Pressure

Definition of pressure

A fluid at rest exerts a force on the surface of contact — a wall, or the bottom of the container. The normal force $F$ exerted by a fluid at rest per unit surface area $A$ of contact is called the pressure $p$ of the fluid.

$$p = \dfrac{F}{A}$$

• SI unit: N m$^{-2}$ = pascal (Pa). Dimensions $[L^{-1}M^{1}T^{-2}]$. Pressure is a scalar quantity.
• Other units: $1$ bar $= 10^{5}$ N m$^{-2}$; $1$ hectapascal (hPa) $= 100$ Pa; $1$ torr $= 1$ mm of mercury column.
• A small weight can exert a large pressure if it acts on a small area — the reason a knife has a sharp edge and a needle a sharp tip. A force of 10 N on 1 cm$^{2}$ gives $10^{5}$ N m$^{-2}$, while the same 10 N on 1 m$^{2}$ gives only 10 N m$^{-2}$.

2.3.1 Pressure due to a liquid column

Consider a vessel filled with a liquid of density $\rho$. Imagine inside it a cylinder of cross sectional area $A$ and height $h$, its top at the free surface. The liquid column exerts on the bottom of the cylinder a force equal to its weight $F = mg$, acting downwards. Therefore

$$p = \dfrac{F}{A} = \dfrac{mg}{A}$$

The mass of the liquid in the cylinder is $m = (\text{volume})\times(\text{density}) = Ah\rho$, so

$$p = \dfrac{Ah\rho g}{A}$$

$$p = h\rho g$$

• The pressure due to a liquid of density $\rho$ at a depth $h$ below the free surface is $h\rho g$.
• The result does not contain $A$: the pressure does not depend on the area of the imaginary cylinder used to derive it.
• Consequently the pressure exerted by a fluid at rest is independent of the shape and size of the container, and $p = h\rho g$ holds for gases as well as liquids.
• For a constant pressure, $h \propto 1/\rho$ — two liquids exerting the same pressure have column heights in the inverse ratio of their densities, $h_{1}/h_{2} = \rho_{2}/\rho_{1}$.

2.3.2 Atmospheric pressure

The Earth's atmosphere is a fluid — air — and exerts a downward force due to its weight. At any point, the atmospheric pressure is the weight of a column of air of unit cross section starting at that point and extending to the top of the atmosphere.

• Atmospheric pressure is highest at sea level and decreases with height, since the air column above becomes shorter.
• The atmospheric pressure at sea level is called normal atmospheric pressure. The density of air becomes negligible beyond about 8 km, so the effective height of the air column may be taken as 8 km.
• A region where the gas pressure is less than atmospheric is called a vacuum. Perfect or absolute vacuum is when no matter at all is present.

2.3.3 Absolute pressure and gauge pressure

Consider a tank of water and, inside it, an imaginary cylinder of horizontal base area $A$ whose top and bottom faces are at heights $x_{1}$ and $x_{2}$ measured upwards from a reference level, with $x_{1} > x_{2}$ and $x_{1} - x_{2} = h$. Three vertical forces act on this cylinder:

• $F_{1}$, downwards on the top face, due to the weight of the water column above it,
• $F_{2}$, upwards on the bottom face, due to the water below it,
• the weight $mg$ of the water enclosed, downwards.

The water is in static equilibrium, so

$$F_{2} = F_{1} + mg$$

With $F_{1} = p_{1}A$, $F_{2} = p_{2}A$ and $m = \rho V = \rho A\,(x_{1} - x_{2})$,

$$p_{2}A = p_{1}A + \rho A g\,(x_{1} - x_{2})$$

$$p_{2} = p_{1} + \rho g\,(x_{1} - x_{2})$$

This one equation gives both the pressure inside a liquid as a function of depth and the atmospheric pressure as a function of height.

Pressure at a depth $h$

Put the top of the cylinder at the free surface: $x_{1} = 0$, $p_{1} = p_{0}$, $x_{2} = -h$, $p_{2} = p$. Then

$$p = p_{0} + h\rho g$$

This is the total or absolute pressure at a depth $h$ below the surface of the liquid. It is the sum of $p_{0}$, the pressure due to the atmosphere acting on the free surface, and $h\rho g$, the pressure due to the liquid.

Gauge pressure

The difference between the absolute pressure and the atmospheric pressure is called the gauge pressure:

$$p - p_{0} = h\rho g$$

The same relation applies above a liquid surface. For a point at a height $d$ above the surface, putting $x_{1} = d$, $p_{1} = p$, $x_{2} = 0$, $p_{2} = p_{0}$ and $\rho = \rho_{\text{air}}$,

$$p = p_{0} - d\rho_{\text{air}}\,g$$

2.3.4 Hydrostatic paradox

When a liquid is poured into any one of a set of interconnected vessels of different shapes, the level of the liquid in all the vessels is found to be the same. Before hydrostatics was understood this was called the hydrostatic paradox: one feels that a wide, slanting vessel holds more liquid and so should push the liquid up into the narrower vessel next to it.

