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Combined Gas Law worksheet

Grades 10–11 · CCSS HSA.CED.A.4

Solve each problem and show your work. Round answers to 2 decimal places where needed.

1.
Problem 1.
- Solve for: Initial volume (V₁)
- Pressure unit (both pressures): Pascals (Pa)
- Volume unit (both volumes): Cubic centimeters (cm³)
- Temperature scale (both temperatures): °F
- Initial pressure (P₁): 1
- Initial temperature (T₁): 28
- Final pressure (P₂): 2
- Final volume (V₂): 1
- Final temperature (T₂): 68
Find: Initial volume (V₁)
Answer
2.
Problem 2.
- Solve for: Initial volume (V₁)
- Pressure unit (both pressures): Millimeters of mercury (mmHg)
- Volume unit (both volumes): Cubic meters (m³)
- Temperature scale (both temperatures): °C
- Initial pressure (P₁): 2
- Initial temperature (T₁): 30
- Final pressure (P₂): 2
- Final volume (V₂): 0.9
- Final temperature (T₂): 82
Find: Initial volume (V₁)
Answer
3.
Problem 3.
- Solve for: Initial volume (V₁)
- Pressure unit (both pressures): Inches of mercury (inHg)
- Volume unit (both volumes): Milliliters (mL)
- Temperature scale (both temperatures): °F
- Initial pressure (P₁): 1
- Initial temperature (T₁): 36
- Final pressure (P₂): 3
- Final volume (V₂): 1.5
- Final temperature (T₂): 110
Find: Initial volume (V₁)
Answer
4.
Problem 4.
- Solve for: Initial temperature (T₁)
- Pressure unit (both pressures): Hectopascals / millibars (hPa)
- Volume unit (both volumes): US gallons (US gal)
- Temperature scale (both temperatures): °C
- Initial pressure (P₁): 1
- Initial volume (V₁): 2
- Final pressure (P₂): 3
- Final volume (V₂): 1.2
- Final temperature (T₂): 100
Find: Initial temperature (T₁)
Answer
5.
Problem 5.
- Solve for: Final volume (V₂)
- Pressure unit (both pressures): Pascals (Pa)
- Volume unit (both volumes): Cubic meters (m³)
- Temperature scale (both temperatures): °F
- Initial pressure (P₁): 1
- Initial volume (V₁): 2
- Initial temperature (T₁): 28
- Final pressure (P₂): 2
- Final temperature (T₂): 90
Find: Final volume (V₂)
Answer
6.
Problem 6.
- Solve for: Initial pressure (P₁)
- Pressure unit (both pressures): Millimeters of mercury (mmHg)
- Volume unit (both volumes): Milliliters (mL)
- Temperature scale (both temperatures): °C
- Initial volume (V₁): 2
- Initial temperature (T₁): 32
- Final pressure (P₂): 2
- Final volume (V₂): 0.9
- Final temperature (T₂): 98
Find: Initial pressure (P₁)
Answer
7.
Problem 7.
- Solve for: Final volume (V₂)
- Pressure unit (both pressures): Millimeters of mercury (mmHg)
- Volume unit (both volumes): Cubic meters (m³)
- Temperature scale (both temperatures): °C
- Initial pressure (P₁): 1
- Initial volume (V₁): 2
- Initial temperature (T₁): 26
- Final pressure (P₂): 2
- Final temperature (T₂): 75
Find: Final volume (V₂)
Answer
8.
Problem 8.
- Solve for: Initial volume (V₁)
- Pressure unit (both pressures): Megapascals (MPa)
- Volume unit (both volumes): Cubic meters (m³)
- Temperature scale (both temperatures): K
- Initial pressure (P₁): 1
- Initial temperature (T₁): 20
- Final pressure (P₂): 2
- Final volume (V₂): 1.3
- Final temperature (T₂): 82
Find: Initial volume (V₁)
Answer
9.
Problem 9.
- Solve for: Initial volume (V₁)
- Pressure unit (both pressures): Megapascals (MPa)
- Volume unit (both volumes): Cubic inches (in³)
- Temperature scale (both temperatures): °C
- Initial pressure (P₁): 2
- Initial temperature (T₁): 24
- Final pressure (P₂): 2
- Final volume (V₂): 1.4
- Final temperature (T₂): 78
Find: Initial volume (V₁)
Answer
10.
Problem 10.
- Solve for: Final volume (V₂)
- Pressure unit (both pressures): Pounds per square inch (psi)
- Volume unit (both volumes): Cubic feet (ft³)
- Temperature scale (both temperatures): °C
- Initial pressure (P₁): 1
- Initial volume (V₁): 3
- Initial temperature (T₁): 28
- Final pressure (P₂): 2
- Final temperature (T₂): 70
Find: Final volume (V₂)
Answer

