The Combined Gas Law Formula

The combined gas law is the formula P1V1/T1 = P2V2/T2, which relates the pressure, volume and absolute temperature of a fixed amount of gas before and after a change. In brief: if 3.0 L of gas at 1.0 atm and 300 K is compressed to 2.0 L at 1.5 atm, the final temperature is 300 × (1.5 × 2.0) / (1.0 × 3.0) = 300 K.

The combined gas law merges Boyle’s, Charles’s and Gay-Lussac’s laws into a single expression. It is useful when pressure, volume and temperature all change at once, for example when a weather balloon rises into thinner, colder air. This article derives the formula, explains how to solve problems and gives several examples and a summary table.

What is the combined gas law formula?

For a fixed amount of gas:

P1V1 / T1 = P2V2 / T2

The quantity PV/T stays the same during any change as long as no gas enters or leaves. Pressure and volume can be in any units, provided each is consistent on both sides. The temperature must be in kelvins.

How is the combined gas law derived?

Boyle’s law says that PV is constant at constant temperature, and Charles’s law says that V/T is constant at constant pressure. Both can be written as one statement: PV is proportional to T for a given amount of gas, so PV/T = constant. The constant equals nR in the ideal gas law, which means the combined gas law is a special case of PV = nRT where n does not change.

How does it contain the other gas laws?

Special cases of the combined gas law
Constant Law Resulting equation
TemperatureBoyleP1V1 = P2V2
PressureCharlesV1/T1 = V2/T2
VolumeGay-LussacP1/T1 = P2/T2

How do you solve combined gas law problems?

  1. Write down P, V and T for the initial and final states and mark the unknown.
  2. Convert the temperatures to kelvins.
  3. Make the units of the two pressures the same, and the units of the two volumes the same.
  4. Solve for the unknown. For example, V2 = V1 × (P1/P2) × (T2/T1).
  5. Check that the direction of change is reasonable for each variable.

Worked example 1: a weather balloon

A weather balloon has a volume of 2.50 m3 at ground level, where the pressure is 101 kPa and the temperature is 20 °C. At a height where the pressure is 35 kPa and the temperature is −30 °C, what is its volume?

  1. T1 = 293 K and T2 = 243 K.
  2. V2 = 2.50 × (101 / 35) × (243 / 293).
  3. 101 / 35 = 2.886 and 243 / 293 = 0.829.
  4. V2 = 2.50 × 2.886 × 0.829 = 5.98 m3.

The volume more than doubles because the effect of the falling pressure is larger than the effect of the cooling.

2.50 m3 101 kPa 293 K 5.98 m3 35 kPa 243 K

Worked example 2: a laboratory sample

A gas occupies 400 mL at 740 mmHg and 25 °C. Find its volume at 760 mmHg and 0 °C.

V2 = 400 × (740 / 760) × (273 / 298) = 400 × 0.9737 × 0.9161 = 357 mL.

400 mL 740 mmHg 298 K 357 mL 760 mmHg 273 K

Worked example 3: finding the temperature

A sealed 5.00 L cylinder is at 2.00 atm and 300 K. The gas is compressed to 2.50 L and the pressure becomes 3.00 atm. What is the new temperature?

T2 = T1 × (P2V2) / (P1V1) = 300 × (3.00 × 2.50) / (2.00 × 5.00) = 300 × 7.50 / 10.0 = 225 K, or −48 °C.

5.00 L 2.00 atm 300 K 2.50 L 3.00 atm 225 K

How do you convert a gas volume to STP?

The combined gas law is often used to convert a measured volume to the volume under standard conditions, which makes it possible to compare gas measurements made on different days. Using STP as the second state, VSTP = V × (P / 760 mmHg) × (273.15 K / T). In the second worked example above, 357 mL is the volume at STP, and it corresponds to 357 / 22,414 = 0.0159 mol of gas.

What are the limits of the law?

The law applies to ideal gases and to a fixed amount of gas. Real gases follow it closely at moderate pressures and temperatures well above the boiling point. At high pressures or low temperatures, deviations appear, and the van der Waals equation gives a better description. If gas enters or leaves the system, the ideal gas law with a changing n must be used instead.

Which mistakes are common?