Ideal Gases

Statement

• At constant volume and for a fixed amount of an ideal gas:

P ∝ T (in kelvin)

• Equivalently, if subscript 1 marks the initial state and subscript 2 marks a later state:

p₁ / T₁ = p₂ / T₂

• where:
  • p₁, p₂ = pressures in Pa (any consistent unit, as long as both are the same)
  • T₁, T₂ = absolute temperatures in kelvin (always, never °C)

Using the equation

• Convert any Celsius readings to kelvin first by adding 273. Forgetting to do this is one of the most common exam mistakes
• Rearranging for p₂: p₂ = p₁ × (T₂ / T₁)
• Rearranging for T₂: T₂ = T₁ × (p₂ / p₁)

Example — the air in a sealed bicycle tyre is at 1.8 × 10⁵ Pa when the temperature is 290 K. After a long ride, friction has warmed the tyre to 320 K. Assuming the tyre's volume is unchanged, calculate the new pressure.

• Use p₁ / T₁ = p₂ / T₂
• p₂ = p₁ × (T₂ / T₁) = 1.8 × 10⁵ × (320 / 290)
• p₂ = 1.8 × 10⁵ × 1.103 = 1.99 × 10⁵ Pa (about 2.0 × 10⁵ Pa, a noticeable rise)

Example — a rigid gas canister contains air at 2.0 × 10⁵ Pa and 27 °C. The canister is taken into a freezer at −13 °C. Calculate the new pressure.

• Convert temperatures: T₁ = 27 + 273 = 300 K; T₂ = −13 + 273 = 260 K
• p₂ = p₁ × (T₂ / T₁) = 2.0 × 10⁵ × (260 / 300) = 1.73 × 10⁵ Pa