# Pressure

Submitted by ChemPRIME Staff on Thu, 12/16/2010 - 13:19

You are probably familiar with the general idea of pressureForce per unit area; in gases arising from the force exerted by collisions of gas molecules with the wall of the container. from experiences in pumping tires or squeezing balloons. A gasA state of matter in which a substance occupies the full volume of its container and changes shape to match the shape of the container. In a gas the distance between particles is much greater than the diameters of the particles themselves; hence the distances between particles can change as necessary so that the matter uniformly occupies its container. exerts force on any surface that it contacts. The force per unitA particular measure of a physical quantity that is used to express the magnitude of the physical quantity; for example, the meter is the unit of the physical quantity, length. surface area is called the pressure and is represented by P. The symbols F and A represent force and area, respectively.

$\text{Pressure}=\frac{\text{force}}{\text{area}}\text{ }P=\frac{F}{A}$      (1)

As a simple example of pressure, consider a rectangular block of lead which measures 20.0 cm by 50.0 cm by 100.0 cm (Fig. 1). The volume V of the block is 1.00 × 105 cm3, and since the densityThe ratio of the mass of a sample of a material to its volume. ρ of Pb is 11.35 g cm–3, the massA measure of the force required to impart unit acceleration to an object; mass is proportional to chemical amount, which represents the quantity of matter in an object. m is

$m=V\rho =\text{1}\text{.00 }\times \text{ 10}^{\text{5}}\text{ cm}^{\text{3}}\text{ }\times \text{ }\frac{\text{11}\text{.35 g}}{\text{1 cm}^{\text{3}}}\text{ }$

$=\text{1}\text{.135 }\times \text{ 10}^{\text{6}}\text{ g}=\text{1}\text{.135 }\times \text{ 10}^{\text{3}}\text{ kg}$

Figure 1 The pressure exerted by a block of lead on the floor. When the block stands upright, the weight of the block (11.1 kN) is distributed over an area of 0.1 m3. If the block is laid flat, this same force is now exerted over an area 5 times larger, namely, 0.5 m3. The pressure exerted by the block on the floor is 5 times as large when the block is upright than when the block is laid flat, even though the force exerted is the same in both cases.

According to the second law of motion, discovered by British physicist Isaac Newton (1643 to 1727), the force on an object is the productA substance produced by a chemical reaction. of the mass of the object and its acceleration a:

F = ma      (2)

At the surface of the earth, the acceleration of gravity is 9.81 m s–2. Substituting into Eq. (2), we have

F = 1.135 × 103 kg × m s–2 = 11.13 × 10 kg m s–2

The units kilogramThe SI unit for mass. meterThe SI unit for distance or length. per square second are given the name newton inthe International System and abbreviated N. Thus the force which gravity exerts on the lead block (the weight of the block) is 11.13 × 103 N. A block that is resting on the floor will always exert a downward force of 11.13 kN. The pressure exerted on the floor depends on the area over which this force is exerted. If the block rests on the 20.0 cm by 50.0 cm side (Fig. 9.2a), its weight is distributed over an area of 20.0 cm × 50.0 cm = 1000 cm3.

Thus

$P=\frac{F}{A}=\frac{\text{11}\text{.13 kN}}{\text{1000 cm}^{\text{2}}}=\frac{\text{11}\text{.13 kN}}{\text{1000 cm}^{\text{2}}}\text{ }\times \text{ }\left( \frac{\text{100 cm}}{\text{1 m}} \right)^{\text{2}}$

$=\frac{\text{11}\text{.13 kN}}{\text{10}^{\text{3}}\text{ cm}^{\text{2}}}=\frac{\text{10}^{\text{4}}\text{ cm}^{\text{2}}}{\text{1 m}^{\text{2}}}=\text{111}\text{.3 }\frac{\text{kN}}{\text{m}^{\text{2}}}$

$=\text{111}\text{.3 }\times \text{ 10}^{\text{3}}\text{ N m}^{-\text{2}}$

Thus we see that pressure can be measured in units of newtons (force) per square meter (area). The units newton per square meter are used in the International System to measure pressure, and they are given the name pascalThe SI unit for pressure, equal to 1 N m-2 or 1 kg s-2m-1 (abbreviated Pa). Like the newton, the pascal honors a famous scientist, in this case Blaise Pascal (1623 to 1662), one of the earliest investigators of the pressure of liquids and gases.

