Fundamentals
F.5 Pressure
Pressure,p, is force, F, divided by the area, A, on which the force is exerted:
Pressure f a
o r r
e c
a e p
A
F (F.4)
When you stand on ice, you generate a pressure on the ice as a result of the grav- itational force acting on your mass and pulling you toward the center of the Earth.
However, the pressure is low because the downward force of your body is spread over the area equal to that of the soles of your shoes. When you stand on skates, the area of the blades in contact with the ice is much smaller, so although your downwardforceis the same, the pressureyou exert is much greater (Fig. F.1).
Pressure can arise in ways other than from the gravitational pull of the Earth on an object. For example, the impact of gas molecules on a surface gives rise to a force and hence to a pressure. If an object is immersed in the gas, it experiences a pressure over its entire surface because molecules collide with it from all directions.
In this way, the atmosphere exerts a pressure on all the objects in it. We are in- cessantly battered by molecules of gas in the atmosphere and experience this bat- tering as the “atmospheric pressure.” The pressure is greatest at sea level because the density of air, and hence the number of colliding molecules, is greatest there.
The atmospheric pressure is very considerable: it is the same as would be exerted by loading 1 kg of lead (or any other material) onto a surface of area 1 cm2. We go through our lives under this heavy burden pressing on every square centimeter of our bodies. Some deep-sea creatures are built to withstand even greater pressures:
at 1000 m below sea level the pressure is 100 times greater than at the surface.
Creatures and submarines that operate at these depths must withstand the equiva- lent of 100 kg of lead loaded onto each square centimeter of their surfaces. The pressure of the air in our lungs helps us withstand the relatively low but still sub- stantial pressures that we experience close to sea level.
When a gas is confined to a cylinder fitted with a movable piston, the posi- tion of the piston adjusts until the pressure of the gas inside the cylinder is equal Fig. F.1 These two blocks of
matter have the same mass.
They exert the same force on the surface on which they are standing, but the block on the right exerts a stronger pressure because it exerts the same force over a smaller area than the block on the left.
to that exerted by the atmosphere. When the pressures on either side of the piston are the same, we say that the two regions on either side are in mechanical equilib- rium. The pressure of the confined gas arises from the impact of the particles: they batter the inside surface of the piston and counter the battering of the molecules in the atmosphere that is pressing on the outside surface of the piston (Fig. F.2).
Provided the piston is weightless (that is, provided we can neglect any gravitational pull on it), the gas is in mechanical equilibrium with the atmosphere whatever the orientation of the piston and cylinder, because the external battering is the same in all directions.
The SI unit of pressure is the pascal, Pa:
1 Pa1 kg m1s2
The pressure of the atmosphere at sea level is about 105Pa (100 kPa). This fact lets us imagine the magnitude of 1 Pa, for we have just seen that 1 kg of lead rest- ing on 1 cm2on the surface of the Earth exerts about the same pressure as the atmo- sphere; so 1/105of that mass, or 0.01 g, will exert about 1 Pa, we see that the pas- cal is rather a small unit of pressure. Table F.1 lists the other units commonly used to report pressure.1One of the most important in modern physical chemistry is the bar, where 1 bar105Pa exactly. Normal atmospheric pressure is close to 1 bar.
EXAMPLE F.1 Converting between units
A scientist was exploring the effect of atmospheric pressure on the rate of growth of a lichen and measured a pressure of 1.115 bar. What is the pressure in atmospheres?
Strategy Write the relation between the “old units” (the units to be replaced) and the “new units” (the units required) in the form
1 old unitxnew units
then replace the “old unit” everywhere it occurs by “xnew units” and multiply out the numerical expression.
Solution From Table F.1 we have 1.013 25 bar1 atm
Table F.1
Pressure units and conversion factors*pascal, Pa 1 Pa1 N m2
bar 1 bar105Pa
atmosphere, atm 1 atm101.325kPa1.013 25bar torr, Torr† 760Torr1 atm
1 Torr133.32 Pa
*Values in bold are exact.
†The name of the unit is torr; its symbol is Torr.
Inside Outside
Fig. F.2 A system is in mechanical equilibrium with its surroundings if it is separated from them by a movable wall and the external pressure is equal to the pressure of the gas in the system.
1SeeAppendix 1for a fuller description of the units.
Vacuum
Hydrostatic pressure
External pressure h
Fig. F.3 The operation of a mercury barometer. The space above the mercury in the vertical tube is a vacuum, so no pressure is exerted on the top of the mercury column;
however, the atmosphere exerts a pressure on the mercury in the reservoir and pushes the column up the tube until the pressure exerted by the mercury column is equal to that exerted by the atmosphere.
The height, h, reached by the column is proportional to the external pressure, so the height can be used as a measure of this pressure.
with atm the “new unit” and bar the “old unit.” As a first step we write 1 bar
1.01 1
3 25atm
Then we replace bar wherever it appears by (1/1.013 25) atm:
p1.115 bar1.115
冢
1.0113 25atm冣
1.100 atmA note on good practice:The number of significant figures in the answer (4) is the same as the number of significant figures in the data; the relation between old and new units in this case is exact.
SELF-TEST F.1 The pressure in the eye of a hurricane was recorded as 723 Torr. What is the pressure in kilopascals?
Answer:96.4 kPa ■
Atmospheric pressure (a property that varies with altitude and the weather) is measured with a barometer, which was invented by Torricelli, a student of Galileo’s.
A mercury barometer consists of an inverted tube of mercury that is sealed at its upper end and stands with its lower end in a bath of mercury. The mercury falls until the pressure it exerts at its base is equal to the atmospheric pressure (Fig. F.3).
We can calculate the atmospheric pressure pby measuring the height hof the mer- cury column and using the relation (see DerivationF.1)
pgh (F.5)
where(rho) is the mass density (commonly just “density”), the mass of a sample divided by the volume it occupies:
(F.6)
With the mass measured in kilograms and the volume in meters cubed, density is reported in kilograms per cubic meter (kg m3); however, it is equally acceptable and often more convenient to report mass density in grams per cubic centimeter (g cm3) or grams per milliliter (g mL1). The relation between these units is
1 g cm31 g mL1103kg m3
Thus, the density of mercury may be reported as either 13.6 g cm3(which is equiv- alent to 13.6 g mL1) or as 1.36 104kg m3.
DERIVATION F.1 Hydrostatic pressure
The strategy of the calculation is to relate the mass of the column to its height, to calculate the downward force exerted by that mass, and then to divide the force by the area over which it is exerted. Consider Fig. F.4. The volume of a cylinder of liquid of height hand cross-sectional area AishA. The mass, m, of this cylinder of liquid is the volume multiplied by the density, , of the liquid, orm hA. The downward force exerted by this mass is mg, where gis the acceleration of free fall, a measure of the Earth’s gravitational pull on an object.
m V
Therefore, the force exerted by the column is hA g. This force acts over the area Aat the foot of the column, so according to eqn F.4, the pressure at the base is hAgdivided by A, which is eqn F.5.
ILLUSTRATION F.1 Calculating a hydrostatic pressure
The pressure at the foot of a column of mercury of height 760 mm (0.760 m) and density 13.6 g cm3(1.36 104kg m3) is
p(9.81 m s2) (1.36 104kg m3) (0.760 m) 1.01 105kg m1s21.01 105Pa
This pressure corresponds to 101 kPa (1.00 atm).
A note on good practice:Write units at everystage of a calculation and do not sim- ply attach them to a final numerical value. Also, it is often sensible to express all numerical quantities in terms of base units when carrying out a calculation. ■