CalculateY
Quick Answer
A gas at 10 L and 300 K heated to 600 K expands to an estimated 20 L. A balloon at 2 L and 293 K (20 C) cooled to 253 K (-20 C) shrinks to approximately 1.73 L.
Common Examples
| Input | Result |
|---|---|
| V₁ = 10 L, T₁ = 300 K, T₂ = 600 K | V₂ = 20 L |
| V₁ = 2 L, T₁ = 293 K, T₂ = 253 K | V₂ = 1.73 L |
| T₁ = 273 K, V₂ = 15 L, T₂ = 373 K | V₁ = 10.98 L |
| V₁ = 5 L, V₂ = 8 L, T₂ = 400 K | T₁ = 250 K |
| V₁ = 22.4 L, T₁ = 273 K, V₂ = 24.5 L | T₂ = 298.79 K (25.64 C) |
How It Works
Charles’s law
Discovered by French physicist Jacques Charles around 1787 and later published by Joseph Louis Gay-Lussac in 1802, Charles’s law describes the relationship between the volume and temperature of a gas held at constant pressure.
V₁/T₁ = V₂/T₂
Where:
- V₁ = initial volume in liters (L)
- T₁ = initial temperature in Kelvin (K)
- V₂ = final volume in liters (L)
- T₂ = final temperature in Kelvin (K)
Rearranged forms for each unknown:
- V₂ = V₁ × T₂ / T₁ (final volume after temperature change)
- V₁ = V₂ × T₁ / T₂ (initial volume before temperature change)
- T₂ = T₁ × V₂ / V₁ (final temperature after volume change)
- T₁ = T₂ × V₁ / V₂ (initial temperature before volume change)
Why Kelvin is required
Charles’s law requires absolute temperature (Kelvin) because gas volume is proportional to temperature only on the absolute scale. At absolute zero (0 K), an ideal gas would theoretically shrink to zero volume. The Celsius and Fahrenheit scales have arbitrary zero points, so using them produces incorrect results.
Converting Celsius to Kelvin: K = C + 273.15
Common reference temperatures: 0 C = 273.15 K (water freezes), 20 C = 293.15 K (room temperature), 100 C = 373.15 K (water boils).
Worked example
A balloon contains 2.5 L of air at 20 C (293.15 K). It is then placed in boiling water and heated to 100 C (373.15 K). What is the new volume?
V₂ = V₁ × T₂ / T₁ = 2.5 × 373.15 / 293.15 = 932.875 / 293.15 = 3.18 L
The balloon expands from 2.5 L to approximately 3.18 L as it heats up.
Assumptions and limitations
Charles’s law applies to ideal gases at constant pressure. It holds well for common gases (nitrogen, oxygen, air) at moderate temperatures and pressures. Near a gas’s condensation point or at very high pressures, the law becomes less accurate. The law also assumes the amount of gas (moles) does not change during the process.