at constant temperature, P₂ = (P₁ × V₁) ÷ V₂ = (60 kPa × 30 dm³) ÷ 10 dm³ = 180 kPa.

  11. 11. A sealed cylinder contains 8 dm³ of ideal gas at an initial pressure of 220 kPa. The gas is compressed at constant temperature to a volume of 4 dm³. Using Boyle’s law (P₁V₁ = P₂V₂), calculate the final pressure.

    440 kPa

    By Boyle’s law at constant temperature, P₂ = (P₁ × V₁) ÷ V₂ = (220 kPa × 8 dm³) ÷ 4 dm³ = 440 kPa.

  12. 12. A sealed cylinder contains 56 dm³ of ideal gas at an initial pressure of 170 kPa. The gas is compressed at constant temperature to a volume of 7 dm³. Using Boyle’s law (P₁V₁ = P₂V₂), calculate the final pressure.

    1360 kPa

    By Boyle’s law at constant temperature, P₂ = (P₁ × V₁) ÷ V₂ = (170 kPa × 56 dm³) ÷ 7 dm³ = 1360 kPa.

12 questions · Seed 104729 · Age guidance, not curriculum certification
Worksheet StudioPractice pack · Ages 13-15

Boyle’s law & gas compression

Take your time. Show your thinking.

  1. 1. A sealed cylinder contains 84 dm³ of ideal gas at an initial pressure of 245 kPa. The gas is compressed at constant temperature to a volume of 12 dm³. Using Boyle’s law (P₁V₁ = P₂V₂), calculate the final pressure.

    P₁ = 245 kPa, V₁ = 84 dm³ · Compressed to V₂ = 12 dm³ → P₂ = ?

  2. 2. A sealed cylinder contains 4 dm³ of ideal gas at an initial pressure of 280 kPa. The gas is compressed at constant temperature to a volume of 2 dm³. Using Boyle’s law (P₁V₁ = P₂V₂), calculate the final pressure.

    P₁ = 280 kPa, V₁ = 4 dm³ · Compressed to V₂ = 2 dm³ → P₂ = ?

  3. 3. A sealed cylinder contains 12 dm³ of ideal gas at an initial pressure of 140 kPa. The gas is compressed at constant temperature to a volume of 4 dm³. Using Boyle’s law (P₁V₁ = P₂V₂), calculate the final pressure.

    P₁ = 140 kPa, V₁ = 12 dm³ · Compressed to V₂ = 4 dm³ → P₂ = ?

  4. 4. A sealed cylinder contains 52 dm³ of ideal gas at an initial pressure of 205 kPa. The gas is compressed at constant temperature to a volume of 13 dm³. Using Boyle’s law (P₁V₁ = P₂V₂), calculate the final pressure.

    P₁ = 205 kPa, V₁ = 52 dm³ · Compressed to V₂ = 13 dm³ → P₂ = ?

  5. 5. A sealed cylinder contains 15 dm³ of ideal gas at an initial pressure of 180 kPa. The gas is compressed at constant temperature to a volume of 5 dm³. Using Boyle’s law (P₁V₁ = P₂V₂), calculate the final pressure.

    P₁ = 180 kPa, V₁ = 15 dm³ · Compressed to V₂ = 5 dm³ → P₂ = ?

  6. 6. A sealed cylinder contains 30 dm³ of ideal gas at an initial pressure of 220 kPa. The gas is compressed at constant temperature to a volume of 10 dm³. Using Boyle’s law (P₁V₁ = P₂V₂), calculate the final pressure.

    P₁ = 220 kPa, V₁ = 30 dm³ · Compressed to V₂ = 10 dm³ → P₂ = ?

  7. 7. A sealed cylinder contains 40 dm³ of ideal gas at an initial pressure of 210 kPa. The gas is compressed at constant temperature to a volume of 8 dm³. Using Boyle’s law (P₁V₁ = P₂V₂), calculate the final pressure.

