Gay-Lussac’s Law: Pressure and Temperature

Gay-Lussac’s law states that for a fixed amount of gas at constant volume, the pressure is directly proportional to the absolute temperature in kelvins: P1/T1 = P2/T2. In brief: a sealed container of gas at 2.00 atm and 300 K will have a pressure of 2.40 atm when heated to 360 K.

This law explains why a sealed aerosol can can explode in a fire and why tire pressure changes with the seasons. It describes how heat affects a gas that cannot expand. This article gives the formula, the method, worked examples, a data table and the history, followed by common mistakes.

What does Gay-Lussac’s law say?

If the volume and the amount of a gas stay fixed, pressure and absolute temperature increase together:

P / T = k, or P1/T1 = P2/T2

The molecular explanation is straightforward. At a higher temperature the particles move faster, so they strike the walls more often and with greater force. Because the container cannot expand, the pressure rises. Temperature must always be in kelvins; a graph of P against T in kelvins is a straight line through the origin.

How do you solve Gay-Lussac’s law problems?

  1. Check that the gas is in a rigid, sealed container.
  2. Convert temperatures to kelvins (K = °C + 273.15).
  3. Rearrange: P2 = P1 × T2 / T1.
  4. Substitute values and calculate.
  5. Check: heating must increase the pressure, and cooling must reduce it.

Worked example 1: a heated cylinder

A steel cylinder holds oxygen at 150 kPa and 20 °C. What is the pressure after it is heated to 85 °C?

T1 = 293 K and T2 = 358 K. P2 = 150 kPa × 358 / 293 = 183 kPa.

150 kPa 293 K 183 kPa 358 K

Worked example 2: a cooled tire

A car tire is filled to an absolute pressure of 320 kPa at 25 °C. After a night at −5 °C, what is the pressure? Assume the volume is unchanged.

T1 = 298 K and T2 = 268 K. P2 = 320 × 268 / 298 = 288 kPa. That is a drop of 32 kPa, or about 10 %, which is enough to trigger a low-pressure warning in some cars.

320 kPa 298 K 288 kPa 268 K

Worked example 3: finding the temperature

A sealed flask of gas has a pressure of 1.00 atm at 300 K. At what temperature will the pressure reach 1.50 atm?

T2 = 300 K × 1.50 / 1.00 = 450 K, or 177 °C.

1.00 atm 300 K 1.50 atm 450 K

Pressure and temperature data table

Pressure of a gas in a rigid container, with P/T = 0.00500 atm/K
Temperature (K) Temperature (°C) Pressure (atm) P/T (atm/K)
200−731.000.00500
250−231.250.00500
300271.500.00500
4001272.000.00500
5002272.500.00500
6003273.000.00500
Pressure against kelvin temperature for a gas at constant volume, a straight line through the origin 0 100 200 300 400 500 600 0 1 2 3 Temperature T (K) Pressure P (atm) P/T = 0.00500 atm/K
Pressure against absolute temperature for a gas in a rigid container, using the table values (P/T = 0.00500 atm/K). The dots lie on a straight line through the origin: 1.50 atm at 300 K becomes 3.00 atm at 600 K. The dashed part extrapolates to zero pressure at 0 K.

Where do you see Gay-Lussac’s law in daily life?

Who was Gay-Lussac?

Joseph Louis Gay-Lussac (1778–1850) was a French chemist and physicist. In 1802 he published measurements of the thermal expansion of gases that confirmed the work of Jacques Charles. In 1808 he stated the law of combining volumes, which says that gases react in simple whole-number volume ratios. Guillaume Amontons had described the pressure–temperature relationship much earlier, in the late 1600s, so the law is sometimes called Amontons’s law. The name Gay-Lussac’s law is applied to the pressure law in most English-language textbooks, but in some countries it refers to the volume–temperature relationship.

Which mistakes are common?