8.1 Gas Pressure

Michele Lansigan

8.1 Gas Pressure

Learning Objectives

By the end of this section, you will be able to:

Define the property of pressure
Define and convert among the units of pressure measurements
Describe the operation of common tools for measuring gas pressure
Calculate pressure from manometer data

The earth’s atmosphere exerts a pressure, as does any other gas. Although we do not normally notice atmospheric pressure, we are sensitive to pressure changes—for example, when your ears “pop” during take-off and landing while flying, or when you dive underwater. Gas pressure is caused by the force exerted by gas molecules colliding with the surfaces of objects (Figure 1). Although the force of each collision is very small, any surface of appreciable area experiences a large number of collisions in a short time, which can result in a high pressure. In fact, normal air pressure is strong enough to crush a metal container when not balanced by equal pressure from inside the container.

**Figure 1.** The atmosphere above us exerts a large pressure on objects at the surface of the earth, roughly equal to the weight of a bowling ball pressing on an area the size of a human thumbnail.

A dramatic illustration of atmospheric pressure is provided in this brief video, which shows a railway tanker car imploding when its internal pressure is decreased.

A smaller scale demonstration of this phenomenon is briefly explained.

Atmospheric pressure is caused by the weight of the column of air molecules in the atmosphere above an object, such as the tanker car. At sea level, this pressure is roughly the same as that exerted by a full-grown African elephant standing on a doormat, or a typical bowling ball resting on your thumbnail. These may seem like huge amounts, and they are, but life on earth has evolved under such atmospheric pressure. If you actually perch a bowling ball on your thumbnail, the pressure experienced is twice the usual pressure, and the sensation is unpleasant.

In general, pressure is defined as the force exerted on a given area: $P=\frac{F}{A}$ . Note that pressure is directly proportional to force and inversely proportional to area. Thus, pressure can be increased either by increasing the amount of force or by decreasing the area over which it is applied; pressure can be decreased by decreasing the force or increasing the area.

Let’s apply this concept to determine which would be more likely to fall through thin ice in Figure 2—the elephant or the figure skater? A large African elephant can weigh 7 tons, supported on four feet, each with a diameter of about 1.5 ft (footprint area of 250 in²), so the pressure exerted by each foot is about 14 lb/in²:

pressure per elephant foot = $14,000 \,\frac{\textup{lb}}{\textup{elephant}}\, \times \frac{1\,\textup{elephant}}{4\,\textup{feet}}\,\times\,\frac{1\,\textup{foot}}{250\,\textup{in}^{2}}\,=\,14\,\textup{lb/in}^{2}$

The figure skater weighs about 120lbs, supported on two skate blades, each with an area of about 2 in², so the pressure exerted by each blade is about 30 lb/in²:

pressure per skate blade = $120 \,\frac{\textup{lb}}{\textup{skater}}\, \times \frac{1\,\textup{skater}}{2\,\textup{blades}}\,\times\,\frac{1\,\textup{blade}}{2\,\textup{in}^{2}}\,=\,30\,\textup{lb/in}^{2}$

Even though the elephant is more than one hundred-times heavier than the skater, it exerts less than one-half of the pressure and would therefore be less likely to fall though thin ice. On the other hand, if the skater removes her skates and stands with bare feet (or regular footwear) on the ice, the larger area over which her weight is applied greatly reduces the pressure exerted:

pressure per human foot = $120 \,\frac{\textup{lb}}{\textup{skater}}\, \times \frac{1\,\textup{skater}}{2\,\textup{feet}}\,\times\,\frac{1\,\textup{foot}}{30\,\textup{in}^{2}}\,=\,2\,\textup{lb/in}^{2}$

**Figure 2.** Although (a) an elephant’s weight is large, creating a very large force on the ground, (b) the figure skater exerts a much higher pressure on the ice due to the small surface area of her skates. (credit a: modification of work by Guido da Rozze; credit b: modification of work by Ryosuke Yagi)

