Lesson 5: The Pressure Law (Gay-Lussac's Law)

5.1 The experiment

Simple experiment.

• A glass flask closed with a stopper through which a U-shaped tube (A, B) containing mercury passes.
• At first the mercury is at the same level (x and y) in both arms, so the gas pressure is atmospheric; temperature t₁.

1. Immerse the flask in warm water at t₂. The mercury falls in arm A (connected to the flask) and rises in B.
2. Add mercury through a funnel until the level in A returns to mark x, so the volume of the gas is again the same (constant volume).
3. The level in B is now higher than in A by h cm: the pressure of the gas has increased by h cm Hg when the temperature rose from t₁ to t₂.
4. Repeat with different gases and the same rise in temperature. Findings: (a) at constant volume, the pressure increases with temperature; (b) equal pressures of different gases increase by equal amounts for equal rises in temperature.

Coefficient of pressure increase of a gas at constant volume (β_P): the increase in the pressure of a unit pressure measured at 0 °C when the temperature rises by 1 °C at constant volume.

ΔP ∝ P₀ Δt → ΔP = β_P P₀ Δt → β_P = ΔP/(P₀ Δt) = (P₁₀₀ − P₀)/(P₀ × 100)

5.2 Jolly's apparatus

• A spherical thin glass bulb (A) connected to a long capillary tube (B) (bent at two right angles), fixed on a vertical board that stands on a base with levelling screws.
• The other end of the capillary is joined by a rubber tube to a wide tube (C) which can be moved up and down along the board; there is a graduated scale.

Procedure

1. Read the atmospheric pressure from a barometer.
2. Put mercury in the bulb's side to take up part of the volume (about 1/7 of the bulb), so that the volume of the remaining air stays practically constant at all temperatures (mercury expands about 7 times as much as glass).
3. Immerse bulb (A) in a beaker of water; pour mercury into (C) until the mercury in the other branch rises to a fixed mark X.
4. Heat the water to boiling, wait for the temperature to become steady.
5. Move (C) until the mercury in the bulb-side arm returns to mark X (so the volume is again constant). The gas pressure P = P_a + h, where h is the height difference between the surfaces in the two arms.
6. Let the water cool to about 90 °C, adjust (C) so that the mercury returns to X, measure the temperature and height difference and calculate P again.
7. Repeat at several temperatures.
8. Plot pressure (vertical) against temperature in °C (horizontal). A straight line is obtained which, when extended backwards, cuts the temperature axis at −273 °C.
9. Calculate β_P = (P₁₀₀ − P₀)/(P₀ × 100). It is found to be 1/273 per °C for air and the same for all gases.

5.3 The pressure law

The pressure law: At constant volume, the pressure of a fixed mass of gas increases by 1/273 of its pressure at 0 °C for every 1 °C rise in temperature.

Second form (Kelvin scale). From similar triangles: P₁/T₁ = P₂/T₂, i.e.

P / T = constant (fixed mass, constant volume), so P ∝ T

At constant volume, the pressure of a fixed mass of gas is directly proportional to its absolute temperature.

Note that the coefficient of volume expansion at constant pressure α_v and the coefficient of pressure increase at constant volume β_P are equal for all gases: both = 1/273 K⁻¹.

5.4 Worked examples

Example 1. A gas has a pressure of 45 cm Hg at 27 °C (300 K). Its volume is constant. At 127 °C (400 K): P₂ = 45 × 400/300 = 60 cm Hg.

Example 2. A sealed container of air is at 0 °C with pressure 80 cm Hg. It is cooled to −91 °C (182 K): P = 80 × 182/273 = 53.3 cm Hg.

Example 3 (car tyre). A tyre has air at 2.0 atm at 7 °C (280 K). After driving, the air is at 47 °C (320 K). Assuming constant volume, P = 2.0 × 320/280 = 2.29 atm. (That is why tyre pressure should be checked when the tyres are cold.)

Example 4 (find the temperature). The pressure of a gas at constant volume rises from 1 atm at 27 °C to 1.5 atm. T₂ = 300 × 1.5 = 450 K = 177 °C.