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Charles's 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)
- Volume unit (both volumes): Liters (L)
- Temperature scale (both temperatures): K
- Initial temperature (T₁): 32
- Final volume (V₂): 4
- Final temperature (T₂): 85
Find: Initial volume (V₁)
Answer
2.
Problem 2.
- Solve for: Final volume (V₂)
- Volume unit (both volumes): Cubic centimeters (cm³)
- Temperature scale (both temperatures): °C
- Initial volume (V₁): 2
- Initial temperature (T₁): 22
- Final temperature (T₂): 90
Find: Final volume (V₂)
Answer
3.
Problem 3.
- Solve for: Initial temperature (T₁)
- Volume unit (both volumes): Milliliters (mL)
- Temperature scale (both temperatures): °F
- Initial volume (V₁): 3
- Final volume (V₂): 3
- Final temperature (T₂): 90
Find: Initial temperature (T₁)
Answer
4.
Problem 4.
- Solve for: Final volume (V₂)
- Volume unit (both volumes): Milliliters (mL)
- Temperature scale (both temperatures): °C
- Initial volume (V₁): 2
- Initial temperature (T₁): 24
- Final temperature (T₂): 140
Find: Final volume (V₂)
Answer
5.
Problem 5.
- Solve for: Initial temperature (T₁)
- Volume unit (both volumes): US gallons (US gal)
- Temperature scale (both temperatures): °C
- Initial volume (V₁): 2
- Final volume (V₂): 4
- Final temperature (T₂): 110
Find: Initial temperature (T₁)
Answer
6.
Problem 6.
- Solve for: Final volume (V₂)
- Volume unit (both volumes): Cubic meters (m³)
- Temperature scale (both temperatures): °C
- Initial volume (V₁): 2
- Initial temperature (T₁): 24
- Final temperature (T₂): 75
Find: Final volume (V₂)
Answer
7.
Problem 7.
- Solve for: Final temperature (T₂)
- Volume unit (both volumes): Milliliters (mL)
- Temperature scale (both temperatures): K
- Initial volume (V₁): 2
- Initial temperature (T₁): 22
- Final volume (V₂): 3
Find: Final temperature (T₂)
Answer
8.
Problem 8.
- Solve for: Initial temperature (T₁)
- Volume unit (both volumes): Milliliters (mL)
- Temperature scale (both temperatures): °F
- Initial volume (V₁): 2
- Final volume (V₂): 3
- Final temperature (T₂): 90
Find: Initial temperature (T₁)
Answer
9.
Problem 9.
- Solve for: Initial temperature (T₁)
- Volume unit (both volumes): Cubic inches (in³)
- Temperature scale (both temperatures): °C
- Initial volume (V₁): 1
- Final volume (V₂): 3
- Final temperature (T₂): 125
Find: Initial temperature (T₁)
Answer
10.
Problem 10.
- Solve for: Final temperature (T₂)
- Volume unit (both volumes): Cubic meters (m³)
- Temperature scale (both temperatures): °C
- Initial volume (V₁): 3
- Initial temperature (T₁): 16
- Final volume (V₂): 2
Find: Final temperature (T₂)
Answer

Answer key: Charles's Law worksheet

1. Initial volume (V₁): 1.51 L
1. Find the starting volume V₁. Volume goes up in step with temperature, so multiply the final volume by T₁ ÷ T₂
  $\frac{4 L \times 32 K}{85 K}$ $= 1.5058823529 L$
2. Final volume (V₂): 2.46 cm³
1. Put the starting temperature in kelvins, a scale that starts at absolute zero: add 273.15
  $22 °C + 273.15$ $= 295.15 K$
2. Put the final temperature in kelvins, a scale that starts at absolute zero: add 273.15
  $90 °C + 273.15$ $= 363.15 K$
3. Find the final volume V₂. Volume goes up in step with temperature, so multiply the starting volume by T₂ ÷ T₁
  $\frac{2 cm³ \times 363.15 K}{295.15 K}$ $= 2.4607826529 cm³$
3. Initial temperature (T₁): 90 °F
1. Put the final temperature in kelvins, a scale that starts at absolute zero: subtract 32, multiply by 5/9, then add 273.15