The SI unit of pressure is the pascal (Pa), with 1 Pa = 1 N/m², where N is the newton, a unit of force defined as 1 kg m/s². One pascal is a small pressure; in many cases, it is more convenient to use units of kilopascal (1 kPa = 1000 Pa) or bar (1 bar = 100,000 Pa). In the United States, pressure is often measured in pounds of force on an area of one square inch—pounds per square inch (psi)—for example, in car tires. Pressure can also be measured using the unit atmosphere (atm), which originally represented the average sea level air pressure at the approximate latitude of Paris (45°). Table 1 provides some information on these and a few other common units for pressure measurements

Example 1: Conversion of Pressure Units

The United States National Weather Service reports pressure in both inches of Hg and millibars. Convert a pressure of 29.2 in. Hg into:
(a) torr
(b) atm
(c) kPa
(d) mbar

Solution:

This is a unit conversion problem. The relationships between the various pressure units are given in Table 1.

(a) $29.2\,\textup{in Hg}\,\times\,\frac{24.5\,\textup{mm}}{1\,\textup{in}}\,\times\,\frac{1\,\textup{torr}}{1\,\textup{mm Hg}}\,=\,742\,\textup{torr}$

(b) $742\,\textup{torr}\,\times\,\frac{1\,\textup{atm}}{760\,\textup{torr}}\,=\,0.976\,\textup{ atm}$

(c) $742\,\textup{torr}\,\times\,\frac{101.325\,\textup{kPa}}{760\,\textup{torr}}\,=\,98.9\textup{ kPa}$

(d) $98.9\,\textup{kPa}\,\times\,\frac{1000\,\textup{Pa}}{1\,\textup{kPa}}\,\times\,\frac{1\,\textup{bar}}{100,000\,\textup{Pa}}\,\times\,\frac{1000\,\textup{mbar}}{1\,\textup{bar}}\,=\,989\,\textup{mbar}$

Check Your Learning:

A typical barometric pressure in Kansas City is 740 torr. What is this pressure in atmospheres, in millimeters of mercury, in kilopascals, and in bar?

Answer:

0.974 atm; 740 mm Hg; 98.7 kPa; 0.987 bar

We can measure atmospheric pressure, the force exerted by the atmosphere on the earth’s surface, with a barometer (Figure 3). A barometer is a glass tube that is closed at one end, filled with a nonvolatile liquid such as mercury, and then inverted and immersed in a container of that liquid. The atmosphere exerts pressure on the liquid outside the tube, the column of liquid exerts pressure inside the tube, and the pressure at the liquid surface is the same inside and outside the tube. The height of the liquid in the tube is therefore proportional to the pressure exerted by the atmosphere.

**Figure 3.** In a barometer, the height, h, of the column of liquid is used as a measurement of the air pressure. Using very dense liquid mercury (left) permits the construction of reasonably sized barometers, whereas using water (right) would require a barometer more than 30 feet tall.

If the liquid is water, normal atmospheric pressure will support a column of water over 10 meters high, which is rather inconvenient for making (and reading) a barometer. Because mercury (Hg) is about 13.6-times denser than water, a mercury barometer only needs to be 1/13.6 as tall as a water barometer—a more suitable size. Standard atmospheric pressure of 1 atm at sea level (101,325 Pa) corresponds to a column of mercury that is about 760 mm (29.92 in.) high. The torr was originally intended to be a unit equal to one millimeter of mercury, but it no longer corresponds exactly. The pressure exerted by a fluid due to gravity is known as hydrostatic pressure, p:

p = Hρg

where h is the height of the fluid, ρ is the density of the fluid, and g is acceleration due to gravity.

Example 2: Calculation of Barometric Pressure

Show the calculation supporting the claim that atmospheric pressure near sea level corresponds to the pressure exerted by a column of mercury that is about 760 mm high. The density of mercury = 13.6 g/cm³.