Chemistry · General chemistry I · Worked example
Use the combined gas law
A weather balloon holds 2.50 L of helium at 1.00 atm and 25.0 °C. What is its volume at 0.500 atm and −15.0 °C?
List both states, in kelvin
Convert both temperatures: 25.0 °C is 298.2 K and −15.0 °C is 258.2 K. The final volume is the unknown.
| State | P (atm) | V (L) | T (K) |
|---|---|---|---|
| Initial | 1.00 | 2.50 | 298.2 |
| Final | 0.500 | ? | 258.2 |
Solve for V₂
The amount of helium is fixed, so P₁V₁/T₁ = P₂V₂/T₂. Multiply both sides by T₂ and divide by P₂.
Substitute and calculate
Halving the pressure doubles the volume, and cooling shrinks it by the factor 258.2/298.2.
Check the direction
Lower pressure expands the gas and lower temperature contracts it. The pressure change is the larger effect, so the volume grows, from 2.50 L to 4.33 L.
Result
The balloon’s volume becomes 4.33 L.
Your turn
A gas occupies 3.00 L at 20.0 °C. At constant pressure, what is its volume at 80.0 °C?
Show the answer and explanation
3.61 L.
At constant pressure, V₂ = V₁T₂/T₁ = 3.00 L × 353.2 K/293.2 K = 3.61 L, which is Charles’s law. Using Celsius would wrongly give 12.0 L.
Keep exploring
Open it in Gas laws & mixtures and set the final temperature to 25.0 °C: the volume exactly doubles, to 5.00 L. That is Boyle’s law.
Return to the concept →Sources and scope
Authored study material. Tool results depend on the stated inputs and model assumptions.
Try in the workspace
Open the example inputs, change a value and keep a useful result on your board.
Solve it in Gas laws & mixtures Check the calculation in a Chemistry box Open worked example on a board Chemistry formulas: gasesYour existing work stays on this device. Examples open as editable copies.