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# 15.3: Gases and Pressure

Difficulty Level: At Grade Created by: CK-12

## Lesson Objectives

The student will:

• define pressure.
• convert requested pressure units.
• read barometers and open and closed-end manometers.

• atmosphere
• barometer
• manometer
• pascal
• torr

## Introduction

As you learned earlier, gases exert a pressure on their surroundings. The pressure is a result of trillions of tiny particles pounding on a surface. This section will define pressure in more details, as well as explore the different ways that pressure is measured.

## Pressure

Pressure is defined as the force exerted divided by the area over which the force is exerted:

\begin{align*}\text{Pressure} = \frac {\text{Force}} {\text{Area}}\end{align*}

The concept of force is quite straightforward. Force can be described as a push or pull that causes an object to change its velocity or to change shape. Pressure, however, is a little trickier to understand. Let's consider an example that will illustrate the concept of pressure. Consider the large man and the smaller woman shown in the figure below.

The man weighs \begin{align*}200\ \mathrm{pounds}\end{align*} and is wearing normal men’s shoes. As he walks, at times all of his weight will be exerted on the heel of one shoe. If that heel is \begin{align*}2 \ \mathrm{inches}\end{align*} long and \begin{align*}2 \ \mathrm{inches}\end{align*} wide, then his weight is exerted over an area of \begin{align*}4\ \mathrm{inches}^2\end{align*}. The pressure exerted by the heel of his shoe would be:

\begin{align*}\text{Pressure} = \frac {\text{Force}} {\text{Area}} = \frac {200. \ \text{pounds}} {4.0 \ \text{inches}^2} = 50. \ \text{lbs}/\text{in}^2\end{align*}

The woman, on the other hand, weighs only \begin{align*}100 \ \mathrm{pounds}\end{align*}, but she is wearing high heels. If the heel on one of her shoes is \begin{align*}1/2 \ \mathrm{inch}\end{align*} by \begin{align*}1/2 \ \mathrm{inch}\end{align*}, then when she walks, there will be times when all her weight is exerted over an area of \begin{align*}1/4 \ \mathrm{inches}^2\end{align*}. The pressure exerted by one of her heels would be:

\begin{align*}\text{Pressure} = \frac {\text{Force}} {\text{Area}} = \frac {100. \ \text{pounds}} {0.250 \ \text{inches}^2} = 400. \ \text{lbs}/\text{in}^2\end{align*}

This huge pressure has little to do with her weight and more to do with the area of her shoe heels. If these two people attempted to walk across the lawn, the 200-pound man would likely have no problem, whereas the 100-pound woman may run into trouble and have her heels sink into the grass. Pressure is not just about the total force exerted, but also about the area over which it exerted. This is the reason that nails and tent pegs are sharpened on the end. If the end were blunt, the force exerted by a hammer would be insufficient to generate enough pressure to cause the object to be pounded into a piece of wood or the ground.

## Atmospheric Pressure

The tremendous pressure that can be exerted by gaseous molecules was once demonstrated by a German physicist named Otto von Guericke, who was the inventor of the air pump. Von Guericke placed two hemispheres about the size of dinner plates together and pumped the air out from between them. Before pumping, the pounding of molecules on both sides of the hemispheres is balanced. When the air between the hemispheres is removed, however, there are no air molecules on the inside of the hemispheres to balance the pounding of air molecules on the outside. As a result, there are only air molecules on the the outside pushing the hemispheres together. The force holding the two hemispheres is so strong, teams of horses were unable to pull the hemispheres apart. When von Guericke opened a valve and allowed air back inside, he could easily separate the hemispheres by hand.

The air molecules in our atmosphere exert pressure on every surface they contact. The air pressure of our atmosphere at sea level is approximately \begin{align*}15 \ \mathrm{lbs/in}^2\end{align*}. This pressure is unnoticed because the air is not only on the outside of the surfaces, it is also inside, which allows the atmospheric air pressure to be balanced. The pressure exerted by our atmosphere quickly will become noticed, however, if the air is removed or reduced inside an object. A common demonstration of air pressure is to use a one-gallon metal can. The can has a few drops of water placed inside the can that is heated. When the water inside the can vaporizes, the water vapor expands to fill the can and pushes air out. The lid is then tightly sealed on the can. As the can cools, the water vapor inside condenses back to liquid water. Because the can is sealed, air molecules cannot get back inside. As a result, the air pressure outside the can slowly crushes the can flat. People, of course, also have atmospheric pressure pressing on them. An average-sized person probably has a total force in excess of \begin{align*}25,000 \ \mathrm{pounds}\end{align*} exerted on them. Fortunately, people also have air inside them to balance the external force exerted by the atmosphere. A device to measure atmospheric pressure, called the barometer, was invented in 1643 by an Italian scientist named Evangelista Torricelli (1608 – 1647), who had been a student of Galileo. Torricelli’s barometer consists of a glass tube filled with liquid mercury placed in a dish of mercury, as shown in the diagram below.

