Pressure in liquids and gases

The same load barely marks a surface on a broad support but presses into it on a narrow one. Underwater, pressure increases even if you hold no load. How are contact area, the weight of liquid and a pressure gauge reading connected?

Divide the support into equal small patches. With a uniformly distributed force, a smaller total area gives each patch a larger load. Pressure p=F/S is normal force per unit area; use newtons for F and m² for S. For nonuniform contact, this gives average pressure. Doubling force doubles pressure; doubling area halves it. Tangential force is not included in the numerator: it describes a different action on the surface.

One pascal is the pressure of a normal force of 1 N over 1 m²: 1\,\text{Па}=1\,\text{Н/м}^2. A 1 cm² patch is 10 000 times smaller than a square metre, so the same force gives 1\,\text{Н/см}^2=10^4\,\text{Па}. Divide by 10^{-4} m², rather than multiply by it. Physical meaning helps catch mistakes: a smaller area should produce greater pressure.

An open water surface is not free of forces: air presses on it. Water transmits that pressure, and its own weight adds more with depth. A standard reference is normal atmospheric pressure p_0=101325\,\text{Па}=1\,\text{атм}, roughly 10^5 Pa. That corresponds to about a hundred thousand newtons per square metre, but pressure usually acts from several sides, so the force difference matters.

Why not substitute exactly 101325 Pa in every problem? That defines the unit standard atmosphere; actual air pressure depends on altitude and conditions. In a sealed vessel, the pressure above a liquid can be quite different. Use the stated surface pressure and round only as the problem permits. The technical atmosphere is a different unit, so the word atmosphere without a symbol can lead to a wrong conversion.

Imagine a vertical column of stationary liquid with area S and height h. An upward force p_{\text{низ}}S supports it, a force p_{\text{верх}}S presses down on top, and its weight is \rho Shg. For no acceleration, the difference between the pressure forces must balance the weight. Here \rho is density in kg/m³ and h is depth in metres. The pressure increase with depth follows from equilibrium, not from the shape of the whole vessel.

Divide (p_{\text{низ}}-p_{\text{верх}})S=\rho Shg by the area to get \Delta p=\rho gh. Area cancels, so narrow and wide columns of equal height give the same increase. Doubling depth or density doubles it; at h=0 there is no increase. We assume rest and constant \rho and g. At the same horizontal level in a connected region of this liquid, pressure is equal; otherwise the liquid would start flowing.

If the pressure above water is p_{\text{пов}}, the absolute pressure at depth is p=p_{\text{пов}}+\rho gh. At h=0 the surface pressure remains, not zero. A gauge may display only the difference from ambient air pressure: gauge pressure. Determine which pressure is required before calculating. If the surface and the outside of the sensor share the same air, its contribution cancels from the reading but does not physically vanish.

Find the absolute water pressure at a depth of 2 m when surface pressure is 100 kPa. ρ=1000 кг/м³, g=10 м/с².

Pressure consists of surface pressure plus the contribution from the liquid column.

The increase is ρgh=1000×10×2=20000 Па=20 кПа.

Absolute pressure is p=100+20=120 кПа.

Check: absolute pressure exceeds surface pressure; 20 kPa is only the gauge contribution.

The absolute pressure is 120 kPa, of which 20 kPa is added by the water's weight. A gauge comparing pressure at this depth with the same air above the surface would read 20 kPa.

Pressure compares normal force with area. One force can therefore produce different pressures on different supports.

The increase of pressure with depth balances the weight of the liquid layer; the column area cancels in the calculation.

ρgh is an increase. Absolute pressure includes pressure above the liquid; gauge pressure depends on the reference used by the instrument.

Next, pressure differences will explain the buoyant force.