Density & Pressure

Mass and volume are extensive: saw an aluminium block in half and each piece has half of both. Their ratio is intensive: both pieces are still 2700 kg m⁻³. Fluids need that kind of quantity, because there is rarely a natural sample to weigh. The definition ρ = dm/dV hides a choice of scale. Sea-level air holds about 2.5 × 10²⁵ molecules per cubic metre, so a sampling cube 1 μm across contains 2.5 × 10⁷ of them. Molecules drifting in and out change that count by about √N = 5000, a wobble of 0.02%, and the ratio of mass to volume is steady. Shrink the cube to 10 nm and it holds about 25 molecules, a count that flickers by 20% from moment to moment, and density has lost its meaning. The dV of fluid mechanics lives between those scales: tiny beside the pipe or the tank, enormous beside the molecular spacing. Wherever that window exists, ρ is a smooth function of position and the fluid can be treated as a continuum.

The SI unit is kg m⁻³, but tables and lab work often use g cm⁻³. Since 1 g = 10⁻³ kg and 1 cm³ = 10⁻⁶ m³, 1 g cm⁻³ = 1000 kg m⁻³. Specific gravity removes the unit entirely: SG = ρ/ρ_w with ρ_w = 1000 kg m⁻³, so ice is 0.917, ethanol 0.79, aluminium 2.70, mercury 13.6, and air at 20 °C a mere 0.0012. Densities are not constants of nature. Water falls from 1000 kg m⁻³ at 4 °C to 998 at 20 °C and 958 at 100 °C, a change of 4%. Squeezing it is far harder: with a bulk modulus of 2.2 × 10⁹ Pa, an extra 1.0 × 10⁷ Pa raises its density by only 1.0 × 10⁷/2.2 × 10⁹ = 0.45%, which is why liquids are modelled as incompressible. A gas at fixed temperature has ρ proportional to P, so the same assumption would be badly wrong for air. When ρ varies inside a body the mass is an integral, m = ∫ρ dV, and only a linear profile lets you replace the integral by the average of the two end values.

A contact force on a patch of area A has a part perpendicular to the surface and a part along it. Pressure is the perpendicular part per unit area, P = F⊥/A, in pascals; the parallel part per unit area is shear stress. Area matters as much as force: a 15 N thumb spread over a drawing pin's 0.80 cm² head gives 1.9 × 10⁵ Pa, but the same 15 N at the 0.010 mm² point gives 1.5 × 10⁹ Pa, enough to enter wood. Direction matters too. A concrete slab of density 2400 kg m⁻³ and thickness 0.20 m lying on level ground presses with W/A = ρgt = 2400 × 9.81 × 0.20 = 4.7 × 10³ Pa, whatever its area. Rest it on a ramp tilted at 30° and the weight per contact area is unchanged, but only the normal component makes pressure: P = ρgt cos 30° = 4.1 × 10³ Pa. The other ρgt sin 30° = 2.4 × 10³ Pa is shear, carried by friction. A solid surface can supply that. A fluid at rest cannot: any shear stress, however small, makes it flow, so the force a static fluid exerts on a surface is purely normal.

Isolate a small wedge of fluid at rest: a vertical face of height Δz, a horizontal bottom face of width Δx, and a sloping top face of length Δs at angle θ to the horizontal, all of depth Δy. The surrounding fluid pushes normally on each face, with pressures P₁ on the vertical face, P₂ on the bottom and P₃ on the slope. Horizontally, the slope's push P₃ΔsΔy has component P₃ΔsΔy sin θ, and Δs sin θ = Δz, so balancing it against P₁ΔzΔy gives P₁ = P₃. Vertically, the bottom face must hold up the slope's push and the wedge's weight: P₂ΔxΔy = P₃ΔsΔy cos θ + ½ρgΔxΔzΔy, and with Δs cos θ = Δx this is P₂ = P₃ + ½ρgΔz. Now shrink the wedge. The weight term carries one more power of length than the surface terms, so it vanishes, leaving P₁ = P₂ = P₃ for every θ. At a point, pressure is a single number, the same on a surface facing any way: a scalar. The leftover ½ρgΔz is no error. Followed over a finite height, it becomes the depth dependence of the next topic.

Pressure is the scalar; force is the vector you rebuild from it. On a flat surface under uniform pressure the force has magnitude F = PA and points along the normal, into the surface being pushed. Atmospheric air at 1.013 × 10⁵ Pa pushes on a 0.30 m × 0.40 m window pane with 1.013 × 10⁵ × 0.12 = 1.2 × 10⁴ N, about the weight of a small car. The pane survives because the air on the other side pushes back just as hard, so those two large forces very nearly cancel; glass gives way only when a pump or a passing storm leaves its two sides at different pressures. One number serves every surface that meets at a point. Inside a sealed gas cylinder the two end caps are pushed apart and the curved wall is pushed outward, every patch of the shell carrying the same P. When the pressure changes across a surface, as on the face of a dam, or the surface is curved so that its normals differ, F = PA no longer applies as it stands. Each patch then contributes dF = P dA along its own normal, and the contributions add as vectors, which is a job for an integral.

Why does a gas at rest push on its container? Model it as n molecules per unit volume, each of mass m₀, bouncing elastically off a wall perpendicular to x. A molecule arriving with velocity component vₓ leaves with −vₓ and hands the wall momentum 2m₀vₓ. In a time Δt the molecules with that vₓ that reach an area A are those within a distance vₓΔt of it, and only half of all molecules are heading toward the wall. Adding up, the momentum delivered per unit area per unit time is P = n m₀⟨vₓ²⟩. With no preferred direction ⟨vₓ²⟩ = ⅓⟨v²⟩, and n m₀ is simply the density, so P = ⅓ρ⟨v²⟩. Test it on air: ρ = 1.20 kg m⁻³ at 1.013 × 10⁵ Pa gives vᵣₘₛ = √(3P/ρ) = 503 m s⁻¹, the right size for air molecules at room temperature. The steady push you measure is a hail of about 3 × 10²⁷ impacts on each square metre every second. The formula fails for liquids: water molecules at 20 °C would give ⅓ρ⟨v²⟩ ≈ 1.4 × 10⁸ Pa, yet an open glass sits at 10⁵ Pa, because the attraction between close-packed neighbours, absent from the gas model, cancels nearly all of that push.