The combined gas law unifies three classical gas relationships — Boyle’s law, Charles’ law, and Gay-Lussac’s law — into a single equation that lets you predict what happens to a fixed amount of gas when pressure, volume, and temperature all change simultaneously. This calculator solves for any one of the six variables, shows full step-by-step working, and also handles the ideal gas law (PV = nRT) so you can move between single-state and two-state gas problems without switching tools.
The formula
The combined gas law is written as:
(P₁ × V₁) / T₁ = (P₂ × V₂) / T₂
Where:
- P is absolute pressure in kilopascals (kPa)
- V is volume in litres (L)
- T is absolute temperature in Kelvin (K = °C + 273.15)
- Subscript 1 = initial state, subscript 2 = final state
The amount of gas (moles, n) is assumed constant. If n changes, use the ideal gas law PV = nRT separately for each state instead. The universal gas constant is R = 8.314 J/(mol·K).
The three sub-laws follow directly by holding one variable constant:
| Law | Fixed | Equation |
|---|---|---|
| Boyle’s | Temperature | P₁V₁ = P₂V₂ |
| Charles’ | Pressure | V₁/T₁ = V₂/T₂ |
| Gay-Lussac’s | Volume | P₁/T₁ = P₂/T₂ |
How it works
Select the gas law mode and choose which variable to find. The calculator hides that field and asks for all remaining known values. For temperatures it accepts either Kelvin or Celsius — select your preferred unit and it converts to Kelvin automatically before computing. Pressures must be in kPa and volumes in litres; if your data is in atm, bar, mmHg, or cm³ use the conversion factors in the FAQ before entering. Once all inputs are filled the answer updates in real time. Click Show working to expand a table of the rearranged formula and the substituted numbers at each step.
Worked example
A gas sample initially has pressure P₁ = 150.0 kPa, volume V₁ = 3.0 L, and temperature T₁ = 300 K. The gas is then compressed and cooled to P₂ = 250.0 kPa and T₂ = 240 K. What is the new volume V₂?
Using the combined gas law rearranged for V₂:
V₂ = (P₁ × V₁ × T₂) / (T₁ × P₂)
Substituting:
V₂ = (150.0 × 3.0 × 240) / (300 × 250.0)
V₂ = 108 000 / 75 000 = 1.44 L
The gas volume dropped from 3.0 L to 1.44 L — the pressure increase more than outweighed the volume expansion that would result from cooling alone.
Formula note
Temperature conversions: K = °C + 273.15. Room temperature ≈ 293 K (20 °C). Standard temperature and pressure (STP) is defined as 273.15 K (0 °C) and 100 kPa. Standard ambient temperature and pressure (SATP) is 298.15 K (25 °C) and 100 kPa. One atmosphere = 101.325 kPa. For the ideal gas law mode the constant R = 8.314 J/(mol·K) is used; n is the number of moles of gas. Every calculation runs client-side — no data is sent to any server.