Answer key: Combined Gas Law worksheet

1. Initial volume (V₁): 1.85 cm³
1. Put the initial temperature in kelvins. The law needs a scale that starts at absolute zero, so subtract 32, multiply by 5/9, then add 273.15
  $(28 °F - 32) \times \frac{5}{9} + 273.15$ $= 270.927778 K$
2. Put the final temperature in kelvins. The law needs a scale that starts at absolute zero, so subtract 32, multiply by 5/9, then add 273.15
  $(68 °F - 32) \times \frac{5}{9} + 273.15$ $= 293.15 K$
3. PV ÷ T stays the same before and after. So rearrange P₁V₁ ÷ T₁ = P₂V₂ ÷ T₂ to get the initial volume
  $\frac{2 Pa \times 1 cm³ \times 270.9277777778 K}{293.15 K \times 1 Pa}$ $= 1.8483900923 cm³$
2. Initial volume (V₁): 0.77 m³
1. Put the initial temperature in kelvins. The law needs a scale that starts at absolute zero, so add 273.15
  $30 °C + 273.15$ $= 303.15 K$
2. Put the final temperature in kelvins. The law needs a scale that starts at absolute zero, so add 273.15
  $82 °C + 273.15$ $= 355.15 K$
3. PV ÷ T stays the same before and after. So rearrange P₁V₁ ÷ T₁ = P₂V₂ ÷ T₂ to get the initial volume
  $\frac{2 mmHg \times 0.9 m³ \times 303.15 K}{355.15 K \times 2 mmHg}$ $= 0.7682246938 m³$
3. Initial volume (V₁): 3.92 mL
1. Put the initial temperature in kelvins. The law needs a scale that starts at absolute zero, so subtract 32, multiply by 5/9, then add 273.15
  $(36 °F - 32) \times \frac{5}{9} + 273.15$ $= 275.372222 K$
2. Put the final temperature in kelvins. The law needs a scale that starts at absolute zero, so subtract 32, multiply by 5/9, then add 273.15
  $(110 °F - 32) \times \frac{5}{9} + 273.15$ $= 316.483333 K$
3. PV ÷ T stays the same before and after. So rearrange P₁V₁ ÷ T₁ = P₂V₂ ÷ T₂ to get the initial volume
  $\frac{3 inHg \times 1.5 mL \times 275.3722222222 K}{316.4833333333 K \times 1 inHg}$ $= 3.9154510506 mL$
4. Initial temperature (T₁): -65.84 °C
1. Put the final temperature in kelvins. The law needs a scale that starts at absolute zero, so add 273.15
  $100 °C + 273.15$ $= 373.15 K$
2. PV ÷ T stays the same before and after. So rearrange P₁V₁ ÷ T₁ = P₂V₂ ÷ T₂ to get the initial temperature, in kelvins
  $\frac{1 hPa \times 2 US gal \times 373.15 K}{3 hPa \times 1.2 US gal}$ $= 207.3055555556 K$
3. Change the answer back to Celsius: subtract 273.15
  $207.305556 K - 273.15$ $= −65.844444 °C$
5. Final volume (V₂): 1.13 m³
1. Put the initial temperature in kelvins. The law needs a scale that starts at absolute zero, so subtract 32, multiply by 5/9, then add 273.15
  $(28 °F - 32) \times \frac{5}{9} + 273.15$ $= 270.927778 K$
