Gas Pressure

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 (CNX_Chem_09_01_Pressure1). 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, 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 exerts a greater pressure in CNX_Chem_09_01_Pressure2—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:

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:

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

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°). fs-idp189967312 provides some information on these and a few other common units for pressure measurements

Conversion of Pressure Units

(a) torr

(b) atm

(c) kPa

(d) mbar

Solution

(a) $29.2\text{in Hg}\times \frac{\text{25.4}\text{mm}}{1\text{in}}\times \frac{\text{1 torr}}{1\text{mm Hg}}=\text{742 torr}$

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

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

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

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We can measure atmospheric pressure, the force exerted by the atmosphere on the earth’s surface, with a barometer (CNX_Chem_09_01_Barometer). 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 $\frac{1}{13.6}$ 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:

where h is the height of the fluid, ρ (lowercase Greek letter rho) is the density of the fluid, and g is acceleration due to gravity.

Calculation of Barometric Pressure

Solution

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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 (CNX_Chem_09_01_Manometer) 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.

Calculation of Pressure Using a Closed-End Manometer

(a) torr

(b) Pa

(c) bar

View original figure (external; image not reproduced in this text edition)

Original figure description: A diagram of a closed-end manometer is shown. To the upper left is a spherical container labeled, “gas.” This container is connected by a valve to a U-shaped tube which is labeled “closed end” at the upper right end. The container and a portion of tube that follows are shaded pink. The lower portion of the U-shaped tube is shaded grey with the height of the gray region being greater on the right side than on the left. The difference in height between the left side and right side is 26.4 c m which is indicated with horizontal line segments and arrows.

Solution

(a) $26.4\text{cm Hg}\times \frac{10\text{mm Hg}}{1\text{cm Hg}}\times \frac{\text{1 torr}}{1\text{mm Hg}}=\text{264 torr}$

(b) $264\text{torr}\times \frac{1\text{atm}}{760\text{torr}}\times \frac{\text{101,325 Pa}}{1\text{atm}}=\text{35,200 Pa}$

(c) $35\text{,200}\text{Pa}\times \frac{\text{1 bar}}{\mathrm{100,000}\text{Pa}}=\text{0.352 bar}$

Check Your Learning

(a) torr

(b) Pa

(c) bar

View original figure (external; image not reproduced in this text edition)

Original figure description: A diagram of a closed-end manometer is shown. To the upper left is a spherical container labeled, “gas.” This container is connected by a valve to a U-shaped tube which is labeled “closed end” at the upper right end. The container and a portion of tube that follows are shaded pink. The lower portion of the U-shaped tube is shaded grey with the height of the gray region being greater on the right side than on the left. The difference in height of 6.0 i n is indicated with horizontal line segments and arrows.

Calculation of Pressure Using an Open-End Manometer

(a) mm Hg

(b) atm

(c) kPa

View original figure (external; image not reproduced in this text edition)

Original figure description: A diagram of an opne-end manometer is shown. To the upper left is a spherical container labeled, “gas.” This container is connected by a valve to a U-shaped tube which is labeled “open end” at the upper right end. The container and a portion of tube that follows are shaded pink. The lower portion of the U-shaped tube is shaded grey with the height of the gray region being greater on the right side than on the left. The difference in height of 13.7 c m is indicated with horizontal line segments and arrows.

Solution

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

(b) $897\text{mm Hg}\times \frac{\text{1 atm}}{760\text{mm Hg}}=\text{1.18 atm}$

(c) $1.18\text{atm}\times \frac{\text{101.325 kPa}}{1\text{atm}}=1.20\times {10}^2\text{kPa}$

Check Your Learning

(a) mm Hg

(b) atm

(c) kPa

View original figure (external; image not reproduced in this text edition)

Original figure description: A diagram of an open-end manometer is shown. To the upper left is a spherical container labeled, “gas.” This container is connected by a valve to a U-shaped tube which is labeled “open end” at the upper right end. The container and a portion of tube that follows are shaded pink. The lower portion of the U-shaped tube is shaded grey with the height of the gray region being greater on the left side than on the right. The difference in height of 4.63 i n is indicated with horizontal line segments and arrows.