Combined gas law calculator

What is the combined gas law?

P₁V₁ ÷ T₁ = P₂V₂ ÷ T₂ for a fixed amount of gas, with temperatures in kelvin. It combines Boyle’s, Charles’s and Gay-Lussac’s laws.

Example

A gas at 1 atm, 2 L and 300 K is squeezed to 1 L and heated to 600 K: P₂ = 1 × 2 × 600 ÷ (300 × 1) = 4 atm.

Why must temperature be in kelvin?

The law uses absolute temperature, which starts at absolute zero. The calculator converts °C and °F to kelvin for you.

Which units can I use for pressure and volume?

Any, as long as both pressures share a unit and both volumes share a unit — the answer comes out in the same unit.

How do I use it?

Fill in five of the six values and leave the unknown empty.

How do I get Boyle’s, Charles’s or Amontons’s law from it?

Keep the quantity that is constant the same on both sides and it cancels. With the temperature unchanged you get P₁V₁ = P₂V₂ (Boyle): a gas at 13.0 psi in 15.0 mL compressed to 7.5 mL reaches 26 psi (OpenStax Chemistry 2e, Example 9.8). With the pressure unchanged you get V₁ ÷ T₁ = V₂ ÷ T₂ (Charles); with the volume unchanged, P₁ ÷ T₁ = P₂ ÷ T₂ (Amontons). In the calculator, just type the same number in both fields.

What if some gas escapes or is added?

Then the law does not apply, because it holds only for a fixed amount of gas (OpenStax Chemistry 2e, section 9.2). If the number of moles changes, use the ideal gas law for each state and compare, or solve PV = nRT for the new amount.

Do I use gauge pressure or absolute pressure?

Absolute. A tire gauge reads the pressure above the surrounding atmosphere, so add the atmospheric pressure before using it here. Using gauge values gives a wrong ratio, and the error is largest at low pressures.

The law and its assumptions

For a fixed amount of an ideal gas, pressure times volume divided by absolute temperature does not change: P₁V₁ ÷ T₁ = P₂V₂ ÷ T₂ (OpenStax Chemistry 2e, section 9.2). It follows from the ideal gas law, because nR is the same before and after. Three conditions matter. The amount of gas stays constant, the temperatures are in kelvin, and each pressure and each volume pair shares a unit, so the answer comes out in that unit. The result is only as good as the ideal gas approximation, which is best at low pressure and moderate temperature.

Worked examples

OpenStax Example 9.10: a 13.2 L scuba tank holds air at 153 atm and 27 °C. How large a volume does that air fill at 3.13 atm and 37 °C, the conditions in the lungs? Converting to kelvin gives 300.15 K and 310.15 K, and V₂ = 153 × 13.2 × 310.15 ÷ (300.15 × 3.13) = 667 L. The textbook points out that this is only a rough estimate because the compressed gas is far from ideal. A smaller case, the practice problem: 0.250 L of ammonia at 27 °C and 0.850 atm, brought to 0 °C and 1.00 atm, occupies 0.193 L. For a constant-volume case (Example 9.5), a 350 mL can at 24 °C and 360 kPa heated to 50 °C reaches 391 kPa (the book rounds to 390).

Common mistakes

The most common one is leaving temperatures in Celsius. In the can example, 24 °C and 50 °C entered as plain numbers would give 750 kPa instead of 391 kPa. The calculator’s unit selector converts for you, but a hand calculation must add 273.15 first. The second is mixing units within a pair, such as P₁ in atm and P₂ in kPa. The third is filling in all six values or fewer than five: exactly one field must be empty. The fourth is using gauge pressure, which leaves out the atmosphere.

Sources: OpenStax Chemistry 2e, section 9.2 “Relating Pressure, Volume, Amount, and Temperature: The Ideal Gas Law” (combined gas law, Examples 9.5, 9.8 and 9.10 and the NH₃ practice problem). Related: ideal gas law calculator, partial pressure calculator, Graham’s law calculator and pressure calculator.