Chapter 9: Gases

# 9.1 Gas Pressure

### Learning Outcomes

• Define the property of pressure
• Define and convert among the units of pressure measurements
• Describe how common tools for measuring gas pressure (i.e. barometer, manometer) work and interpret the results obtained with them

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. 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.

Atmospheric pressure is caused by the weight of the column of air molecules in the atmosphere above an object (Figure 9.1.1). 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: $P=\dfrac{F}{A}.$ 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.

Let’s apply this concept to determine which would be more likely to fall through thin ice in Figure 9.1.2 — 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:

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

The figure skater weighs about 120 lbs, 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:

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

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:

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

## Pressure Units

The SI unit of pressure is the pascal, 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—for example, in car tires. Pressure can also be measured using the unit atmosphere, which originally represented the average sea level air pressure at the approximate latitude of Paris (45°). Table 9.1.1 provides some information on these and a few other common units for pressure measurements

Table 9.1.1. Pressure Units
Unit Name and Abbreviation Definition or Relation to Other Unit
pascal (Pa) 1 Pa = 1 N/m2
recommended IUPAC unit
kilopascal (kPa) 1 kPa = 1000 Pa
pounds per square inch (psi) air pressure at sea level is ~14.7 psi
atmosphere (atm) 1 atm = 101,325 Pa
air pressure at sea level is ~1 atm
bar (bar, or b) 1 bar = 100,000 Pa (exactly)
commonly used in meteorology
millibar (mbar, or mb) 1000 mbar = 1 bar
inches of mercury (in. Hg) 1 in. Hg = 3386 Pa
used by aviation industry, also some weather reports
torr $\text{1 torr}=\dfrac{\text{1}}{\text{760}}\text{atm}$
named after Evangelista Torricelli, inventor of the barometer
millimeters of mercury (mm Hg) $1$ mm Hg $\text{~}1$ torr

### Example 9.1.1: Conversion of Pressure Units

The United States National Weather Service reports pressure in both inches of $\ce{Hg}$ and millibars. Convert a pressure of 29.2 in. $\ce{Hg}$ into:

1. torr
2. atm
3. kPa
4. mbar
Show Solution

This is a unit conversion problem. The relationships between the various pressure units are given in Table 9.1.1.

1. $29.2\cancel{\text{in Hg}}\times \dfrac{\text{760 torr}}{29.92\cancel{\text{in Hg}}}=\text{742 torr}$
2. $742\cancel{\text{torr}}\times \dfrac{\text{1 atm}}{760\cancel{\text{torr}}}=\text{0.976 atm}$
3. $742\cancel{\text{torr}}\times \dfrac{\text{101.325 kPa}}{760\cancel{\text{torr}}}=\text{98.9 kPa}$
4. $98.9\cancel{\text{kPa}}\times \dfrac{1000\cancel{\text{Pa}}}{1\cancel{\text{kPa}}}\times \dfrac{1\cancel{\text{bar}}}{100,000\cancel{\text{Pa}}}\times \dfrac{\text{1000 mbar}}{1\cancel{\text{bar}}}=\text{989 mbar}$

## Measuring Pressure

We can measure atmospheric pressure, the force exerted by the atmosphere on the earth’s surface, with a barometer (Figure 9.1.3). 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.

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 ($\ce{Hg}$) is about 13.6-times denser than water, a mercury barometer only needs to be $\dfrac{1}{13.6}$ 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, $p$:

$p=h\rho g$

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

### Example 9.1.2: 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.

Show Solution

The hydrostatic pressure is given by p = hρg, with h = 760 mm, ρ = 13.6 g/cm3, and g = 9.81 m/s2. Plugging these values into the equation and doing the necessary unit conversions will give us the value we seek. (Note: We are expecting to find a pressure of ~101,325 Pa:)

$101,325N{\text{/m}}^{2}=101,325\dfrac{{\text{kg}\cdot\text{m/s}}^{2}}{{\text{m}}^{2}}=101,325\dfrac{\text{kg}}{{\text{m·s}}^{2}}$

$\begin{array}{rcl}p&=&\left(\text{760 mm}\times \dfrac{\text{1 m}}{\text{1000 mm}}\right)\times \left(\dfrac{\text{13.6 g}}{1{\text{cm}}^{3}}\times \dfrac{\text{1 kg}}{\text{1000 g}}\times \dfrac{{\left(\text{100 cm}\right)}^{3}}{{\left(\text{1 m}\right)}^{3}}\right)\times \left(\dfrac{\text{9.81 m}}{1{\text{s}}^{2}}\right)\\{}&=&\left(\text{0.760 m}\right)\left(13,600{\text{kg/m}}^{3}\right)\left(9.81{\text{m/s}}^{2}\right)=1.01\times {10}^{5}{\text{kg/ms}}^{2}=1.01\times {10}^{5}{N\text{/m}}^{2}\\{}&=&1.01\times {10}^{5}\text{Pa}\end{array}$

#### Check Your Learning

A manometer is a device similar to a barometer that can be used to measure the pressure of a gas trapped in a container. A closed-end manometer is a U-shaped tube with one closed arm, one arm that connects to the gas to be measured, and a nonvolatile liquid (usually mercury) in between. As with a barometer, the distance between the liquid levels in the two arms of the tube (h in the diagram) is proportional to the pressure of the gas in the container. An open-end manometer (Figure 9.1.4) is similar to a closed-end manometer, but one of its arms is open to the atmosphere. In this case, the distance between the liquid levels corresponds to the difference in pressure between the gas in the container and the atmosphere.

