Difficulty: Introductory to Intermediate | Prerequisites: Basic algebra, temperature unit conversions (Celsius to Kelvin)
Gay-Lussac's Law describes how the pressure of a gas changes when you heat or cool it in a sealed, rigid container. It sits alongside Boyle's Law (pressure and volume) and Charles's Law (volume and temperature) as one of the foundational gas laws leading to the combined and ideal gas laws. If you are comfortable converting Celsius to Kelvin and rearranging a simple ratio, you have what you need. This is bread-and-butter exam material in introductory chemistry and physics courses.
When volume is held constant, the pressure of a gas is directly proportional to its absolute temperature (in Kelvin). Double the Kelvin temperature, double the pressure. The working equation is P₁/T₁ = P₂/T₂, and the single most common mistake is forgetting to convert Celsius to Kelvin before plugging in.
Gay-Lussac's Law (Pressure-Temperature Law)
At constant volume, the pressure of a fixed amount of gas is directly proportional to its absolute temperature. In simple terms, this means that heating a sealed container makes the gas push harder on the walls, and cooling it makes the gas push less.
Absolute temperature
Temperature measured in Kelvin (K), starting from absolute zero. Think of it as the "true" temperature scale for gas law calculations, because zero Kelvin means zero molecular motion (and therefore zero pressure).
Kelvin (K)
The SI unit of temperature. Convert from Celsius by adding 273.15: T(K) = T(°C) + 273.15. You must use Kelvin in every gas law equation. No exceptions.
Pressure
The force a gas exerts per unit area on the walls of its container. Common units include atmospheres (atm), millimetres of mercury (mmHg, also called torr), kilopascals (kPa), and pascals (Pa).
Directly proportional
Two quantities are directly proportional when one increases, the other increases by the same factor (and vice versa). If you plot pressure against Kelvin temperature at constant volume, you get a straight line through the origin.
Constant volume (isochoric process)
A process in which the container does not expand or contract. A rigid steel tank or a sealed glass bulb are typical exam examples. Sometimes called an isovolumetric or isochoric process.
For a fixed amount of gas held at constant volume, pressure and absolute temperature are directly proportional. Raise the temperature, the pressure rises by the same factor. Lower the temperature, the pressure falls.
P₁ / T₁ = P₂ / T₂
where P₁ and T₁ are the initial pressure and absolute temperature, and P₂ and T₂ are the final values. You can rearrange to solve for any one unknown:
Solving for P₂: P₂ = P₁ × (T₂ / T₁)
Solving for T₂: T₂ = T₁ × (P₂ / P₁)
Identify which quantities you know (P₁, T₁, and one of P₂ or T₂).
Convert every temperature to Kelvin: T(K) = T(°C) + 273.15. (For most worksheet-level problems, adding 273 is acceptable.)
Rearrange P₁/T₁ = P₂/T₂ for the unknown.
Substitute, compute, and check that your answer makes physical sense (heating should raise pressure; cooling should lower it).
Example 1: Heating from 20.0 °C to 30.0 °C at 1.00 atm
T₁ = 20.0 + 273.15 = 293.15 K
T₂ = 30.0 + 273.15 = 303.15 K
P₂ = 1.00 × (303.15 / 293.15) = 1.03 atm
A modest 10-degree rise yields only a small pressure increase, which is physically reasonable.
Example 2: 0.500 atm at 25 °C, heated to 125 °C
T₁ = 25 + 273.15 = 298.15 K
T₂ = 125 + 273.15 = 398.15 K
P₂ = 0.500 × (398.15 / 298.15) = 0.668 atm
Example 3: From liquid-nitrogen temperature (77 K) to room temperature (25 °C)
T₁ = 77 K (already in Kelvin)
T₂ = 25 + 273.15 = 298.15 K
P₂ = 47 × (298.15 / 77) = 182 mmHg
A large temperature jump produces a large pressure change. The ratio 298/77 is roughly 3.9, so the pressure nearly quadruples.
Example 4: Finding temperature from pressure (gas thermometer)
P₁ = 248 kPa, T₁ = 0 + 273.15 = 273.15 K
P₂ = 345 kPa
T₂ = 273.15 × (345 / 248) = 380 K
Notice the answer is in Kelvin. If the question asks for Celsius, subtract 273.15.
Example 5: Cooling from 22.0 °C to 0 °C at 745.0 mmHg
T₁ = 22.0 + 273.15 = 295.15 K
T₂ = 0 + 273.15 = 273.15 K
P₂ = 745.0 × (273.15 / 295.15) = 689 mmHg
Example 6: Finding temperature at standard pressure
P₁ = 699.0 mmHg, T₁ = 40.0 + 273.15 = 313.15 K
P₂ = 760 mmHg
T₂ = 313.15 × (760 / 699.0) = 340 K
Note: the worksheet answer says "340. mmHg" but the quantity being solved for is temperature. The answer is 340 K (approximately 67 °C).
Example 7: Cooling from 323.0 K to 273.15 K at 750.0 mmHg
Both temperatures are already in Kelvin.
