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. A sealed cylinder contains 104 dm³ of ideal gas at an initial pressure of 120 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 = 120 kPa, V₁ = 104 dm³ · Compressed to V₂ = 13 dm³ → P₂ = ?
2. A sealed cylinder contains 30 dm³ of ideal gas at an initial pressure of 105 kPa. The gas is compressed at constant temperature to a volume of 15 dm³. Using Boyle’s law (P₁V₁ = P₂V₂), calculate the final pressure.
P₁ = 105 kPa, V₁ = 30 dm³ · Compressed to V₂ = 15 dm³ → P₂ = ?
3. A sealed cylinder contains 18 dm³ of ideal gas at an initial pressure of 180 kPa. The gas is compressed at constant temperature to a volume of 9 dm³. Using Boyle’s law (P₁V₁ = P₂V₂), calculate the final pressure.
P₁ = 180 kPa, V₁ = 18 dm³ · Compressed to V₂ = 9 dm³ → P₂ = ?
4. A sealed cylinder contains 54 dm³ of ideal gas at an initial pressure of 145 kPa. The gas is compressed at constant temperature to a volume of 9 dm³. Using Boyle’s law (P₁V₁ = P₂V₂), calculate the final pressure.
P₁ = 145 kPa, V₁ = 54 dm³ · Compressed to V₂ = 9 dm³ → P₂ = ?
5. A sealed cylinder contains 90 dm³ of ideal gas at an initial pressure of 135 kPa. The gas is compressed at constant temperature to a volume of 15 dm³. Using Boyle’s law (P₁V₁ = P₂V₂), calculate the final pressure.
P₁ = 135 kPa, V₁ = 90 dm³ · Compressed to V₂ = 15 dm³ → P₂ = ?
6. A sealed cylinder contains 28 dm³ of ideal gas at an initial pressure of 190 kPa. The gas is compressed at constant temperature to a volume of 14 dm³. Using Boyle’s law (P₁V₁ = P₂V₂), calculate the final pressure.
P₁ = 190 kPa, V₁ = 28 dm³ · Compressed to V₂ = 14 dm³ → P₂ = ?
7. A sealed cylinder contains 104 dm³ of ideal gas at an initial pressure of 235 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 = 235 kPa, V₁ = 104 dm³ · Compressed to V₂ = 13 dm³ → P₂ = ?
8. A sealed cylinder contains 26 dm³ of ideal gas at an initial pressure of 80 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 = 80 kPa, V₁ = 26 dm³ · Compressed to V₂ = 13 dm³ → P₂ = ?
9. A sealed cylinder contains 18 dm³ of ideal gas at an initial pressure of 170 kPa. The gas is compressed at constant temperature to a volume of 6 dm³. Using Boyle’s law (P₁V₁ = P₂V₂), calculate the final pressure.
P₁ = 170 kPa, V₁ = 18 dm³ · Compressed to V₂ = 6 dm³ → P₂ = ?
10. A sealed cylinder contains 48 dm³ of ideal gas at an initial pressure of 220 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 = 220 kPa, V₁ = 48 dm³ · Compressed to V₂ = 12 dm³ → P₂ = ?
11. A sealed cylinder contains 104 dm³ of ideal gas at an initial pressure of 150 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 = 150 kPa, V₁ = 104 dm³ · Compressed to V₂ = 13 dm³ → P₂ = ?
12. A sealed cylinder contains 72 dm³ of ideal gas at an initial pressure of 55 kPa. The gas is compressed at constant temperature to a volume of 9 dm³. Using Boyle’s law (P₁V₁ = P₂V₂), calculate the final pressure.
P₁ = 55 kPa, V₁ = 72 dm³ · Compressed to V₂ = 9 dm³ → P₂ = ?