The resolution: the pressure at a point depends only on the height of the liquid column above it, and not on the shape of the vessel. The height of the liquid column is the same in all the vessels, so the pressure at the base of each is the same and the system is in equilibrium.

Looked at through forces: the walls of a slanted vessel push against the liquid with forces perpendicular to the wall at each point. Resolved, the vertical components of these forces act upwards and support the extra liquid, so that only the weight of the liquid in the vertical section directly above the base contributes to the pressure at the base.

2.3.5 Pascal's law

Statement: the pressure applied at any point of an enclosed fluid at rest is transmitted equally and undiminished to every point of the fluid and also to the walls of the container, provided the effect of gravity is neglected.

Experimental proof

Take a vessel with four arms A, B, C and D fitted with frictionless, watertight pistons of cross sectional areas $a$, $2a$, $3a$ and $a/2$, filled with an incompressible fluid. Apply a force $F$ on piston A; the pressure exerted on the liquid is $p = F/a$. The other three pistons are observed to move outward, and to hold them in their original positions forces $2F$, $3F$ and $F/2$ respectively are required. Hence

$$p_{B} = \dfrac{2F}{2a}, \qquad p_{C} = \dfrac{3F}{3a}, \qquad p_{D} = \dfrac{F/2}{a/2}$$

so that $p_{B} = p_{C} = p_{D} = p$. The pressure applied on A is transmitted equally and undiminished to all parts of the fluid and to the walls.

Application 1: Hydraulic lift

A tank containing a fluid is fitted with two pistons $S_{1}$ of small area $A_{1}$ and $S_{2}$ of much larger area $A_{2}$, with $A_{2} \gg A_{1}$. A force $F_{1}$ applied downwards on $S_{1}$ generates a pressure $p = F_{1}/A_{1}$, which is transmitted undiminished to $S_{2}$, where an upward force $F_{2} = pA_{2}$ acts:

$$F_{2} = \left(\dfrac{A_{2}}{A_{1}}\right)F_{1}$$

Since $A_{2} \gg A_{1}$, $F_{2}$ is much larger than $F_{1}$: a heavy load placed on $S_{2}$ can be lifted by a small force on $S_{1}$.

Application 2: Hydraulic brakes

Pressing the brake pedal pushes the piston of the master cylinder forward. The slave cylinder, whose cross section $A_{2}$ is much larger than that of the master cylinder $A_{1}$, has its piston pushed forward too, so as to keep the volume of oil constant. The slave piston pushes the friction pads against the rotating disc connected to the wheel, slowing or stopping the vehicle. The force on the slave cylinder is again

$$F_{2} = p A_{2} = \dfrac{F_{1}}{A_{1}}A_{2} = \left(\dfrac{A_{2}}{A_{1}}\right)F_{1}$$

so a small force on the pedal becomes a large force at the wheel.

2.3.6 Measurement of pressure

Instruments used to measure pressure are called pressure meters, pressure gauges or vacuum gauges.

Mercury barometer

A glass tube about 1 m long and 1 cm in diameter is filled with mercury to the brim and quickly inverted into a dish of mercury. Some mercury spills out, the level in the tube falls, and the gap left at the closed top contains no air: it is Torricelli's vacuum (it does hold a little mercury vapour).

• The pressure at the top of the mercury column, point A, is zero: $p_{A} = 0$.
• Take a point C on the mercury surface in the dish and a point B inside the tube at the same horizontal level. The pressure at C is atmospheric, since it is open to the atmosphere, hence $p_{B} = p_{0}$ too.
• If B is at a depth $h$ below A and $\rho$ is the density of mercury, $p_{B} = p_{A} + h\rho g$, and therefore

$$p_{0} = h\rho g$$

Atmospheric pressure is therefore quoted as the length of a mercury column: $p_{\text{atm}} = 76$ cm of Hg $= 760$ mm of Hg $= 760$ torr.

Open tube manometer

A manometer is a U-shaped tube partly filled with a low density liquid such as water or kerosene — a low density gives a larger level difference and hence a more sensitive reading. One arm is open to the atmosphere and the other is connected to the container D whose pressure $p$ is to be measured.

• Point A in the open arm is at atmospheric pressure $p_{0}$.
• Take B in the open arm at the same level as C, the point exposed to the gas: $p_{C} = p_{B}$.
• $p_{B} = p_{0} + h\rho g$, where $h$ is the height of the liquid column above B and $\rho$ its density.
• By Pascal's principle the pressure at C equals that at D, inside the chamber, so $p = p_{C}$ and

$$p = p_{0} + h\rho g, \qquad p - p_{0} = h\rho g$$

The manometer therefore measures the gauge pressure of the gas in the container.

Standard results

• $p = F/A$; $p = h\rho g$; $p = p_{0} + h\rho g$; gauge pressure $= p - p_{0} = h\rho g$.
• $p_{2} = p_{1} + \rho g (x_{1} - x_{2})$, heights measured upwards.
• Pascal's law and $F_{2} = (A_{2}/A_{1})F_{1}$ for hydraulic lift and brakes.
• Barometer: $p_{0} = h\rho g$; normal atmospheric pressure $= 1.013\times10^{5}$ Pa $=$ 76 cm of Hg.