F. Gases and Atmospheric Chemistry · Gases and Atmospheric ChemistryCharles's Law: Volume, Temperature and Absolute Zero
F3.5 F3.4 F2.3About 65 minutes
Learning goals- I can state Charles's law and write it as V₁/T₁ = V₂/T₂.
- I can explain how extrapolating volume–temperature data leads to absolute zero.
- I can explain Charles's law using the kinetic molecular theory.
- I can solve Charles's law problems using Kelvin temperatures.
Words to know
Charles's law
At constant pressure and amount of gas, volume is directly proportional to Kelvin temperature: V₁/T₁ = V₂/T₂.
direct proportion
A relationship in which one quantity divided by the other is constant; when one doubles, so does the other.
extrapolate
To extend a graph beyond the measured data to predict values.
absolute zero
0 K (−273.15 °C), the temperature at which extrapolated gas volumes reach zero.
The shrinking balloon
Bring a party balloon in from a −20 °C January evening and it looks wrinkled and small. Leave it near a warm window and it plumps up again. Nothing got in or out. The gas inside simply takes up more room when it's warm.
In the 1780s the French scientist and balloonist Jacques Charles studied how gases expand when heated. In 1802 Joseph Louis Gay-Lussac published careful measurements. The relationship is known as Charles's law.
Charles's law
For a fixed amount of gas at constant pressure, the volume is directly proportional to the absolute (Kelvin) temperature.
- Double the Kelvin temperature → the volume doubles.
- Halve the Kelvin temperature → the volume halves.
So V ÷ T is constant, and for a change from state 1 to state 2:
V₁ ÷ T₁ = V₂ ÷ T₂ (constant P and n)
Temperatures must be in kelvins. A gas heated from 10 °C to 20 °C does not double in volume. In kelvins that is 283 K to 293 K, an increase of only about 3.5 %.
Discovering absolute zero
When scientists measured the volume of different gas samples at different Celsius temperatures and drew the graphs, each was a straight line. Then they extrapolated each line (extended it beyond the measured data) to find where the volume would reach zero. Every line, for every gas and every sample size, hit zero volume at the same temperature: about −273 °C.
A real gas would turn into a liquid long before reaching this point, so its volume never actually becomes zero. But the pattern showed that −273 °C is a natural zero for temperature. In 1848 William Thomson (Lord Kelvin) proposed an absolute temperature scale starting there, now called the Kelvin scale.
The KMT explanation
Heating a gas raises the average kinetic energy of its particles. They move faster and hit the walls harder and more often. If the container can expand (a balloon, or a cylinder with a free-moving piston), the gas pushes the walls outward until the pressure inside again equals the pressure outside. The result: same pressure, bigger volume.
Worked exampleExample 1: A balloon in the freezer
A balloon has a volume of 2.00 L at 22 °C. It is placed in a freezer at −18 °C. Assuming constant pressure, what is its new volume?
- Convert to kelvins: T₁ = 22 + 273 = 295 K; T₂ = −18 + 273 = 255 K
- Predict: colder → smaller volume.
- Rearrange V₁/T₁ = V₂/T₂: V₂ = V₁ × T₂ ÷ T₁
- V₂ = 2.00 L × 255 K ÷ 295 K = 1.73 L
- Check: smaller, as predicted. The temperature ratio is 255/295 = 0.86, so the volume is about 86 % of the original.
Worked exampleExample 2: Finding a temperature
A 500 mL sample of gas at 27 °C is heated at constant pressure until its volume is 750 mL. What is the final temperature in °C?
- T₁ = 27 + 273 = 300 K
- Rearrange: T₂ = T₁ × V₂ ÷ V₁ = 300 K × 750 mL ÷ 500 mL = 450 K
- Convert back: t₂ = 450 − 273 = 177 °C
- Check: the volume rose by a factor of 1.5, so the Kelvin temperature must also rise by 1.5. ✓
A problem-solving routine for every gas law
- List every given value with its symbol and units; mark the unknown.
- Convert temperatures to kelvins. Make pressure units match each other, and volume units match each other.
- Decide which quantities are constant; choose the law.
- Predict whether the unknown gets bigger or smaller.
- Rearrange, substitute with units, calculate, and round to the right number of significant figures.
- Check the answer against your prediction.
Hands-on activityHow much does a balloon shrink in the freezer?
You need: a grown-up, a round latex balloon, a piece of string, a ruler or measuring tape, a kitchen thermometer (or use your freezer's setting, usually about −18 °C), a freezer.
Safety: keep balloons and broken balloon pieces away from young children (choking hazard). Anyone with a latex allergy should use a non-latex balloon.
- Blow up the balloon to a medium size and tie it. Leave it at room temperature for 10 minutes. Record the room temperature.
- Wrap the string around the fattest part of the balloon and mark it. Measure the circumference, C₁, with the ruler.
- Put the balloon in the freezer for 30 minutes. Take it out and immediately measure the circumference again, C₂.
- Calculate: for a sphere, V ∝ C³, so V₂/V₁ = (C₂/C₁)³. Compare this ratio with T₂/T₁ (in kelvins).
What to look for: if the room is 21 °C (294 K) and the freezer −18 °C (255 K), Charles's law predicts V₂/V₁ = 255/294 = 0.87. That means C₂/C₁ = 0.87^(1/3) = 0.95: the circumference should shrink by about 5 % (for example, from 60 cm to 57 cm). Your result may differ a bit because the balloon warms up while you measure, the rubber pushes back on the gas, and a balloon isn't a perfect sphere. Those are sources of error worth writing down.