12 questions · Seed 42 · 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. A sealed cylinder contains 104 dm³ of ideal gas at an initial pressure of 120 kPa. The gas is compressed at constant temperature to a volume of 13 dm³. Using Boyle’s law (P₁V₁ = P₂V₂), calculate the final pressure.
960 kPa
By Boyle’s law at constant temperature, P₂ = (P₁ × V₁) ÷ V₂ = (120 kPa × 104 dm³) ÷ 13 dm³ = 960 kPa.
2. A sealed cylinder contains 30 dm³ of ideal gas at an initial pressure of 105 kPa. The gas is compressed at constant temperature to a volume of 15 dm³. Using Boyle’s law (P₁V₁ = P₂V₂), calculate the final pressure.
210 kPa
By Boyle’s law at constant temperature, P₂ = (P₁ × V₁) ÷ V₂ = (105 kPa × 30 dm³) ÷ 15 dm³ = 210 kPa.
3. A sealed cylinder contains 18 dm³ of ideal gas at an initial pressure of 180 kPa. The gas is compressed at constant temperature to a volume of 9 dm³. Using Boyle’s law (P₁V₁ = P₂V₂), calculate the final pressure.
360 kPa
By Boyle’s law at constant temperature, P₂ = (P₁ × V₁) ÷ V₂ = (180 kPa × 18 dm³) ÷ 9 dm³ = 360 kPa.
4. A sealed cylinder contains 54 dm³ of ideal gas at an initial pressure of 145 kPa. The gas is compressed at constant temperature to a volume of 9 dm³. Using Boyle’s law (P₁V₁ = P₂V₂), calculate the final pressure.
870 kPa
By Boyle’s law at constant temperature, P₂ = (P₁ × V₁) ÷ V₂ = (145 kPa × 54 dm³) ÷ 9 dm³ = 870 kPa.
5. A sealed cylinder contains 90 dm³ of ideal gas at an initial pressure of 135 kPa. The gas is compressed at constant temperature to a volume of 15 dm³. Using Boyle’s law (P₁V₁ = P₂V₂), calculate the final pressure.
810 kPa
By Boyle’s law at constant temperature, P₂ = (P₁ × V₁) ÷ V₂ = (135 kPa × 90 dm³) ÷ 15 dm³ = 810 kPa.
6. A sealed cylinder contains 28 dm³ of ideal gas at an initial pressure of 190 kPa. The gas is compressed at constant temperature to a volume of 14 dm³. Using Boyle’s law (P₁V₁ = P₂V₂), calculate the final pressure.
380 kPa
By Boyle’s law at constant temperature, P₂ = (P₁ × V₁) ÷ V₂ = (190 kPa × 28 dm³) ÷ 14 dm³ = 380 kPa.
7. A sealed cylinder contains 104 dm³ of ideal gas at an initial pressure of 235 kPa. The gas is compressed at constant temperature to a volume of 13 dm³. Using Boyle’s law (P₁V₁ = P₂V₂), calculate the final pressure.
1880 kPa
By Boyle’s law at constant temperature, P₂ = (P₁ × V₁) ÷ V₂ = (235 kPa × 104 dm³) ÷ 13 dm³ = 1880 kPa.
8. A sealed cylinder contains 26 dm³ of ideal gas at an initial pressure of 80 kPa. The gas is compressed at constant temperature to a volume of 13 dm³. Using Boyle’s law (P₁V₁ = P₂V₂), calculate the final pressure.
160 kPa
By Boyle’s law at constant temperature, P₂ = (P₁ × V₁) ÷ V₂ = (80 kPa × 26 dm³) ÷ 13 dm³ = 160 kPa.
9. A sealed cylinder contains 18 dm³ of ideal gas at an initial pressure of 170 kPa. The gas is compressed at constant temperature to a volume of 6 dm³. Using Boyle’s law (P₁V₁ = P₂V₂), calculate the final pressure.
510 kPa
By Boyle’s law at constant temperature, P₂ = (P₁ × V₁) ÷ V₂ = (170 kPa × 18 dm³) ÷ 6 dm³ = 510 kPa.
10. A sealed cylinder contains 48 dm³ of ideal gas at an initial pressure of 220 kPa. The gas is compressed at constant temperature to a volume of 12 dm³. Using Boyle’s law (P₁V₁ = P₂V₂), calculate the final pressure.
880 kPa
By Boyle’s law at constant temperature, P₂ = (P₁ × V₁) ÷ V₂ = (220 kPa × 48 dm³) ÷ 12 dm³ = 880 kPa.
11. A sealed cylinder contains 104 dm³ of ideal gas at an initial pressure of 150 kPa. The gas is compressed at constant temperature to a volume of 13 dm³. Using Boyle’s law (P₁V₁ = P₂V₂), calculate the final pressure.
1200 kPa
By Boyle’s law at constant temperature, P₂ = (P₁ × V₁) ÷ V₂ = (150 kPa × 104 dm³) ÷ 13 dm³ = 1200 kPa.
12. A sealed cylinder contains 72 dm³ of ideal gas at an initial pressure of 55 kPa. The gas is compressed at constant temperature to a volume of 9 dm³. Using Boyle’s law (P₁V₁ = P₂V₂), calculate the final pressure.
440 kPa
By Boyle’s law at constant temperature, P₂ = (P₁ × V₁) ÷ V₂ = (55 kPa × 72 dm³) ÷ 9 dm³ = 440 kPa.

12 questions · Seed 42 · Age guidance, not curriculum certification

Worksheet StudioPractice pack · Ages 13-15

Boyle’s law & gas compression

Take your time. Show your thinking.

