## Gas Pressure

Contents

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 (see the figure below).

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.

### Optional Videos:

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 in the video below.

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: \(P=\phantom{\rule{0.2em}{0ex}}\frac{F}{A}.\) 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 the figure below—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 in^{2}), so the pressure exerted by each foot is about 14 lb/in^{2}:

\(\text{pressure per elephant foot}=\text{14,000}\phantom{\rule{0.2em}{0ex}}\cfrac{\text{lb}}{\text{elephant}}\phantom{\rule{0.4em}{0ex}}×\phantom{\rule{0.4em}{0ex}}\cfrac{\text{1 elephant}}{\text{4 feet}}\phantom{\rule{0.4em}{0ex}}×\phantom{\rule{0.4em}{0ex}}\cfrac{\text{1 foot}}{250\phantom{\rule{0.2em}{0ex}}{\text{in}}^{2}}\phantom{\rule{0.2em}{0ex}}=14\phantom{\rule{0.2em}{0ex}}{\text{lb/in}}^{2}\)

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

\(\text{pressure per skate blade}=120\phantom{\rule{0.2em}{0ex}}\cfrac{\text{lb}}{\text{skater}}\phantom{\rule{0.4em}{0ex}}×\phantom{\rule{0.4em}{0ex}}\cfrac{\text{1 skater}}{\text{2 blades}}\phantom{\rule{0.4em}{0ex}}×\phantom{\rule{0.4em}{0ex}}\cfrac{\text{1 blade}}{2\phantom{\rule{0.2em}{0ex}}{\text{in}}^{2}}\phantom{\rule{0.2em}{0ex}}=30\phantom{\rule{0.2em}{0ex}}{\text{lb/in}}^{2}\)

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:

\(\text{pressure per human foot}=120\phantom{\rule{0.2em}{0ex}}\cfrac{\text{lb}}{\text{skater}}\phantom{\rule{0.4em}{0ex}}×\phantom{\rule{0.4em}{0ex}}\cfrac{\text{1 skater}}{\text{2 feet}}\phantom{\rule{0.4em}{0ex}}×\phantom{\rule{0.4em}{0ex}}\cfrac{\text{1 foot}}{30\phantom{\rule{0.2em}{0ex}}{\text{in}}^{2}}\phantom{\rule{0.2em}{0ex}}=2\phantom{\rule{0.2em}{0ex}}{\text{lb/in}}^{2}\)

The SI unit of pressure is the **pascal (Pa)**, with 1 Pa = 1 N/m^{2}, where N is the newton, a unit of force defined as 1 kg m/s^{2}. 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°). The table below provides some information on these and a few other common units for pressure measurements

Pressure Units | |
---|---|

Unit Name and Abbreviation | Definition or Relation to Other Unit |

pascal (Pa) | 1 Pa = 1 N/m^{2}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 Paair 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 Paused by aviation industry, also some weather reports |

torr | \(\text{1 torr}=\phantom{\rule{0.2em}{0ex}}\cfrac{\text{1}}{\text{760}}\phantom{\rule{0.2em}{0ex}}\text{atm}\)named after Evangelista Torricelli, inventor of the barometer |

millimeters of mercury (mm Hg) | 1 mm Hg ~1 torr |

## Example

### 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 the table above.

(a) \(\require{cancel}29.2\phantom{\rule{0.2em}{0ex}}\cancel{\text{in Hg}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\cfrac{\text{25.4}\cancel{\text{mm}}}{1\phantom{\rule{0.2em}{0ex}}\cancel{\text{in}}}\phantom{\rule{0.2em}{0ex}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\cfrac{\text{1 torr}}{1\phantom{\rule{0.2em}{0ex}}\cancel{\text{mm Hg}}}\phantom{\rule{0.2em}{0ex}}=\text{742 torr}\)

(b) \(742\phantom{\rule{0.2em}{0ex}}\cancel{\text{torr}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\cfrac{\text{1 atm}}{760\phantom{\rule{0.2em}{0ex}}\cancel{\text{torr}}}\phantom{\rule{0.2em}{0ex}}=\text{0.976 atm}\)

(c) \(742\phantom{\rule{0.2em}{0ex}}\cancel{\text{torr}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\cfrac{\text{101.325 kPa}}{760\phantom{\rule{0.2em}{0ex}}\cancel{\text{torr}}}\phantom{\rule{0.2em}{0ex}}=\text{98.9 kPa}\)

(d) \(98.9\phantom{\rule{0.2em}{0ex}}\cancel{\text{kPa}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\cfrac{1000\phantom{\rule{0.2em}{0ex}}\cancel{\text{Pa}}}{1\phantom{\rule{0.2em}{0ex}}\cancel{\text{kPa}}}\phantom{\rule{0.4em}{0ex}}×\phantom{\rule{0.4em}{0ex}}\cfrac{1\phantom{\rule{0.2em}{0ex}}\cancel{\text{bar}}}{100,000\phantom{\rule{0.2em}{0ex}}\cancel{\text{Pa}}}\phantom{\rule{0.4em}{0ex}}×\phantom{\rule{0.4em}{0ex}}\cfrac{\text{1000 mbar}}{1\phantom{\rule{0.2em}{0ex}}\cancel{\text{bar}}}\phantom{\rule{0.2em}{0ex}}=\text{989 mbar}\)

We can measure atmospheric pressure, the force exerted by the atmosphere on the earth’s surface, with a **barometer** (see the figure below). 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 (Hg) is about 13.6-times denser than water, a mercury barometer only needs to be \(\cfrac{1}{13.6}\) 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, *ρ* is the density of the fluid, and *g* is acceleration due to gravity.