    P₁ = 210 kPa, V₁ = 40 dm³ · Compressed to V₂ = 8 dm³ → P₂ = ?

  8. 8. A sealed cylinder contains 64 dm³ of ideal gas at an initial pressure of 175 kPa. The gas is compressed at constant temperature to a volume of 8 dm³. Using Boyle’s law (P₁V₁ = P₂V₂), calculate the final pressure.

    P₁ = 175 kPa, V₁ = 64 dm³ · Compressed to V₂ = 8 dm³ → P₂ = ?

  9. 9. A sealed cylinder contains 18 dm³ of ideal gas at an initial pressure of 285 kPa. The gas is compressed at constant temperature to a volume of 3 dm³. Using Boyle’s law (P₁V₁ = P₂V₂), calculate the final pressure.

    P₁ = 285 kPa, V₁ = 18 dm³ · Compressed to V₂ = 3 dm³ → P₂ = ?

  10. 10. A sealed cylinder contains 30 dm³ of ideal gas at an initial pressure of 60 kPa. The gas is compressed at constant temperature to a volume of 10 dm³. Using Boyle’s law (P₁V₁ = P₂V₂), calculate the final pressure.

    P₁ = 60 kPa, V₁ = 30 dm³ · Compressed to V₂ = 10 dm³ → P₂ = ?

  11. 11. A sealed cylinder contains 8 dm³ of ideal gas at an initial pressure of 220 kPa. The gas is compressed at constant temperature to a volume of 4 dm³. Using Boyle’s law (P₁V₁ = P₂V₂), calculate the final pressure.

    P₁ = 220 kPa, V₁ = 8 dm³ · Compressed to V₂ = 4 dm³ → P₂ = ?

  12. 12. A sealed cylinder contains 56 dm³ of ideal gas at an initial pressure of 170 kPa. The gas is compressed at constant temperature to a volume of 7 dm³. Using Boyle’s law (P₁V₁ = P₂V₂), calculate the final pressure.

    P₁ = 170 kPa, V₁ = 56 dm³ · Compressed to V₂ = 7 dm³ → P₂ = ?
12 questions · Seed 104729 · Age guidance, not curriculum certification
Worksheet StudioAnswer key · Ages 13-15

Boyle’s law & gas compression

Answers and reasoning. Review before teaching.

  1. 1. A sealed cylinder contains 84 dm³ of ideal gas at an initial pressure of 245 kPa. The gas is compressed at constant temperature to a volume of 12 dm³. Using Boyle’s law (P₁V₁ = P₂V₂), calculate the final pressure.

    1715 kPa

    By Boyle’s law at constant temperature, P₂ = (P₁ × V₁) ÷ V₂ = (245 kPa × 84 dm³) ÷ 12 dm³ = 1715 kPa.

  2. 2. A sealed cylinder contains 4 dm³ of ideal gas at an initial pressure of 280 kPa. The gas is compressed at constant temperature to a volume of 2 dm³. Using Boyle’s law (P₁V₁ = P₂V₂), calculate the final pressure.

    560 kPa

    By Boyle’s law at constant temperature, P₂ = (P₁ × V₁) ÷ V₂ = (280 kPa × 4 dm³) ÷ 2 dm³ = 560 kPa.

  3. 3. A sealed cylinder contains 12 dm³ of ideal gas at an initial pressure of 140 kPa. The gas is compressed at constant temperature to a volume of 4 dm³. Using Boyle’s law (P₁V₁ = P₂V₂), calculate the final pressure.

    420 kPa

    By Boyle’s law at constant temperature, P₂ = (P₁ × V₁) ÷ V₂ = (140 kPa × 12 dm³) ÷ 4 dm³ = 420 kPa.

  4. 4. A sealed cylinder contains 52 dm³ of ideal gas at an initial pressure of 205 kPa. The gas is compressed at constant temperature to a volume of 13 dm³. Using Boyle’s law (P₁V₁ = P₂V₂), calculate the final pressure.

