Combined Gas Law — One Formula for Boyle, Charles and Gay-Lussac
Most students memorise three separate gas laws and then panic when a question changes two things at once. There is no need. Boyle's law, Charles's law and Gay-Lussac's law are three special cases of a single equation. Learn that one equation and the other three come free — and so does every "a gas at these conditions is moved to those conditions" problem in your syllabus.
The formula
Read it as: for a fixed amount of gas, the quantity PV/T does not change. Subscript 1 is the starting state, subscript 2 is the final state. Nothing else is needed.
What each symbol means
| Symbol | Meaning | Unit rule |
|---|---|---|
| P₁, P₂ | Initial and final pressure | Any unit — but the SAME unit on both sides |
| V₁, V₂ | Initial and final volume | Any unit — but the SAME unit on both sides |
| T₁, T₂ | Initial and final temperature | Kelvin only. K = °C + 273.15 |
Pressure and volume units cancel, so atm with atm or mL with mL is perfectly fine. Temperature does not cancel in the same way, because the Celsius scale has an arbitrary zero. Kelvin is compulsory.
How the three named laws fall out
| Hold constant | What remains | Name |
|---|---|---|
| Temperature (T₁ = T₂) | P₁V₁ = P₂V₂ | Boyle's law |
| Pressure (P₁ = P₂) | V₁/T₁ = V₂/T₂ | Charles's law |
| Volume (V₁ = V₂) | P₁/T₁ = P₂/T₂ | Gay-Lussac's law |
So you are not learning four formulas. You are learning one, and cancelling whatever is written as "constant" in the question.
The rearrangements you will actually use
Worked example 1 — volume at new pressure and temperature
A gas occupies 250 mL at 25 °C and 1.00 atm. What volume will it occupy at 100 °C and 0.80 atm?
Convert temperatures first: T₁ = 25 + 273 = 298 K, T₂ = 100 + 273 = 373 K.
V₂ = P₁V₁T₂ ÷ (T₁P₂)
Numerator: 1.00 × 250 × 373 = 93 250
Denominator: 298 × 0.80 = 238.4
V₂ = 93 250 ÷ 238.4 = 391 mL
Sanity check: heating expands the gas and lowering the pressure expands it further, so the volume must increase. 391 mL > 250 mL. Correct direction.
Worked example 2 — a weather balloon
A balloon holds 20.0 L at 1.00 atm and 300 K at ground level. It rises to an altitude where the pressure is 0.40 atm and the temperature is 250 K. Find the new volume.
V₂ = P₁V₁T₂ ÷ (T₁P₂)
Numerator: 1.00 × 20.0 × 250 = 5000
Denominator: 300 × 0.40 = 120
V₂ = 5000 ÷ 120 = 41.7 L
Two effects fight each other here. The pressure drop alone would give 20.0 × (1.00/0.40) = 50.0 L; the cooling then shrinks that by a factor 250/300, giving 50.0 × 0.8333 = 41.7 L. Pressure won, so the balloon still expands — which is why weather balloons are launched only partly inflated.
Worked example 3 — solving for pressure
A gas at 5.0 atm occupies 3.0 L at 400 K. It is compressed to 1.5 L and cooled to 300 K. What is the new pressure?
P₂ = P₁V₁T₂ ÷ (T₁V₂)
Numerator: 5.0 × 3.0 × 300 = 4500
Denominator: 400 × 1.5 = 600
P₂ = 4500 ÷ 600 = 7.5 atm
Halving the volume alone would double the pressure to 10 atm; cooling from 400 K to 300 K then multiplies by 300/400 = 0.75, giving 7.5 atm. Splitting a problem into two single-effect steps like this is a good way to check any combined-gas-law answer.
Common mistakes
- Leaving temperature in Celsius. This is the single biggest source of wrong answers in this chapter. Convert before you substitute, not after.
- Mixing volume units across the equation. 250 mL on one side and litres on the other silently changes the answer by 1000.
- Using it when the amount of gas changes. P₁V₁/T₁ = P₂V₂/T₂ assumes n is fixed. If gas is added, removed or produced by a reaction, you need PV = nRT instead.
- Forgetting gauge vs absolute pressure. Tyre and cylinder gauges read pressure above atmospheric. Add roughly 1 atm to get absolute pressure before using any gas law.
- Not sanity-checking the direction. Heat expands, compression raises pressure. If your answer moves the wrong way, you have inverted a ratio.
- Assuming it holds at extreme conditions. It is an ideal-gas result. Near condensation or at very high pressure, real gases deviate.
Where it appears in exams
| Exam | Typical question |
|---|---|
| CBSE/ICSE Class 9–11 | Direct P–V–T conversions; deriving Boyle and Charles as special cases |
| JEE / NEET | STP↔NTP conversions, balloon and syringe problems, gas collected over water |
| IIT-JAM / CUET-PG | Link to PV = nRT and to the kinetic theory of gases |
| GATE / CSIR-NET | Real-gas corrections, compressibility factor as a departure from this law |
Check every substitution instantly. Enter any five of the six quantities and the Combined Gas Law calculator returns the sixth, with the temperature conversion done for you — the step where most marks are lost.
Open the Combined Gas Law Calculator →Gas laws are the foundation for thermodynamics and physical chemistry later on, so it pays to get them solid early. ABC Chemistry runs Class 11–12 chemistry coaching at the Gurugram centre and online classes across India, with home tuition available in Delhi-NCR — details at abcchemistry.in.