How do I use the combined gas law?
Type the pressure, volume and temperature of a gas before a change, and two of them after it. The combined gas law calculator finds the missing value, whichever one it is.
- Volume V₂
- 666.7 L
A gas at 153 atm, 13.2 L and 27 °C is at 3.13 atm, 666.7 L and 37 °C after the change.
Volume V₂: 666.7 L. A gas at 153 atm, 13.2 L and 27 °C is at 3.13 atm, 666.7 L and 37 °C after the change.
How to calculate
Solves the combined gas law P₁V₁/T₁ = P₂V₂/T₂ for any one of pressure, volume or temperature before or after a change, in atm, kPa, mmHg, L, °C or K.
Example with the default inputs (Pressure P₁ 153 atm, Volume V₁ 13.2 L, Temperature T₁ 27 °C, Pressure P₂ 3.13 atm, Temperature T₂ 37 °C): A gas at 153 atm, 13.2 L and 27 °C is at 3.13 atm, 666.7 L and 37 °C after the change.
Formula: P₁V₁ ÷ T₁ = P₂V₂ ÷ T₂, with T in kelvins. For example V₂ = P₁V₁T₂ ÷ (T₁P₂).
- The gas is ideal and its amount does not change (no gas leaks in or out).
- Temperatures are absolute: °C + 273.15 = K, and (°F + 459.67) × 5/9 = K.
- Pressures are absolute pressures, not gauge readings.
Worked examples
Each example is checked against the calculator on every build.
- Pressure P₁ 153 atm, Volume V₁ 13.2 L, Temperature T₁ 27 °C, Pressure P₂ 3.13 atm, Temperature T₂ 37 °C gives Volume V₂ 666.7 L.Source: OpenStax, Chemistry 2e, §9.2 Relating Pressure, Volume, Amount, and Temperature: The Ideal Gas Law, https://openstax.org/books/chemistry-2e/pages/9-2-relating-pressure-volume-amount-and-temperature-the-ideal-gas-law, Example 9.10 Using the Combined Gas Law: about 667 L (the book rounds to 300 K and 310 K)
- Pressure P₁ 3.553 atm, Volume V₁ 0.35 L, Temperature T₁ 24 °C, Volume V₂ 0.35 L, Temperature T₂ 50 °C gives Pressure P₂ 3.864 atm.Source: OpenStax, Chemistry 2e, §9.2 Relating Pressure, Volume, Amount, and Temperature: The Ideal Gas Law, https://openstax.org/books/chemistry-2e/pages/9-2-relating-pressure-volume-amount-and-temperature-the-ideal-gas-law, Example 9.5 Predicting Change in Pressure with Temperature: about 390 kPa
- Pressure P₁ 1 atm, Volume V₁ 0.15 L, Temperature T₁ 0 °C, Pressure P₂ 1 atm, Volume V₂ 0.1317 L gives Temperature T₂ -33.32 °C.Source: OpenStax, Chemistry 2e, §9.2 Relating Pressure, Volume, Amount, and Temperature: The Ideal Gas Law, https://openstax.org/books/chemistry-2e/pages/9-2-relating-pressure-volume-amount-and-temperature-the-ideal-gas-law, Example 9.7 Measuring Temperature with a Volume Change: 239.8 K (−33.4 °C)
- Pressure P₁ 0.9868 atm, Volume V₁ 0.3 L, Temperature T₁ 10 °C, Pressure P₂ 0.9868 atm, Temperature T₂ 30 °C gives Volume V₂ 0.3212 L.Source: OpenStax, Chemistry 2e, §9.2 Relating Pressure, Volume, Amount, and Temperature: The Ideal Gas Law, https://openstax.org/books/chemistry-2e/pages/9-2-relating-pressure-volume-amount-and-temperature-the-ideal-gas-law, Example 9.6 Predicting Change in Volume with Temperature: 0.321 L
How it works
Type five of the six values; the calculator finds the sixth from
P₁V₁ ÷ T₁ = P₂V₂ ÷ T₂
where P is the absolute pressure, V the volume and T the absolute temperature, before (1) and after (2) the change. Every value is first converted to SI base units: pascals, cubic meters and kelvins (°C + 273.15; (°F + 459.67) × 5/9; 1 atm = 101,325 Pa; 1 kPa = 1,000 Pa; 1 bar = 100,000 Pa; 1 mmHg = 13.5951 × 9.80665 ≈ 133.3224 Pa; 1 torr = 101,325/760 Pa; 1 psi = 4.4482216152605 N ÷ 0.00064516 m² ≈ 6,894.757 Pa; 1 L = 0.001 m³; 1 mL = 1 cm³ = 10⁻⁶ m³; 1 ft³ = 0.028316846592 m³; 1 gal = 0.003785411784 m³). The six rearrangements are:
- P₁ = P₂V₂T₁ ÷ (T₂V₁), V₁ = P₂V₂T₁ ÷ (T₂P₁), T₁ = P₁V₁T₂ ÷ (P₂V₂)
- P₂ = P₁V₁T₂ ÷ (T₁V₂), V₂ = P₁V₁T₂ ÷ (T₁P₂), T₂ = P₂V₂T₁ ÷ (P₁V₁)
The maths runs in floating point; the answer rounds for display only.