An empty glass tube with one end opened and the other end closed was completely filled with liquid mercury. The dish was also filled two-thirds of the way full with mercury. The open end of the tube was covered by a finger before it was inverted and submerged into the dish of mercury. Since the open end was covered, no air could get into the tube. When the finger was removed, the mercury in the tube fell to a height such that the difference between the surface of the mercury in the dish and the top of the mercury column in the tube was \begin{align*}760\ \mathrm{millimeters.}\end{align*}

The reason why mercury stays in the tube is because there are air molecules pounding on the surface of the mercury in the dish but not on the top of the mercury in the tube. The volume of empty space in the tube is a vacuum, so there are no air molecules available to exert a balancing pressure. The weight of the mercury in the tube divided by the area of the opening in the tube is exactly equal to the atmospheric pressure. The diameter of the tube makes no difference in determining the atmospheric pressure because doubling the diameter of the tube doubles the volume of mercury in the tube and the weight of the mercury. It also doubles the area over which the force is exerted, so the pressure will be the same for all tubes. No matter the size of the tube you might choose, the air pressure will hold the mercury to the same height.

The height to which the mercury is held would only be \begin{align*}760.\ \mathrm{millimeters}\end{align*} under standard conditions at sea level. The atmospheric pressure changes with the weather conditions, and the height of the mercury in the barometer will reflect the change. Atmospheric pressure also varies with altitude. Higher altitudes have lower air pressure because the air is “thinner” – fewer air molecules per unit volume. In the mountains, at an altitude of \begin{align*}9600\ \mathrm{feet}\end{align*}, the normal atmospheric pressure will only support a mercury column of \begin{align*}520\ \mathrm{millimeters}\end{align*}.

## Measuring Gas Pressure

There are many different units for measuring and expressing gas pressure. You will need to be familiar with most of them so that you can convert between them easily. Because instruments for measuring pressure often contain a column of mercury, the most commonly used units for pressure are based on the height of the mercury column that the gas can support. As a result, one unit for gas pressure is \begin{align*}\mathrm{mm \ of \ Hg}\end{align*} (millimeters of mercury). Standard atmospheric pressure at sea level is \begin{align*}760. \ \mathrm{mm \ of \ Hg}\end{align*}. This unit can be somewhat of a problem. Students, in particular, occasionally leave off the “of Hg,” causing the unit to look like a length unit. To eliminate this problem, the unit was given another name called the torr, named in honor of Torricelli. One torr is equivalent to one mm of Hg, so \begin{align*}760\ \mathrm{torr}\end{align*} is exactly the same as \begin{align*}760\ \mathrm{mm \ of \ Hg}\end{align*}. For certain works, it is more convenient to express gas pressure in terms of multiples of normal atmospheric pressure at sea level. The unit atmosphere (atm) was introduced. Pressures expressed in mm in Hg can be converted to atmospheres using the conversion factor \begin{align*}1.00\end{align*} \begin{align*}\ \mathrm{atmosphere} = 760. \ \mathrm{mm \ of \ Hg}\end{align*}.

\begin{align*}1.00\ \text{atm} = 760.\ \text{mm of Hg} = 760.\ \text{torr}\end{align*}

Recall that pressure is defined as force divided by area. In physics, force is expressed in a unit called newton (N), and area is expressed in meters2 (m2). Therefore, pressure in physics is expressed in \begin{align*}\mathrm{Newtons/meter}^2\end{align*} \begin{align*}(\text{N/m}^2)\end{align*}. This has also been renamed the pascal (Pa) and is the SI unit for pressure.

\begin{align*}1.00\ \text{atm} = 101,325\ \text{N/m}^2 = 101,325\ \text{Pa} = 760\ \text{mm of Hg} = 760\ \text{torr}\end{align*}

As it happens, one Pascal is an extremely small pressure, so it is convenient to use kilopascals (kPa) when expressing gas pressure. Therefore, \begin{align*}1.00\ \mathrm{atm}= 101.325\ \mathrm{kPa}\end{align*}.

