Gay-Lussac’s Law: Pressure–Temperature Numericals
A gas cylinder left in the sun, a sealed can thrown in a fire, a car tyre that feels harder on a hot day: all of these are one idea. When a gas is kept in a closed rigid container, its pressure goes up when you heat it. This is Gay-Lussac’s law of pressure and temperature. It is a short topic in Class 11, but students lose marks because they forget one rule: the temperature must be in kelvin.
The formula
P1/T1 = P2/T2
P1, P2 are the first and second pressures. Use any pressure unit (atm, mmHg, kPa, bar), but use the same unit on both sides. T1, T2 are the absolute temperatures in kelvin: T(K) = °C + 273.15. Most school questions accept 273. The word “constant” matters: the amount of gas (moles) and the volume must not change. A sealed, rigid container satisfies both.
Where it comes from
Start from the ideal gas law, PV = nRT. If n and V are fixed, then n, R and V are all constants. Write P = (nR/V) × T. The bracket is a constant, so P/T = constant. For two states of the same gas, P1/T1 = P2/T2. Notice that this law is the same type as Charles’s law (V/T constant at fixed P) and Boyle’s law (PV constant at fixed T). All three are special cases of PV = nRT in which two of the four quantities P, V, n, T are fixed.
The picture behind it is simple. At higher temperature, gas molecules move faster. In a fixed volume they hit the walls more often and with more force, so the pressure rises. If you could cool an ideal gas down to 0 K, the pressure would fall to zero. This is why the Kelvin scale starts at absolute zero and why a Celsius temperature cannot be used in the ratio.
Worked example 1: a heated cylinder
Question. A rigid cylinder holds a gas at 2.00 atm and 27°C. It is heated to 127°C. Find the new pressure.
Step 1. T1 = 27 + 273.15 = 300.15 K. T2 = 127 + 273.15 = 400.15 K.
Step 2. P2 = P1 × T2/T1 = 2.00 × 400.15 ÷ 300.15 = 2.67 atm.
Check. With the rounded values 300 K and 400 K: 2.00 × 400/300 = 2.67 atm. The same answer to three figures. The pressure went up by a factor 1.333, the same factor as the temperature in kelvin.
Worked example 2: finding the temperature for a given pressure
Question. A sealed can holds air at 1.00 atm and 20°C. At what temperature will the pressure reach 1.50 atm?
Step 1. T1 = 20 + 273.15 = 293.15 K.
Step 2. T2 = T1 × P2/P1 = 293.15 × 1.50 ÷ 1.00 = 439.7 K.
Step 3. Convert back: 439.7 − 273.15 = 166.6°C.
Answer. About 167°C. Notice that going from 20°C to 167°C is not a “50% rise” in Celsius. The pressure ratio equals the kelvin ratio, so the temperature in kelvin rises by 50%, from 293 K to 440 K.
Worked example 3: cooling
Question. A gas sample in a rigid flask is at 760 mmHg and 25°C. It is cooled to −20°C. Find the pressure in mmHg.
Step 1. T1 = 25 + 273.15 = 298.15 K. T2 = −20 + 273.15 = 253.15 K.
Step 2. P2 = 760 × 253.15 ÷ 298.15 = 645.3 mmHg.
Check. 253.15/298.15 = 0.8491, so the pressure fell by about 15%. 760 × 0.8491 = 645.3. Same.
Worked example 4: what happens if you use °C by mistake
Question. Repeat example 1 using Celsius values directly, and compare.
Wrong working. P2 = 2.00 × 127 ÷ 27 = 9.41 atm.
Correct answer. 2.67 atm (example 1).
The wrong method gives an answer more than three times too large. Even a small rise in temperature cannot multiply the pressure by 4.7. A quick sense check like this catches the mistake before you lose the marks.
A second use: pressure rise as a percentage
Sometimes a question asks “by what percent does the pressure rise?” Take a gas at 5.00 atm and 30°C heated to 80°C. T1 = 303.15 K, T2 = 353.15 K. P2 = 5.00 × 353.15 ÷ 303.15 = 5.82 atm. The rise is (353.15/303.15 − 1) × 100 = 16.5%. A 50°C heating gives a 16.5% rise here, because the starting temperature is already high in kelvin. Heating from a lower start would give a larger percent.
Gauge pressure and absolute pressure. Tyre pressure gauges and cylinder gauges often show gauge pressure, which is pressure above the atmosphere. The law needs absolute pressure. If a problem gives a gauge reading, add the atmospheric pressure first. Most textbook questions give absolute pressure directly, so check the wording.
Common mistakes
- Using °C. Always convert to kelvin before dividing.
- Inverting the ratio. Heating raises pressure, so T2/T1 must be greater than 1 when T2 is higher. If your pressure goes down on heating, you flipped the fraction.
- Applying it to an open container. If the gas can escape or the volume can change (a balloon, a piston), then n or V is not constant and P/T is not constant.
- Mixing units. P1 in atm and P2 in mmHg gives nonsense. Convert so both match.
- Forgetting gauge vs absolute pressure. Add atmospheric pressure to a gauge reading first.
Exam relevance
| Question wording | What to do |
|---|---|
| Rigid / sealed container heated | Use P1/T1 = P2/T2, T in kelvin |
| Find the final temperature | T2 = T1 × P2/P1, then convert to °C if asked |
| Both P and V change | Use the combined gas law |
| Amount of gas also changes | Go back to PV = nRT |
| Gauge pressure given | Convert to absolute first |
Check the current syllabus of your board for how many figures to keep and whether 273 or 273.15 is expected. The method does not change.
Check P–T numericals with the Ideal Gas Law tool. Keep n and V the same in both states and compare the pressures.
Open the Ideal Gas Law Calculator →Previous-year CSIR-NET questions and timed quizzes to check what you just read.
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