## Example

### 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/cm^{3}.

### Solution

The hydrostatic pressure is given by *p* = *hρg*, with *h* = 760 mm, *ρ* = 13.6 g/cm^{3}, and *g* = 9.81 m/s^{2}. 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:)

\(101,325\phantom{\rule{0.2em}{0ex}}N{\text{/m}}^{2}=101,325\phantom{\rule{0.2em}{0ex}}\cfrac{{\text{kg·m/s}}^{2}}{{\text{m}}^{2}}\phantom{\rule{0.2em}{0ex}}=101,325\phantom{\rule{0.2em}{0ex}}\cfrac{\text{kg}}{{\text{m·s}}^{2}}\)

\(p=\left(\text{760 mm}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\cfrac{\text{1 m}}{\text{1000 mm}}\right)\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\left(\cfrac{\text{13.6 g}}{1\phantom{\rule{0.2em}{0ex}}{\text{cm}}^{3}}\phantom{\rule{0.4em}{0ex}}×\phantom{\rule{0.4em}{0ex}}\cfrac{\text{1 kg}}{\text{1000 g}}\phantom{\rule{0.4em}{0ex}}×\phantom{\rule{0.4em}{0ex}}\cfrac{{\left(\text{100 cm}\right)}^{3}}{{\left(\text{1 m}\right)}^{3}}\right)\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\left(\cfrac{\text{9.81 m}}{1\phantom{\rule{0.2em}{0ex}}{\text{s}}^{2}}\right)\)

\(=\left(\text{0.760 m}\right)\left(13,600\phantom{\rule{0.2em}{0ex}}{\text{kg/m}}^{3}\right)\left(9.81\phantom{\rule{0.2em}{0ex}}{\text{m/s}}^{2}\right)=1.01\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{5}\phantom{\rule{0.2em}{0ex}}{\text{kg/ms}}^{2}=1.01\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{5}\phantom{\rule{0.2em}{0ex}}{N\text{/m}}^{2}\)

\(=1.01\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{5}\phantom{\rule{0.2em}{0ex}}\text{Pa}\)

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 (see the figure below) is the same as 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

### Calculation of Pressure Using a Closed-End Manometer

The pressure of a sample of gas is measured with a closed-end manometer, as shown to the right. The liquid in the manometer is mercury. Determine the pressure of the gas in:

(a) torr

(b) Pa

(c) bar

### 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 Hg, or mercury.) We could use the equation *p* = *hρg* as in the example above, but it is simpler to just convert between units using the table above.

(a) \(\require{cancel}26.4\phantom{\rule{0.2em}{0ex}}\cancel{\text{cm Hg}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\cfrac{10\phantom{\rule{0.2em}{0ex}}\cancel{\text{mm Hg}}}{1\phantom{\rule{0.2em}{0ex}}\cancel{\text{cm Hg}}}\phantom{\rule{0.4em}{0ex}}×\phantom{\rule{0.4em}{0ex}}\cfrac{\text{1 torr}}{1\phantom{\rule{0.2em}{0ex}}\cancel{\text{mm Hg}}}\phantom{\rule{0.2em}{0ex}}=\text{264 torr}\)

(b) \(264\phantom{\rule{0.2em}{0ex}}\cancel{\text{torr}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\cfrac{1\phantom{\rule{0.2em}{0ex}}\cancel{\text{atm}}}{760\phantom{\rule{0.2em}{0ex}}\cancel{\text{torr}}}\phantom{\rule{0.4em}{0ex}}×\phantom{\rule{0.4em}{0ex}}\cfrac{\text{101,325 Pa}}{1\phantom{\rule{0.2em}{0ex}}\cancel{\text{atm}}}\phantom{\rule{0.2em}{0ex}}=\text{35,200 Pa}\)

(c) \(35\text{,200}\phantom{\rule{0.2em}{0ex}}\cancel{\text{Pa}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\cfrac{\text{1 bar}}{100,000\phantom{\rule{0.2em}{0ex}}\cancel{\text{Pa}}}\phantom{\rule{0.2em}{0ex}}=\text{0.352 bar}\)

## Example

### 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 to the right. Determine the pressure of the gas in:

(a) mm Hg

(b) atm

(c) kPa

### 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.)

(a) In mm Hg, this is: 137 mm Hg + 760 mm Hg = 897 mm Hg

(b) \(897\phantom{\rule{0.2em}{0ex}}\cancel{\text{mm Hg}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\cfrac{\text{1 atm}}{760\phantom{\rule{0.2em}{0ex}}\cancel{\text{mm Hg}}}\phantom{\rule{0.2em}{0ex}}=\text{1.18 atm}\)

(c) \(1.18\phantom{\rule{0.2em}{0ex}}\cancel{\text{atm}}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\cfrac{\text{101.325 kPa}}{1\phantom{\rule{0.2em}{0ex}}\cancel{\text{atm}}}\phantom{\rule{0.2em}{0ex}}=1.20\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{2}\phantom{\rule{0.2em}{0ex}}\text{kPa}\)