1. A sealed cylinder contains 104 dm³ of ideal gas at an initial pressure of 120 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 = 120 kPa, V₁ = 104 dm³ · Compressed to V₂ = 13 dm³ → P₂ = ?
2. A sealed cylinder contains 30 dm³ of ideal gas at an initial pressure of 105 kPa. The gas is compressed at constant temperature to a volume of 15 dm³. Using Boyle’s law (P₁V₁ = P₂V₂), calculate the final pressure.
P₁ = 105 kPa, V₁ = 30 dm³ · Compressed to V₂ = 15 dm³ → P₂ = ?
3. A sealed cylinder contains 18 dm³ of ideal gas at an initial pressure of 180 kPa. The gas is compressed at constant temperature to a volume of 9 dm³. Using Boyle’s law (P₁V₁ = P₂V₂), calculate the final pressure.
P₁ = 180 kPa, V₁ = 18 dm³ · Compressed to V₂ = 9 dm³ → P₂ = ?
4. A sealed cylinder contains 54 dm³ of ideal gas at an initial pressure of 145 kPa. The gas is compressed at constant temperature to a volume of 9 dm³. Using Boyle’s law (P₁V₁ = P₂V₂), calculate the final pressure.
P₁ = 145 kPa, V₁ = 54 dm³ · Compressed to V₂ = 9 dm³ → P₂ = ?
5. A sealed cylinder contains 90 dm³ of ideal gas at an initial pressure of 135 kPa. The gas is compressed at constant temperature to a volume of 15 dm³. Using Boyle’s law (P₁V₁ = P₂V₂), calculate the final pressure.
P₁ = 135 kPa, V₁ = 90 dm³ · Compressed to V₂ = 15 dm³ → P₂ = ?
6. A sealed cylinder contains 28 dm³ of ideal gas at an initial pressure of 190 kPa. The gas is compressed at constant temperature to a volume of 14 dm³. Using Boyle’s law (P₁V₁ = P₂V₂), calculate the final pressure.
P₁ = 190 kPa, V₁ = 28 dm³ · Compressed to V₂ = 14 dm³ → P₂ = ?
7. A sealed cylinder contains 104 dm³ of ideal gas at an initial pressure of 235 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 = 235 kPa, V₁ = 104 dm³ · Compressed to V₂ = 13 dm³ → P₂ = ?
8. A sealed cylinder contains 26 dm³ of ideal gas at an initial pressure of 80 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 = 80 kPa, V₁ = 26 dm³ · Compressed to V₂ = 13 dm³ → P₂ = ?
9. A sealed cylinder contains 18 dm³ of ideal gas at an initial pressure of 170 kPa. The gas is compressed at constant temperature to a volume of 6 dm³. Using Boyle’s law (P₁V₁ = P₂V₂), calculate the final pressure.
P₁ = 170 kPa, V₁ = 18 dm³ · Compressed to V₂ = 6 dm³ → P₂ = ?
10. A sealed cylinder contains 48 dm³ of ideal gas at an initial pressure of 220 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 = 220 kPa, V₁ = 48 dm³ · Compressed to V₂ = 12 dm³ → P₂ = ?
11. A sealed cylinder contains 104 dm³ of ideal gas at an initial pressure of 150 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 = 150 kPa, V₁ = 104 dm³ · Compressed to V₂ = 13 dm³ → P₂ = ?
12. A sealed cylinder contains 72 dm³ of ideal gas at an initial pressure of 55 kPa. The gas is compressed at constant temperature to a volume of 9 dm³. Using Boyle’s law (P₁V₁ = P₂V₂), calculate the final pressure.
P₁ = 55 kPa, V₁ = 72 dm³ · Compressed to V₂ = 9 dm³ → P₂ = ?

12 questions · Seed 42 · Age guidance, not curriculum certification

Worksheet StudioAnswer key · Ages 13-15

Boyle’s law & gas compression

Answers and reasoning. Review before teaching.