2. Put the final temperature in kelvins. The law needs a scale that starts at absolute zero, so subtract 32, multiply by 5/9, then add 273.15
  $(90 °F - 32) \times \frac{5}{9} + 273.15$ $= 305.372222 K$
3. PV ÷ T stays the same before and after. So rearrange P₁V₁ ÷ T₁ = P₂V₂ ÷ T₂ to get the final volume
  $\frac{1 Pa \times 2 m³ \times 305.3722222222 K}{270.9277777778 K \times 2 Pa}$ $= 1.1271351529 m³$
6. Initial pressure (P₁): 0.74 mmHg
1. Put the initial temperature in kelvins. The law needs a scale that starts at absolute zero, so add 273.15
  $32 °C + 273.15$ $= 305.15 K$
2. Put the final temperature in kelvins. The law needs a scale that starts at absolute zero, so add 273.15
  $98 °C + 273.15$ $= 371.15 K$
3. PV ÷ T stays the same before and after. So rearrange P₁V₁ ÷ T₁ = P₂V₂ ÷ T₂ to get the initial pressure
  $\frac{2 mmHg \times 0.9 mL \times 305.15 K}{371.15 K \times 2 mL}$ $= 0.7399568907 mmHg$
7. Final volume (V₂): 1.16 m³
1. Put the initial temperature in kelvins. The law needs a scale that starts at absolute zero, so add 273.15
  $26 °C + 273.15$ $= 299.15 K$
2. Put the final temperature in kelvins. The law needs a scale that starts at absolute zero, so add 273.15
  $75 °C + 273.15$ $= 348.15 K$
3. PV ÷ T stays the same before and after. So rearrange P₁V₁ ÷ T₁ = P₂V₂ ÷ T₂ to get the final volume
  $\frac{1 mmHg \times 2 m³ \times 348.15 K}{299.15 K \times 2 mmHg}$ $= 1.163797426 m³$
8. Initial volume (V₁): 0.63 m³
1. PV ÷ T stays the same before and after. So rearrange P₁V₁ ÷ T₁ = P₂V₂ ÷ T₂ to get the initial volume
  $\frac{2 MPa \times 1.3 m³ \times 20 K}{82 K \times 1 MPa}$ $= 0.6341463415 m³$
9. Initial volume (V₁): 1.18 in³
1. Put the initial temperature in kelvins. The law needs a scale that starts at absolute zero, so add 273.15
  $24 °C + 273.15$ $= 297.15 K$
2. Put the final temperature in kelvins. The law needs a scale that starts at absolute zero, so add 273.15
  $78 °C + 273.15$ $= 351.15 K$
3. PV ÷ T stays the same before and after. So rearrange P₁V₁ ÷ T₁ = P₂V₂ ÷ T₂ to get the initial volume
  $\frac{2 MPa \times 1.4 in³ \times 297.15 K}{351.15 K \times 2 MPa}$ $= 1.18470739 in³$
10. Final volume (V₂): 1.71 ft³
1. Put the initial temperature in kelvins. The law needs a scale that starts at absolute zero, so add 273.15
  $28 °C + 273.15$ $= 301.15 K$
2. Put the final temperature in kelvins. The law needs a scale that starts at absolute zero, so add 273.15
  $70 °C + 273.15$ $= 343.15 K$
3. PV ÷ T stays the same before and after. So rearrange P₁V₁ ÷ T₁ = P₂V₂ ÷ T₂ to get the final volume
  $\frac{1 psi \times 3 ft³ \times 343.15 K}{301.15 K \times 2 psi}$ $= 1.709198074 ft³$