  $(90 °F - 32) \times \frac{5}{9} + 273.15$ $= 305.372222 K$
2. Find the starting temperature T₁ in kelvins. Temperature goes up in step with volume, so multiply the final temperature by V₁ ÷ V₂
  $\frac{305.3722222222 K \times 3 mL}{3 mL}$ $= 305.3722222222 K$
3. Change the answer back to Fahrenheit: subtract 273.15, multiply by 9/5, then add 32
  $(305.372222 K - 273.15) \times \frac{9}{5} + 32$ $= 90 °F$
4. Final volume (V₂): 2.78 mL
1. Put the starting temperature in kelvins, a scale that starts at absolute zero: add 273.15
  $24 °C + 273.15$ $= 297.15 K$
2. Put the final temperature in kelvins, a scale that starts at absolute zero: add 273.15
  $140 °C + 273.15$ $= 413.15 K$
3. Find the final volume V₂. Volume goes up in step with temperature, so multiply the starting volume by T₂ ÷ T₁
  $\frac{2 mL \times 413.15 K}{297.15 K}$ $= 2.7807504627 mL$
5. Initial temperature (T₁): -81.57 °C
1. Put the final temperature in kelvins, a scale that starts at absolute zero: add 273.15
  $110 °C + 273.15$ $= 383.15 K$
2. Find the starting temperature T₁ in kelvins. Temperature goes up in step with volume, so multiply the final temperature by V₁ ÷ V₂
  $\frac{383.15 K \times 2 US gal}{4 US gal}$ $= 191.575 K$
3. Change the answer back to Celsius: subtract 273.15
  $191.575 K - 273.15$ $= −81.575 °C$
6. Final volume (V₂): 2.34 m³
1. Put the starting temperature in kelvins, a scale that starts at absolute zero: add 273.15
  $24 °C + 273.15$ $= 297.15 K$
2. Put the final temperature in kelvins, a scale that starts at absolute zero: add 273.15
  $75 °C + 273.15$ $= 348.15 K$
3. Find the final volume V₂. Volume goes up in step with temperature, so multiply the starting volume by T₂ ÷ T₁
  $\frac{2 m³ \times 348.15 K}{297.15 K}$ $= 2.3432609793 m³$
7. Final temperature (T₂): 33 K
1. Find the final temperature T₂ in kelvins. Temperature goes up in step with volume, so multiply the starting temperature by V₂ ÷ V₁
  $\frac{22 K \times 3 mL}{2 mL}$ $= 33 K$
8. Initial temperature (T₁): -93.22 °F
1. Put the final temperature in kelvins, a scale that starts at absolute zero: subtract 32, multiply by 5/9, then add 273.15
  $(90 °F - 32) \times \frac{5}{9} + 273.15$ $= 305.372222 K$
2. Find the starting temperature T₁ in kelvins. Temperature goes up in step with volume, so multiply the final temperature by V₁ ÷ V₂
  $\frac{305.3722222222 K \times 2 mL}{3 mL}$ $= 203.5814814815 K$
3. Change the answer back to Fahrenheit: subtract 273.15, multiply by 9/5, then add 32
  $(203.581481 K - 273.15) \times \frac{9}{5} + 32$ $= −93.223333 °F$
9. Initial temperature (T₁): -140.43 °C
1. Put the final temperature in kelvins, a scale that starts at absolute zero: add 273.15
  $125 °C + 273.15$ $= 398.15 K$
2. Find the starting temperature T₁ in kelvins. Temperature goes up in step with volume, so multiply the final temperature by V₁ ÷ V₂
  $\frac{398.15 K \times 1 in³}{3 in³}$ $= 132.7166666667 K$
3. Change the answer back to Celsius: subtract 273.15
  $132.716667 K - 273.15$ $= −140.433333 °C$
10. Final temperature (T₂): -80.38 °C
1. Put the starting temperature in kelvins, a scale that starts at absolute zero: add 273.15
  $16 °C + 273.15$ $= 289.15 K$
2. Find the final temperature T₂ in kelvins. Temperature goes up in step with volume, so multiply the starting temperature by V₂ ÷ V₁
  $\frac{289.15 K \times 2 m³}{3 m³}$ $= 192.7666666667 K$
3. Change the answer back to Celsius: subtract 273.15
  $192.766667 K - 273.15$ $= −80.383333 °C$