Example:

Convert \begin{align*}425\ \mathrm{torr}\end{align*} to \begin{align*}\mathrm{atm}\end{align*}.

Solution:

The conversion factor is \begin{align*}760.\ \mathrm{torr} = 1.00\ \mathrm{atm}\end{align*}.

\begin{align*}(425\ \text{torr}) \cdot \left (\frac {1.00\ \text{atm}} {760.\ \text{torr}}\right ) = 0.559\ \text{atm}\end{align*}

Example:

Convert \begin{align*}425\ \mathrm{torr}\end{align*} to \begin{align*}\mathrm{kPa}\end{align*}.

Solution:

The conversion factor is \begin{align*}760.\ \mathrm{torr} = 101.325\ \mathrm{kPa}\end{align*}.

\begin{align*}(425\ \text{torr}) \cdot \left (\frac {101.325\ \text{kPa}} {760.\ \text{torr}}\right ) = 56.7\ \text{kPa}\end{align*}

Example:

Convert \begin{align*}0.500\ \mathrm{atm}\end{align*} to \begin{align*}\mathrm{mm}\end{align*} of \begin{align*}\mathrm{Hg}\end{align*}.

Solution:

The conversion factor is \begin{align*}1.00\ \mathrm{atm}= 760.\ \mathrm{mm}\end{align*} of \begin{align*}\mathrm{Hg}\end{align*}.

\begin{align*}(0.500\ \text{atm}) \cdot \left (\frac {760.\ \text{mm of Hg}} {1.00\ \text{atm}}\right ) = 380.\ \text{mm of Hg}\end{align*}

Example:

Convert \begin{align*}0.500\ \mathrm{atm}\end{align*} to \begin{align*}\mathrm{kPa}\end{align*}.

Solution:

The conversion factor is \begin{align*}1.00\ \mathrm{atm}= 101.325\ \mathrm{kPa}\end{align*}.

\begin{align*}(0.500\ \text{atm}) \cdot \left (\frac {101.325\ \text{kPa}} {1.00\ \text{atm}}\right ) = 50.7\ \text{kPa}\end{align*}

You might notice that if you want to measure a gas pressure around \begin{align*}2.0\ \mathrm{atm}\end{align*} with a barometer, you would need a glass column filled with mercury that was over \begin{align*}1.5\ \mathrm{meters}\end{align*} high. That would be a fragile and dangerous instrument, as mercury fumes are toxic. If we used water (which is one-thirteenth as dense as mercury) instead, the column would have to be \begin{align*}50\ \mathrm{feet}\end{align*} high. As a more practical alternative, instruments called manometers have been designed to measure gas pressure in flasks. There are two kinds of manometers used: open-end manometers and closed-end manometers.

We will look at closed-end manometers (illustrated below) first, as they are easier to read. As indicated in the diagram below, the empty space above the mercury level in the tube is a vacuum. Therefore, there are no molecules pounding on the surface of the mercury in the tube. In manometer A, the flask does not contain any gas, so there are no molecules to exert a pressure. In other words, \begin{align*}\text{P}_{gas} = 0\end{align*}. The mercury level in the outside arm balances the mercury level in the inside arm, so the two mercury levels will be exactly even.

We let the flask of manometer B contain a gas at \begin{align*}1.00\ \mathrm{atm}\end{align*} pressure. The mercury level in the outside tube (the arm further from the flask) will rise to a height of \begin{align*}760\ \mathrm{mm \ of \ Hg}\end{align*}. The excess mercury in the outside tube balances the gas pressure in the flask. In manometer C, we would read the gas pressure in the flask as \begin{align*}200.\ \mathrm{mm \ of \ Hg}\end{align*}. In closed-end manometers, the excess mercury is always in the outside tube, and the height difference in mercury levels will equal the gas pressure in the flask.

