Lesson 3: Atmospheric Pressure and the Barometer

3.1 Atmospheric pressure

Atmospheric pressure

• is the pressure produced by the weight of the column of air above a unit area of the Earth's surface.
• Going higher, the column of air above you gets shorter, so the atmospheric pressure decreases with height.

3.2 The mercury barometer (Torricelli)

• Torricelli took a glass tube about 1 m long, closed at one end, completely filled it with mercury, and turned it upside down with its open end under the surface of mercury in a dish.
• The mercury in the tube dropped until the vertical height of its surface above the surface of mercury in the dish was about 0.76 m.

• The space above the mercury is a vacuum (the Torricelli vacuum), apart from a little mercury vapour whose pressure is negligible.

• The vertical height h does not change if the tube is tilted or if the tube is wider or narrower — only the length of mercury in the tube changes when it is tilted.

• Take two points A (outside the tube, at the mercury surface in the dish) and B (inside the tube, at the same horizontal level). The pressure at B = the pressure at A = atmospheric pressure. The pressure at B is produced by the mercury column above it. So

  P_a = ρ_Hg g h

Standard atmospheric pressure is the pressure that supports a column of mercury 0.76 m high at 0 °C (cross-section 1 m²) at sea level; it is the pressure at STP (standard temperature and pressure: 0.76 m Hg and 0 °C), a term often used with gas laws.

Using ρ_Hg = 13 595 kg/m³ at 0 °C and g = 9.8 m/s²:

P_a = 1 atm = 0.76 × 13 595 × 9.8 = 1.013 × 10⁵ N/m²

3.3 Units of pressure

3.4 Factors that do not affect the barometer height

• The height of the mercury column does not depend on the cross-section of the tube, nor on tilting it, nor on the amount of mercury in the dish.
• It depends on the atmospheric pressure, the density of the liquid used and g.
• (A water barometer would need a column 13.6 times taller: about 10.3 m.)

3.5 Worked examples

Example 1. The barometer height on a hilltop is 0.72 m Hg. P_a = 0.72 × 13 600 × 9.8 = 9.60 × 10⁴ N/m².

Example 2. Water barometer height: P_a = ρ_w g h → h = 1.013 × 10⁵/(1000 × 9.8) = 10.3 m.

Example 3. Convert 1.5 atm: in cm Hg = 1.5 × 76 = 114 cm Hg; in Pa = 1.5 × 1.013 × 10⁵ = 1.52 × 10⁵ Pa.

Example 4 (change of height). A mercury barometer reads 76 cm Hg at the bottom of a tower and 74 cm Hg at the top. The pressure difference = 2 cm Hg = 0.02 × 13 600 × 9.8 = 2666 N/m². If the air density is about 1.2 kg/m³ then ρ_air g H = 2666 → H = 2666/(1.2 × 9.8) ≈ 227 m.

3.6 Enrichment: the ear at heights

• The pressure on the two sides of the eardrum is normally equal.
• When we rise quickly (in a lift, aeroplane or climbing a mountain), the outside pressure decreases, and the higher pressure inside the middle ear pushes the eardrum outward, producing a feeling of tension.
• Swallowing or chewing opens the Eustachian tube and equalises the pressure.

Why don't we feel the huge atmospheric pressure (more than 10 tonnes per square metre)? Because the pressure of the fluids inside our bodies equals the outside pressure, so the forces on the surfaces of the body are balanced.