Just for funHot-air balloons work by Charles's law. Heating the air inside the envelope makes it expand, so some air spills out the bottom. The air left inside is less dense than the cooler air outside, so the balloon floats. It is chemistry, with a picnic basket.
Your turn
1State Charles's law in words.
For a fixed amount of gas at constant pressure, volume is directly proportional to the Kelvin temperature.
2A 3.0 L sample of gas at 0 °C is heated to 100 °C at constant pressure. Find the new volume.
T₁ = 273 K, T₂ = 373 K. V₂ = 3.0 × 373 ÷ 273 = 4.1 L
3250 mL of air at 20 °C is warmed to 60 °C at constant pressure. What is its new volume?
T₁ = 293 K, T₂ = 333 K. V₂ = 250 × 333 ÷ 293 = 284 mL
4A 1.50 L balloon at 25 °C is taken outside where it is −30 °C. Find its new volume (constant pressure).
T₁ = 298 K, T₂ = 243 K. V₂ = 1.50 × 243 ÷ 298 = 1.22 L
5A gas occupies 4.0 L at 300 K. At what temperature (in K and °C) will it occupy 6.0 L at the same pressure?
T₂ = 300 × 6.0 ÷ 4.0 = 450 K = 177 °C
6Explain using the KMT why a balloon expands when heated.
The particles gain kinetic energy and hit the inside of the balloon harder and more often. The stretchy balloon expands until the pressure inside again matches the pressure outside.
More practice: warm-up
1If the Kelvin temperature of a gas doubles at constant pressure, what happens to its volume?
2Convert 25 °C to kelvins.
3Write the equation for Charles's law.
4What must stay constant for Charles's law?
Pressure and the amount of gas.
5V₁ = 2.0 L at 200 K. Find V₂ at 400 K (constant P).
More practice: on your own
1A sealed syringe with a freely moving plunger holds 40.0 mL of air at 20 °C. It is placed in 80 °C water. What is the new volume?
T₁ = 293 K, T₂ = 353 K. V₂ = 40.0 × 353 ÷ 293 = 48.2 mL
2A gas is at 25 °C. To what Celsius temperature must it be cooled at constant pressure for its volume to drop to 85 % of the original?
T₂ = 0.85 × 298 K = 253 K, which is −20 °C.
3Explain how extrapolating volume–temperature graphs led to the idea of absolute zero.
V–t graphs for every gas are straight lines. Extended to zero volume, they all meet at about −273 °C, suggesting a natural lowest temperature. This became 0 K on the Kelvin scale.
4A student calculates that heating a gas from 10 °C to 30 °C triples its volume. What mistake did she make, and what is the correct factor?
She used Celsius temperatures (30 ÷ 10 = 3). In kelvins: 303 ÷ 283 = 1.07, so the volume increases by only about 7 %.
5A sealed bag of air is taken from a 22 °C kitchen into a 35 °C car. Will it expand or shrink, and by roughly what percentage?
Expand. 308 ÷ 295 = 1.044, about 4 % larger (if the bag can stretch and the pressure stays the same).
Challenge
1A 2.0 L balloon at 22 °C is dipped into liquid nitrogen (−196 °C). Use Charles's law to predict its volume, then explain why the real balloon shrinks even more.
T₁ = 295 K, T₂ = 77 K. V₂ = 2.0 × 77 ÷ 295 = 0.52 L. In reality the balloon collapses further because at 77 K the gases are not ideal: oxygen (boiling point about 90 K) condenses to a liquid, and nitrogen is at its boiling point, so much of the gas turns to liquid with a tiny volume. When the balloon warms up, it re-inflates.
2By what factor does the volume of a gas change when it is heated from 0 °C to 273 °C at constant pressure?
273 K → 546 K: the Kelvin temperature doubles, so the volume doubles (factor of 2).
3A 2.50 L sample at 15 °C is heated to 100 °C at constant pressure. Find the new volume and the percentage increase.
T₁ = 288 K, T₂ = 373 K. V₂ = 2.50 × 373 ÷ 288 = 3.24 L. Increase = 373/288 = 1.295, about 30 % (29.5 %).
Quiz
1. Charles's law relates which two variables, holding pressure and amount constant?
2. A gas at 200 K is heated to 400 K at constant pressure. Its volume:
3. A 1.0 L sample at 27 °C is heated to 327 °C at constant pressure. The new volume is:
4. Extrapolated V–t lines for all gases meet the temperature axis at about:
5. Why can't a real gas actually reach zero volume when cooled?
6. Heating 2.0 L of gas from 10 °C to 20 °C at constant pressure gives a volume of about:
Big idea (F3.5, F3.4, F2.3): Charles's law, V₁/T₁ = V₂/T₂ at constant pressure and amount, with temperatures in kelvins. The extrapolation to absolute zero shows why the Kelvin scale exists.
How to help: ask "What happens to the volume if the temperature in kelvins doubles?" until it is automatic. Check every answer against a prediction (hotter = bigger).
Common difficulties:
- Using Celsius. Insist on converting first. A negative or wildly large answer is a clue.
- Confusing direct and inverse proportion. Boyle: P up, V down. Charles: T up, V up.
Safety: latex balloons are a choking hazard for young children; supervise and dispose of pieces promptly. Check for latex allergies.