When the gas changes state but stays the same gas

The combined gas law is what you reach for when you have a fixed quantity of gas and want to know what happens when two or three of its conditions change simultaneously. It folds Boyle’s law (P↔V at fixed T), Charles’s law (V↔T at fixed P), and Gay-Lussac’s law (P↔T at fixed V) into one equation:

(P₁V₁) / T₁ = (P₂V₂) / T₂

Notice what’s missing: no R, no n. The number of moles cancels out because it doesn’t change between state 1 and state 2 — the gas is sealed in a container, the same molecules in both snapshots. As long as the amount of gas is constant and behavior stays roughly ideal, this one ratio is conserved.

The procedure

Step 1: Identify all six variables and which one is unknown

Six slots: P₁, V₁, T₁, P₂, V₂, T₂. Five are given, one is what you’re solving for.

Setup: A gas occupies 2.50 L at 25.0 °C and 1.00 atm. What volume at 50.0 °C and 0.500 atm?

P₁ = 1.00 atm, V₁ = 2.50 L, T₁ = 25.0 °C; P₂ = 0.500 atm, V₂ = ?, T₂ = 50.0 °C.

Step 2: Convert every temperature to Kelvin

T(K) = T(°C) + 273.15. Both temperatures, every time. Celsius will give you nonsense — at 0 °C you’d be dividing by zero, and at any other temperature the ratio T₂/T₁ is wrong.

T₁ = 298.15 K, T₂ = 323.15 K.

Step 3: Match units across both sides

P₁ and P₂ have to share a unit (both atm, both kPa, both mmHg). Same for V. The unit choice doesn’t matter as long as it’s consistent — the equation is a ratio, and matching units cancel cleanly.

Step 4: Rearrange algebraically before plugging in numbers

For V₂: V₂ = (P₁V₁T₂) / (T₁P₂) = (1.00 × 2.50 × 323.15) / (298.15 × 0.500) = 807.875 / 149.075 = 5.42 L

Step 5: Sanity-check the direction

Pressure dropped to half → volume should roughly double. Temperature rose ~25 K → volume should grow a bit more. Combined effect: V should be a bit more than double, which 5.42 L vs. 2.50 L is. Direction confirmed.

When variables hold steady, the equation collapses

Constant T: Boyle’s law, P₁V₁ = P₂V₂
Constant P: Charles’s law, V₁/T₁ = V₂/T₂
Constant V: Gay-Lussac’s law, P₁/T₁ = P₂/T₂

You don’t have to memorize three equations — just drop whichever pair stays constant from the combined form.

Worked example: solving for final temperature

A gas at 750.0 mmHg and 300.0 K occupies 4.00 L. Compress it to 2.00 L at 1500.0 mmHg. Final temperature?

T₂ = (P₂V₂T₁) / (P₁V₁) = (1500.0 × 2.00 × 300.0) / (750.0 × 4.00) = 900,000 / 3,000 = 300.0 K

Doubling the pressure exactly compensated for halving the volume, so temperature is unchanged. The product PV stayed constant — Boyle’s law in disguise.

Common mistakes

Celsius in the equation. This is the error. T(K) = T(°C) + 273.15, every temperature, every problem. Combined gas law in Celsius gives wrong ratios at every temperature except 0 °C, where it gives infinity.

Mismatched units across states. P₁ in atm with P₂ in kPa? Convert one before substituting. The equation only works when units match.

Plugging numbers in before rearranging. Solve algebraically first, then substitute. It cuts arithmetic errors and makes it obvious which terms go in the numerator vs. denominator.

Using the combined gas law when n changes. If you’re adding gas, removing gas, or running a reaction that changes the moles of gas, this equation doesn’t apply — switch to PV = nRT for both states separately.

Practice

Try these and verify with our Combined Gas Law Calculator:

A gas at STP (0 °C, 1.00 atm) occupies 22.4 L. What volume at 100 °C and 2.00 atm?
A balloon holds 1.50 L at 20.0 °C and 101.3 kPa. It rises to where the pressure is 50.0 kPa and the temperature is −10.0 °C. New volume?
A sealed syringe contains 10.0 mL at 25.0 °C and 760 mmHg. Heated to 100 °C and compressed to 5.0 mL. New pressure?
500.0 mL at 100.0 °C and 600.0 mmHg → 250.0 mL at 800.0 mmHg. Final temperature?
Nitrogen at 2.00 atm, 200 K, 3.00 L. Pressure needed for 1.50 L at 400 K?

How to Use the Combined Gas Law