### Example 9.1.3: Calculation of Pressure Using a Closed-End Manometer

The pressure of a sample of gas is measured with a closed-end manometer, as shown below.

The liquid in the manometer is mercury. Determine the pressure of the gas in:

1. torr
2. Pa
3. bar
Show Solution

The pressure of the gas is equal to a column of mercury of height 26.4 cm. (The pressure at the bottom horizontal line is equal on both sides of the tube. The pressure on the left is due to the gas and the pressure on the right is due to 26.4 cm $\ce{Hg}$, or mercury.) We could use the equation p = hρg as in Example 9.1.2, but it is simpler to just convert between units using Figure 9.1.1.

1. $26.4\cancel{\text{cm Hg}}\times \dfrac{10\cancel{\text{mm Hg}}}{1\cancel{\text{cm Hg}}}\times \dfrac{\text{1 torr}}{1\cancel{\text{mm Hg}}}=\text{264 torr}$
2. $264\cancel{\text{torr}}\times \dfrac{1\cancel{\text{atm}}}{760\cancel{\text{torr}}}\times \dfrac{\text{101,325 Pa}}{1\cancel{\text{atm}}}=\text{35,200 Pa}$
3. $35\text{,200}\cancel{\text{Pa}}\times \dfrac{\text{1 bar}}{100,000\cancel{\text{Pa}}}=\text{0.352 bar}$

### Example 9.1.4: Calculation of Pressure Using an Open-End Manometer

The pressure of a sample of gas is measured at sea level with an open-end Hg (mercury) manometer, as shown below.

Determine the pressure of the gas in:

1. mm Hg
2. atm
3. kPa
Show Solution

The pressure of the gas equals the hydrostatic pressure due to a column of mercury of height 13.7 cm plus the pressure of the atmosphere at sea level. (The pressure at the bottom horizontal line is equal on both sides of the tube. The pressure on the left is due to the gas and the pressure on the right is due to 13.7 cm of Hg plus atmospheric pressure.)

1. In mm $\ce{Hg}$, this is: 137 mm $\ce{Hg}$ + 760 mm Hg = 897 mm $\ce{Hg}$
2. $897\cancel{\text{mm Hg}}\times \dfrac{\text{1 atm}}{760\cancel{\text{mm Hg}}}=\text{1.18 atm}$
3. $1.18\cancel{\text{atm}}\times \dfrac{\text{101.325 kPa}}{1\cancel{\text{atm}}}=1.20\times {10}^{2}\text{kPa}$

### Key Concepts and Summary

Gases exert pressure, which is force per unit area. The pressure of a gas may be expressed in the SI unit of pascal or kilopascal, as well as in many other units including torr, atmosphere, and bar. Atmospheric pressure is measured using a barometer; other gas pressures can be measured using one of several types of manometers.

#### Key Equations

• $P=\dfrac{F}{A}$
• $p = h\rho{g}$

### Try It

1. The pressure of a sample of gas is measured at sea level with a closed-end manometer. The liquid in the manometer is mercury.

Determine the pressure of the gas in:

1. torr
2. Pa
3. bar
2. A typical barometric pressure in Denver, Colorado, is 615 mm $\ce{Hg}$. What is this pressure in atmospheres and kilopascals?
Show Selected Solutions
1. The pressure of the gas is:
1. $26.4\cancel{\text{cm}}\times \dfrac{10\cancel{\text{mm}}}{1\cancel{\text{cm}}}\times \dfrac{\text{1 torr}}{1\cancel{\text{mm}}}=\text{264 torr}$
2. $\text{264 torr}\times \dfrac{101,\text{325 Pa}}{\text{760 torr}}=35,\text{200 Pa}$
3. $264\cancel{\text{torr}}\times \dfrac{\text{1.01325 bar}}{760\cancel{\text{torr}}}=\text{0.352 bar}$
2. Convert 615 mm Hg to atmospheres using 760 mm Hg = 1 atm. Use 1 atm = 101.325 kPa in the second part

$\text{615 mm Hg}\times \dfrac{1\text{atm}}{760\text{mmHg}}=0.809\text{atm}$

$\text{0.809 atm}\times \dfrac{101.325\text{kPa}}{1\text{atm}}=82.0\text{kPa}$

## Glossary

atmosphere (atm): unit of pressure; 1 atm = 101,325 Pa

bar: (bar or b) unit of pressure; 1 bar = 100,000 Pa

barometer: device used to measure atmospheric pressure

hydrostatic pressure: pressure exerted by a fluid due to gravity

manometer: device used to measure the pressure of a gas trapped in a container

pascal (Pa): SI unit of pressure; 1 Pa = 1 N/m2

pounds per square inch (psi): unit of pressure common in the US

pressure: force exerted per unit area

torr: unit of pressure; $\text{1 torr}=\frac{1}{760}\text{atm}$