• After studying fluids in motion (hydrodynamics), we study fluids at rest (hydrostatics): how a liquid exerts pressure at every point inside it, how we measure the pressure of the atmosphere and of gases, and how pressure applied to a liquid is transmitted.

• Imagine a horizontal plate of area A at depth h below the free surface of a liquid of density ρ.
• The plate carries a vertical column of liquid above it, of height h and cross-section A.

1. Volume of the column = A h
2. Mass of the column = ρ A h
3. Weight of the column = F_g = ρ A h g

1. Because the liquid is at rest and incompressible
2. the force on the plate due to the liquid equals the weight of this column.

So the pressure due to the liquid alone is

P = F/A = ρ A h g / A → P = ρ g h (N/m²)

• If the liquid surface is open to the atmosphere (atmospheric pressure P_a), this pressure is added, so the total (absolute) pressure at depth h is

P = Pa + ρ g h

• The quantity ρgh is the gauge pressure — the pressure due to the liquid only.
• The pressure at a point depends on:

the depth h (greater depth → greater pressure),

• the density of the liquid ρ (greater density → greater pressure at the same depth),
• g and the pressure at the surface P_a.
  It does not depend on the shape or width of the container nor on the area of its base, nor on the amount of liquid.

• 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.

1. Using ρ_Hg = 13 595 kg/m³ at 0 °C and g = 9.8 m/s²:
2. P_a = 1 atm = 0.76 × 13 595 × 9.8 = 1.013 × 10⁵ N/m²

• A manometer measures the difference between the pressure of a gas in a container and the atmospheric pressure.
• It is a U-shaped tube containing a liquid of known density (usually mercury, or water/oil for small pressures).
• One arm is connected to the gas container; the other is open to the air.

• Gas pressure greater than atmospheric the liquid is lower in the arm connected to the gas. Using two points at the same level in the same liquid: P = Pa + ρ g h
• Gas pressure less than atmospheric the liquid in the open arm is lower: P = Pa − ρ g h
• The pressure difference (gauge pressure) is ΔP = P − P_a = ρ g h (positive or negative).
• Choice of manometer liquid.

• Mercury for large pressures (dense, so the height difference is small); a light liquid (water, oil) for small pressure differences, which gives a large and readable height difference.
• The liquid must not react with or dissolve the gas and should have a low vapour pressure.
• Example 1. A mercury manometer connected to a gas container shows the mercury in the open arm 24 cm higher than in the arm connected to the container. The gas pressure P = 76 + 24 = 100 cm Hg = 100/76 = 1.316 atm = 1.316 × 1.013 × 10⁵ = 1.333 × 10⁵ N/m².
• Example 2. If the mercury in the open arm were 18 cm lower, then P = 76 − 18 = 58 cm Hg (below atmospheric: a partial vacuum).