# 9.2: Pressure

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You are probably familiar with the general idea of pressure from experiences in pumping tires or squeezing balloons. A gas exerts force on any surface that it contacts. *The force per unit surface area *is called the **pressure** and is represented by *P*. The symbols *F* and *A* represent force and area, respectively. On the image below, a force is pushing down on the circular area of a barometer. The pressure is then the amount of force pushing on a unit area of the circle of the barometer.

*V*of the block is 1.00 × 10

^{5}cm

^{3}, and since the density ρ of Pb is 11.35 g cm

^{–3}, the mass

*m*is

\[m=V\rho =\text{1}\text{.00 }\times \text{ 10}^{\text{5}}\text{ cm}^{\text{3}}\text{ }\times \text{ }\frac{\text{11}\text{.35 g}}{\text{1 cm}^{\text{3}}}\text{ }\label{1} =\text{1}\text{.135 }\times \text{ 10}^{\text{6}}\text{ g}=\text{1}\text{.135 }\times \text{ 10}^{\text{3}}\text{ kg}\]

**Figure **\(\PageIndex{1}\) *When the block stands upright, the weight of the block (11.1 kN) is distributed over an area of 0.1 m ^{3}. Laid flat, this same force is now exerted over an area 5 times larger (0.5 m^{3}), exerting a pressure 5 times smaller than before, even though the weight of the block remains the same.*

According to the second law of motion, discovered by British physicist Isaac Newton, the force on an object is the product of the mass of the object and its acceleration *a*:

*\[F = ma\label{4}\]*

At the surface of the earth, the acceleration of gravity is 9.81 m s^{–2}. Substituting the mass of the lead block into Eq. \(\ref{4}\), we have

**newton**in the International System and abbreviated N. Thus the force which gravity exerts on the lead block (the

**weight**of the block) is 11.13 × 10

^{3}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. 9.2

*a*), its weight is distributed over an area of 20.0 cm × 50.0 cm = 1000 cm

^{3}. 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**pascal**(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. 1*b*), the pressure is altered. The area of contact with the floor is now 50.0 cm × 100.0 cm = 5000 cm^{2}, and so

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**.

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

Example \(\PageIndex{1}\): Atmospheric Pressure

The total mass of air directly above a 30 cm by 140 cm section of the Atlantic Ocean was 4.34 × 10^{3} kg on July 27, 1977. Calculate the pressure exerted on the surface of the water by the atmosphere.

**Solution** First calculate the force of gravitational attraction on the air:

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

Thus the pressure is \[P=\frac{F}{A}=\frac{\text{4}\text{.26 }\times \text{ 10}^{\text{4}}\text{ N}}{\text{0}\text{.42 m}^{\text{2}}}=\text{1}\text{.01 }\times \text{ 10}^{\text{5}}\text{ Pa}=\text{101 kPa}\]## Contributors
Ed Vitz (Kutztown University), John W. Moore (UW-Madison), Justin Shorb (Hope College), Xavier Prat-Resina (University of Minnesota Rochester), Tim Wendorff, and Adam Hahn.

Ed Vitz (Kutztown University), John W. Moore (UW-Madison), Justin Shorb (Hope College), Xavier Prat-Resina (University of Minnesota Rochester), Tim Wendorff, and Adam Hahn.