P₂ = 750.0 × (273.15 / 323.0) = 634 mmHg
Example 8: Heating from -100.0 °C to 25.0 °C at 30.0 kPa
T₁ = -100.0 + 273.15 = 173.15 K
T₂ = 25.0 + 273.15 = 298.15 K
P₂ = 30.0 × (298.15 / 173.15) ≈ 51.7 kPa
Note: the worksheet lists the answer as 517 kPa, which would require a starting pressure of 300 kPa rather than 30.0 kPa. Double-check the original source. At 30.0 kPa the correct answer is approximately 51.7 kPa.
\frac{P_1}{T_1} = \frac{P_2}{T_2}Rearranged forms:
P_2 = P_1 \times \frac{T_2}{T_1}T_2 = T_1 \times \frac{P_2}{P_1}Temperature conversion:
T(K) = T(°C) + 273.15All temperatures must be in Kelvin. Pressure units do not need converting so long as both P₁ and P₂ use the same unit (atm with atm, mmHg with mmHg, kPa with kPa).
Gay-Lussac's Law is the reason a sealed aerosol can warns you not to expose it to heat: the rigid container cannot expand, so heating the gas inside drives the pressure up until the can ruptures. The same principle is at work in a pressure cooker, where sealing the lid and raising the temperature increases pressure above 1 atm, which raises the boiling point of water and cooks food faster.
Gas thermometers (see worksheet question 4) exploit this relationship directly: they measure the pressure of a gas at constant volume and read it as a temperature, which is how very precise temperature standards were historically calibrated.
Students often plug Celsius values straight into the equation. Gay-Lussac's Law requires Kelvin. Using Celsius gives wildly wrong answers, and if either temperature is negative in Celsius you can end up with a negative or zero denominator, which is physically meaningless.
Some students assume pressure and temperature are related by addition ("if I add 10 degrees, I add some pressure"). The relationship is multiplicative: pressure scales with the ratio of the temperatures.
Confusing Gay-Lussac's Law with Charles's Law. Charles's Law relates volume and temperature at constant pressure. Gay-Lussac's Law relates pressure and temperature at constant volume. Both are direct proportionalities, but the held-constant variable is different.
Forgetting to check whether the answer makes physical sense. If you heat a gas and your calculated pressure is lower than the starting pressure, something has gone wrong.
⚠️ Temperature conversion is the number-one source of lost marks. Always convert to Kelvin first, before touching the equation.
⚠️ Units of pressure must match on both sides. You do not need to convert mmHg to atm (or vice versa), but P₁ and P₂ must be in the same unit.
⚠️ Some questions ask for the final temperature rather than the final pressure. Read carefully: you may need to rearrange for T₂ instead of P₂.
⚠️ "Determine the pressure change" (as in question 1) may mean the new pressure, or it may mean the difference (P₂ minus P₁). Read the wording and check what your instructor expects.
⚠️ Questions involving gas thermometers (question 4) are a favourite way to test whether students can rearrange the formula for temperature.
True or false: You can use Celsius in the Gay-Lussac's Law equation as long as both temperatures are in Celsius. False. You must use Kelvin.
Fill in the blank: Gay-Lussac's Law applies when ______ is held constant. Volume.
True or false: If you double the Kelvin temperature of a gas in a sealed rigid container, the pressure doubles. True.
Fill in the blank: To convert 25 °C to Kelvin, you add ______. 273.15 (or 273 for worksheet purposes).
True or false: If a gas is cooled in a sealed container, its pressure increases. False. Cooling lowers the pressure.
Q: A sealed container holds a gas at 2.00 atm and 300 K. The temperature is raised to 450 K. What is the new pressure?
A: P₂ = 2.00 × (450 / 300) = 3.00 atm.
Q: A gas in a rigid tank has a pressure of 101.3 kPa at 20.0 °C. What is the pressure at 100.0 °C?
A: T₁ = 293.15 K, T₂ = 373.15 K. P₂ = 101.3 × (373.15 / 293.15) = 128.9 kPa.
Q: A gas thermometer reads 150 kPa at 0 °C. At what Kelvin temperature will it read 200 kPa?
A: T₂ = 273.15 × (200 / 150) = 364.2 K.
Q: A container of gas is at 500 mmHg and 350 K. It is cooled to 250 K. What is the final pressure?
A: P₂ = 500 × (250 / 350) = 357 mmHg.
Q: Why does Gay-Lussac's Law not apply to a balloon being heated in open air?
A: A balloon can expand, so its volume is not constant. Gay-Lussac's Law requires constant volume. In a balloon, Charles's Law (volume and temperature at constant pressure) is the more relevant relationship.
Gay-Lussac's Law is one piece of the combined gas law, P₁V₁/T₁ = P₂V₂/T₂, which collapses to Gay-Lussac's equation when V is constant, to Charles's Law when P is constant, and to Boyle's Law when T is constant. Understanding all three individual laws makes the combined law feel like a natural extension rather than a new formula to memorise.
This also feeds into the ideal gas law, PV = nRT. Gay-Lussac's Law is what you get when you hold n, R, and V constant and watch how P responds to changes in T. If you go on to study thermodynamics, the constant-volume (isochoric) process reappears in heat-engine cycles and in the analysis of bomb calorimeters.
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