1. A sealed cylinder contains 104 dm³ of ideal gas at an initial pressure of 120 kPa. The gas is compressed at constant temperature to a volume of 13 dm³. Using Boyle’s law (P₁V₁ = P₂V₂), calculate the final pressure.
960 kPa
By Boyle’s law at constant temperature, P₂ = (P₁ × V₁) ÷ V₂ = (120 kPa × 104 dm³) ÷ 13 dm³ = 960 kPa.
2. A sealed cylinder contains 30 dm³ of ideal gas at an initial pressure of 105 kPa. The gas is compressed at constant temperature to a volume of 15 dm³. Using Boyle’s law (P₁V₁ = P₂V₂), calculate the final pressure.
210 kPa
By Boyle’s law at constant temperature, P₂ = (P₁ × V₁) ÷ V₂ = (105 kPa × 30 dm³) ÷ 15 dm³ = 210 kPa.
3. A sealed cylinder contains 18 dm³ of ideal gas at an initial pressure of 180 kPa. The gas is compressed at constant temperature to a volume of 9 dm³. Using Boyle’s law (P₁V₁ = P₂V₂), calculate the final pressure.
360 kPa
By Boyle’s law at constant temperature, P₂ = (P₁ × V₁) ÷ V₂ = (180 kPa × 18 dm³) ÷ 9 dm³ = 360 kPa.
4. A sealed cylinder contains 54 dm³ of ideal gas at an initial pressure of 145 kPa. The gas is compressed at constant temperature to a volume of 9 dm³. Using Boyle’s law (P₁V₁ = P₂V₂), calculate the final pressure.
870 kPa
By Boyle’s law at constant temperature, P₂ = (P₁ × V₁) ÷ V₂ = (145 kPa × 54 dm³) ÷ 9 dm³ = 870 kPa.
5. A sealed cylinder contains 90 dm³ of ideal gas at an initial pressure of 135 kPa. The gas is compressed at constant temperature to a volume of 15 dm³. Using Boyle’s law (P₁V₁ = P₂V₂), calculate the final pressure.
810 kPa
By Boyle’s law at constant temperature, P₂ = (P₁ × V₁) ÷ V₂ = (135 kPa × 90 dm³) ÷ 15 dm³ = 810 kPa.
6. A sealed cylinder contains 28 dm³ of ideal gas at an initial pressure of 190 kPa. The gas is compressed at constant temperature to a volume of 14 dm³. Using Boyle’s law (P₁V₁ = P₂V₂), calculate the final pressure.
380 kPa
By Boyle’s law at constant temperature, P₂ = (P₁ × V₁) ÷ V₂ = (190 kPa × 28 dm³) ÷ 14 dm³ = 380 kPa.
7. A sealed cylinder contains 104 dm³ of ideal gas at an initial pressure of 235 kPa. The gas is compressed at constant temperature to a volume of 13 dm³. Using Boyle’s law (P₁V₁ = P₂V₂), calculate the final pressure.
1880 kPa
By Boyle’s law at constant temperature, P₂ = (P₁ × V₁) ÷ V₂ = (235 kPa × 104 dm³) ÷ 13 dm³ = 1880 kPa.
8. A sealed cylinder contains 26 dm³ of ideal gas at an initial pressure of 80 kPa. The gas is compressed at constant temperature to a volume of 13 dm³. Using Boyle’s law (P₁V₁ = P₂V₂), calculate the final pressure.
160 kPa
By Boyle’s law at constant temperature, P₂ = (P₁ × V₁) ÷ V₂ = (80 kPa × 26 dm³) ÷ 13 dm³ = 160 kPa.
9. A sealed cylinder contains 18 dm³ of ideal gas at an initial pressure of 170 kPa. The gas is compressed at constant temperature to a volume of 6 dm³. Using Boyle’s law (P₁V₁ = P₂V₂), calculate the final pressure.
510 kPa
By Boyle’s law at constant temperature, P₂ = (P₁ × V₁) ÷ V₂ = (170 kPa × 18 dm³) ÷ 6 dm³ = 510 kPa.
10. A sealed cylinder contains 48 dm³ of ideal gas at an initial pressure of 220 kPa. The gas is compressed at constant temperature to a volume of 12 dm³. Using Boyle’s law (P₁V₁ = P₂V₂), calculate the final pressure.
880 kPa
By Boyle’s law at constant temperature, P₂ = (P₁ × V₁) ÷ V₂ = (220 kPa × 48 dm³) ÷ 12 dm³ = 880 kPa.
11. A sealed cylinder contains 104 dm³ of ideal gas at an initial pressure of 150 kPa. The gas is compressed at constant temperature to a volume of 13 dm³. Using Boyle’s law (P₁V₁ = P₂V₂), calculate the final pressure.
1200 kPa
By Boyle’s law at constant temperature, P₂ = (P₁ × V₁) ÷ V₂ = (150 kPa × 104 dm³) ÷ 13 dm³ = 1200 kPa.
12. A sealed cylinder contains 72 dm³ of ideal gas at an initial pressure of 55 kPa. The gas is compressed at constant temperature to a volume of 9 dm³. Using Boyle’s law (P₁V₁ = P₂V₂), calculate the final pressure.
440 kPa
By Boyle’s law at constant temperature, P₂ = (P₁ × V₁) ÷ V₂ = (55 kPa × 72 dm³) ÷ 9 dm³ = 440 kPa.

12 questions · Seed 42 · Age guidance, not curriculum certification