Assumptions: the gas behaves as an ideal gas, and no gas enters or leaves (the amount n is fixed). Real gases at very high pressure or near condensing do not follow the law exactly.
Rules
- Every pressure, volume and temperature must be above 0 (temperatures above absolute zero). Each pressure is at most 10¹² Pa, each volume at most 10⁹ m³, each temperature at most 10⁶ K; a typed value outside that shows a message on its field, and a solved value outside it gives no answer.
- With all six values typed, they must fit the equation; otherwise the page says the values do not agree.
Worked examples by hand
Scuba tank (OpenStax Example 9.10). P₁ = 153 atm, V₁ = 13.2 L, T₁ = 27 °C = 300.15 K, P₂ = 3.13 atm, T₂ = 37 °C = 310.15 K. V₂ = 153 × 13.2 × 310.15 ÷ (300.15 × 3.13) = 666.74 L. The book rounds the temperatures to 300 K and 310 K and gets 667 L.
Heated can (OpenStax Example 9.5). V₁ = V₂ = 350 mL, P₁ = 360 kPa, T₁ = 24 °C = 297.15 K, T₂ = 50 °C = 323.15 K. P₂ = 360 × 323.15 ÷ 297.15 = 391.50 kPa (the book: about 390 kPa).
Gas thermometer (OpenStax Example 9.7). P₁ = P₂, V₁ = 150.0 cm³, T₁ = 0.00 °C = 273.15 K, V₂ = 131.7 cm³. T₂ = 273.15 × 131.7 ÷ 150.0 = 239.83 K, or −33.32 °C.
Charles's law (OpenStax Example 9.6). P₁ = P₂ = 750 torr, V₁ = 0.300 L, T₁ = 10 °C = 283.15 K, T₂ = 30 °C = 303.15 K. V₂ = 0.300 × 303.15 ÷ 283.15 = 0.3212 L.
Other questions people ask
What is the combined gas law?
For a fixed amount of gas, pressure × volume ÷ temperature stays the same: P₁V₁ ÷ T₁ = P₂V₂ ÷ T₂. It joins Boyle’s law (P and V), Charles’s law (V and T) and Gay-Lussac’s law (P and T) in one equation.
Why must the temperature be in kelvins?
The law works with absolute temperature, which starts at absolute zero. 27 °C is 300.15 K. You can type °C or °F here; the calculator turns every temperature into kelvins before it divides.
How do I solve for the final volume?
Rearrange to V₂ = P₁V₁T₂ ÷ (T₁P₂). A 13.2 L scuba tank at 153 atm and 27 °C gives about 667 L of air at 3.13 atm and 37 °C.
How do I use it when one value stays the same?
Type the same value before and after. With equal volumes it becomes Gay-Lussac’s law: a can at 360 kPa and 24 °C reaches about 391 kPa at 50 °C.
Do the units have to match?
No. Each value can be in its own unit (atm, kPa, mmHg, psi, L, mL, °C, K, °F). The calculator converts everything to pascals, cubic meters and kelvins, and shows the answer in the unit you pick.
Should I use gauge pressure from a tire or tank gauge?
No. The law needs absolute pressure. Add the air pressure around you (about 1 atm, 101.325 kPa or 14.7 psi) to a gauge reading first.