In the open-end manometers illustrated below, the open-end of the tube allows atmospheric pressure to push down on the top of the column of mercury. In manometer A, the pressure inside the flask is equal to atmospheric pressure. The two columns of mercury balance each other, so they are at the same height. Therefore, the atmospheric pressure pushing on the outside column of mercury must equal the gas pressure in the flask pushing on the inside column of mercury.

In order to properly read an open-end manometer, you must know the actual air pressure in the room because atmospheric pressure is not always \begin{align*}760\ \mathrm{mm \ of \ Hg}\end{align*}. In manometer B, the pressure inside the flask balances the atmospheric pressure plus an additional pressure of \begin{align*}300.\ \mathrm{mm \ of \ Hg}\end{align*}. If the actual atmospheric pressure is \begin{align*}750.\ \mathrm{mm \ of \ Hg}\end{align*}, then the pressure in the flask is \begin{align*}1050\ \mathrm{mm \ of \ Hg}\end{align*}. On the other hand, the pressure in the flask of manometer C is less than atmospheric pressure, so the excess mercury is in the inside arm of the manometer (the arm closer to the flask). If atmospheric pressure is \begin{align*}750.\ \mathrm{mm \ of \ Hg}\end{align*}, then the pressure in the flask is \begin{align*}650.\ \mathrm{mm \ of \ Hg}\end{align*}. For open-end manometers, when the excess mercury is in the outside arm, the height difference is added to atmospheric pressure. When excess mercury is in the inside arm, the height difference is subtracted from atmospheric pressure.

Scientists also use mechanical pressure gauges on occasion. These instruments use the stretching or compression of springs to turn dials, or something similar. While such instruments seem to be less trouble, they must all be calibrated against mercury column instruments and are more susceptible to reactive gases.

## Lesson Summary

• Pressure is defined as the force exerted divided by the area over which the force is exerted: \begin{align*}\text{Pressure} = \frac {\text{Force}} {\text{Area}}\end{align*}.
• The air molecules in our atmosphere exert pressure on every surface they contact.
• There are many different units for measuring and expressing gas pressure, including mm of Hg, torr, atmosphere (atm), and pascal (Pa).
• At sea level, atmospheric pressure is approximately \begin{align*}15 \ \mathrm{lbs/in}^2\end{align*}, \begin{align*}760 \ \mathrm{mm \ of \ Hg}\end{align*}, \begin{align*}760 \ \mathrm{torr}\end{align*}, \begin{align*}1 \ \mathrm{atm}\end{align*}, and \begin{align*}101,325 \ \mathrm{Pa}\end{align*}.
• The barometer is a device to measure atmospheric pressure that consists of a glass tube filled with liquid mercury placed in a dish of mercury.
• A manometer is designed to measure gas pressure in flasks.
• There are two kinds of manometers: an open-end manometer and a closed-end manometer.

This video provides an introduction to gases and gas pressure.

## Review Questions

1. The manometer shown is a closed-end manometer filled with mercury. If the atmospheric pressure in the room is \begin{align*}760.\ \mathrm{mm \ of \ Hg}\end{align*} and \begin{align*}\triangle h\end{align*} is \begin{align*}65\ \mathrm{mm \ of \ Hg}\end{align*}, what is the pressure in the flask?
2. The manometer shown is a closed-end manometer filled with mercury. If the atmospheric pressure in the room is \begin{align*}750.\ \mathrm{mm \ of \ Hg}\end{align*} and \begin{align*}\triangle h\end{align*} is \begin{align*}0\ \mathrm{mm \ of \ Hg}\end{align*}, what is the pressure in the flask?
3. The manometer shown is an open-end manometer filled with mercury. If the atmospheric pressure in the room is \begin{align*}750.\ \mathrm{mm \ of \ Hg}\end{align*} and \begin{align*}\triangle h\end{align*} is \begin{align*}65\ \mathrm{mm \ of \ Hg}\end{align*}, what is the pressure in the flask?
4. The manometer shown is an open-end manometer filled with mercury. If the atmospheric pressure in the room is \begin{align*}760.\ \mathrm{mm \ of \ Hg}\end{align*} and \begin{align*}\triangle h\end{align*} is \begin{align*}0\ \mathrm{mm \ of \ Hg}\end{align*}, what is the pressure in the flask?
5. Explain why at constant volume, the pressure of a gas decreases by half when its Kelvin temperature is reduced by half.

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