    820 kPa

    By Boyle’s law at constant temperature, P₂ = (P₁ × V₁) ÷ V₂ = (205 kPa × 52 dm³) ÷ 13 dm³ = 820 kPa.

  5. 5. A sealed cylinder contains 15 dm³ of ideal gas at an initial pressure of 180 kPa. The gas is compressed at constant temperature to a volume of 5 dm³. Using Boyle’s law (P₁V₁ = P₂V₂), calculate the final pressure.

    540 kPa

    By Boyle’s law at constant temperature, P₂ = (P₁ × V₁) ÷ V₂ = (180 kPa × 15 dm³) ÷ 5 dm³ = 540 kPa.

  6. 6. A sealed cylinder contains 30 dm³ of ideal gas at an initial pressure of 220 kPa. The gas is compressed at constant temperature to a volume of 10 dm³. Using Boyle’s law (P₁V₁ = P₂V₂), calculate the final pressure.

    660 kPa

    By Boyle’s law at constant temperature, P₂ = (P₁ × V₁) ÷ V₂ = (220 kPa × 30 dm³) ÷ 10 dm³ = 660 kPa.

  7. 7. A sealed cylinder contains 40 dm³ of ideal gas at an initial pressure of 210 kPa. The gas is compressed at constant temperature to a volume of 8 dm³. Using Boyle’s law (P₁V₁ = P₂V₂), calculate the final pressure.

    1050 kPa

    By Boyle’s law at constant temperature, P₂ = (P₁ × V₁) ÷ V₂ = (210 kPa × 40 dm³) ÷ 8 dm³ = 1050 kPa.

  8. 8. A sealed cylinder contains 64 dm³ of ideal gas at an initial pressure of 175 kPa. The gas is compressed at constant temperature to a volume of 8 dm³. Using Boyle’s law (P₁V₁ = P₂V₂), calculate the final pressure.

    1400 kPa

    By Boyle’s law at constant temperature, P₂ = (P₁ × V₁) ÷ V₂ = (175 kPa × 64 dm³) ÷ 8 dm³ = 1400 kPa.

  9. 9. A sealed cylinder contains 18 dm³ of ideal gas at an initial pressure of 285 kPa. The gas is compressed at constant temperature to a volume of 3 dm³. Using Boyle’s law (P₁V₁ = P₂V₂), calculate the final pressure.

    1710 kPa

    By Boyle’s law at constant temperature, P₂ = (P₁ × V₁) ÷ V₂ = (285 kPa × 18 dm³) ÷ 3 dm³ = 1710 kPa.

  10. 10. A sealed cylinder contains 30 dm³ of ideal gas at an initial pressure of 60 kPa. The gas is compressed at constant temperature to a volume of 10 dm³. Using Boyle’s law (P₁V₁ = P₂V₂), calculate the final pressure.

    180 kPa

    By Boyle’s law at constant temperature, P₂ = (P₁ × V₁) ÷ V₂ = (60 kPa × 30 dm³) ÷ 10 dm³ = 180 kPa.

  11. 11. A sealed cylinder contains 8 dm³ of ideal gas at an initial pressure of 220 kPa. The gas is compressed at constant temperature to a volume of 4 dm³. Using Boyle’s law (P₁V₁ = P₂V₂), calculate the final pressure.

    440 kPa

    By Boyle’s law at constant temperature, P₂ = (P₁ × V₁) ÷ V₂ = (220 kPa × 8 dm³) ÷ 4 dm³ = 440 kPa.

  12. 12. A sealed cylinder contains 56 dm³ of ideal gas at an initial pressure of 170 kPa. The gas is compressed at constant temperature to a volume of 7 dm³. Using Boyle’s law (P₁V₁ = P₂V₂), calculate the final pressure.

    1360 kPa

    By Boyle’s law at constant temperature, P₂ = (P₁ × V₁) ÷ V₂ = (170 kPa × 56 dm³) ÷ 7 dm³ = 1360 kPa.

12 questions · Seed 104729 · Age guidance, not curriculum certification