If the lead block is laid on its side (Fig. 1b), the pressure is altered. The area of contact with the floor is now 50.0 cm × 100.0 cm = 5000 cm2, and so

$P=\frac{F}{A}=\frac{\text{11}\text{.13 }\times \text{ 10}^{\text{3}}\text{ N}}{\text{5000 cm}^{\text{2}}}=\frac{\text{11}\text{.13 }\times \text{ 10}^{\text{3}}\text{ N}}{\text{0}\text{.500 m}^{\text{2}}}$

$=\text{22}\text{.26 }\times \text{ 10}^{\text{3}}\text{ N m}^{-\text{2}}=\text{22}\text{.26 kPa}$

When the block is lying flat, its pressure on the floor (22.26 kPa) is only one-fifth as great as the pressure (111.3 kPa) when it stands on end. This is because the area of contact is 5 times larger.

The air surrounding the earth is pulled toward the surface by gravity in the same way as the lead block we have been discussing. Consequently the air also exerts a pressure on the surface. This is called atmospheric pressure.

The following video shows the "power" of atmospheric pressure. An aspirator is used to pump air out of a metalAn element characterized by a glossy surface, high thermal and electrical conductivity, malleability, and ductility. can. This creates low pressure inside the can, and the pressure due to the atmosphereA unit of pressure equal to 101.325 kPa or 760 mmHg; abbreviated atm. Also, the mixture of gases surrounding the earth. then crushes the can.

EXAMPLE 1 The total mass of air directly above a 30 cm by 140 cm section of the Atlantic Ocean was 4.34 × 103 kg on July 27, 1977. Calculate the pressure exerted on the surface of the water by the atmosphere.

SolutionA mixture of one or more substances dissolved in a solvent to give a homogeneous mixture. First calculate the force of gravitational attraction on the air:

F = ma = 4.34 × 103 kg × 9.81 m s–2 = 4.26 × 104 kg m s–2 = 4.26 × 104 N

The area is

$A=\text{30 cm }\times \text{ 140 cm}=\text{4200 cm}^{\text{2}}\text{ }\times \text{ }\left( \frac{\text{1 m}}{\text{100 cm}} \right)^{\text{2}}\text{ }=\text{0}\text{.42 m}^{\text{2}}$

Thus the pressure is

$P=\frac{F}{A}=\frac{\text{4}\text{.26 }\times \text{ 10}^{\text{4}}\text{ N}}{\text{0}\text{.42 m}^{\text{2}}}=\text{1}\text{.01 }\times \text{ 10}^{\text{5}}\text{ Pa}=\text{101 kPa}$

Because winds may add more air or take some away from the vertical column above a given area on the surface, atmospheric pressure will vary above and below the result obtained in Example 9.1. Pressure also decreases as one moves to higher altitudes. The tops of the Himalayas, the highest mountains in the world at about 8000 m (almost 5 miles), are above more than half the atmosphere. The lower pressure at such heights makes breathing very difficult—even the slightest exertion leaves one panting and weak. For this reason jet aircraft, which routinely fly at altitudes of 8 to 10 km, have equipment to maintain air pressure in their cabins artificially.

It is often convenient to express pressure using a unit which is about the same as the average atmospheric pressure at sea level. As we saw in Example 1, atmospheric pressure is about 101 kPa, and the standard atmosphere (abbreviated atm) is defined as exactly 101.325 kPa. Since this unit is often used, it is useful to remember that

1 atm = 101.325 kPa