University High School Chemistry: Unit II Reading I

The Three States of Matter

Lesson Objectives

The student will describe molecular arrangement differences among solids, liquids, and gases.
The student will describe the basic characteristic differences among solids, liquids, and gases.

Introduction

The Kinetic Molecular Theory allows us to explain the existence of the three phases of matter. In addition, it helps explain the physical characteristics of each phase and how phases change from one to another. The Kinetic Molecular Theory is essential for the explanations of gas pressure, compressibility, diffusion, and mixing. Our explanations for reaction rates and equilibrium also rest on the concepts of the Kinetic-Molecular Theory.

The Assumptions of the Kinetic Molecular Theory

According to the Kinetic Molecular Theory, all matter is composed of tiny particles that are in constant, random, straight-line motion. This motion is constantly interrupted by collisions between the particles and between the particles and surfaces. The rate of motion of the particles is related to their temperature. The velocity of the particles is greater at higher temperatures and lower at lower temperatures.

In our discussions of gases, we will be referring to what are called ideal gases. In real gases, there is some slight attraction between the gas molecules and the molecules themselves do take up a small amount of space. However, in an ideal gas, we assume there are no attractions between molecules and we assume the molecules themselves take up no space. Later in this chapter, real and ideal gases will be discussed again in more detail.

The Characteristics of Solids

In a solid, the molecules are held in a tightly packed pattern (see Figure above), in which molecules hold a set position in spite of random motion. Molecular motion is reduced to vibrating in place.

In a liquid, the molecules touch each other but are not held in a pattern. The liquid structure has holes in it, which allow molecules to pass each other and change position in the structure. In a gaseous substance, the molecules are completely separated from each other and move around independently. Most of the volume of a gas is empty space. The molecular arrangement in the three phases accounts for the various characteristics of the phases of matter.

For example, mixing of the particles in solids is almost non-existent. This is because the molecules cannot pass by one another in the tightly packed pattern. Solids are essentially incompressible because when a substance is compressed, it is the spaces between molecules that are compressed, not the molecules themselves. Since solids have almost no empty space in their structure, they do not compress. Solids have their own shape and volume as shown in Figure below. A $25 \;\mathrm{mL}$ rectangular piece of copper has the same shape and volume when it is resting on the table top as it does inside a beaker. The volume and shape of a solid is maintained by the particle structure of the solid which is a tightly held pattern of atoms.

The Characteristics of Liquids

In liquids, mixing occurs more readily because there are spaces between the molecules that allow molecules to pass each other. The spaces between the molecules in liquids are small and so liquids have very little compressibility. Liquids maintain their own volume but take their shape from the shape of the container (Figure below).

A $25 \;\mathrm{mL}$ sample of liquid in a graduated cylinder has a volume of $25-\;\mathrm{mL}$ and has the shape of a cylinder. If the $25 \;\mathrm{mL}$ sample is placed in a beaker, it still has a volume of $25 \;\mathrm{mL}$ but now it has the shape of the beaker. The structure of the liquid keeps the particles in touch with each other so the volume does not change but the particles can slide by each other so they can flow to the shape of the container.

The Characteristics of Gases

Mixing in gases is almost instantaneous because there are no inhibitions for particles to pass one another. The volume of a gas is nearly all empty space and so particles move completely freely. Gases are highly compressible because there is a great deal of empty space in gaseous structures which allows the particles to be pushed closer together. Gases do not have either their own volume or their own shape. They take both volume and shape from their container.

Gases mix readily. (Source: Richard Parsons. CC-BY-SA)

Matter is examined in its three principle states – gases, liquids, and solids – relating the visible world to the submicroscopic in an Annenberg video at Video on Demand – The World of Chemistry – A Matter of State (http://www.learner.org/resources/series61.html?pop=yes&pid=793#)

Lesson Summary

All matter is composed of tiny particles called atoms or molecules.
These particles are in constant random motion at all temperatures above absolute zero.
In the solid phase, the molecules are held in a highly organized, tightly-packed pattern.
Due to the tightly-packed pattern of molecules in a solid, solids maintain their own shape and volume and do not mix readily.
In the liquid phase, molecules are in touch with each other but they loosely packed and may move past each other easily.
Due to the loosely packed structure of a liquid, liquids maintain their own volume but take the shape of their container and they are able to mix readily.
In the gaseous phase, molecule are completely separate from each other.
The volume of a gas is mostly empty space.
Due to the structure of gases, they take both the volume and the shape of their container and they mix almost instantaneously.

Review Questions

Automobile brakes have a hose full of liquid connected to your brake pedal on one end and to the brake pads on the other end. When you press on the brake pedal, the force is transferred through the liquid and presses the brake pads against the wheels to slow or stop them. Brakes that use liquid in this fashion are called hydraulic brakes and the liquid is called hydraulic brake fluid. What don't they use air in the brake lines instead of liquid?
Why would it cause a problem if some air got into your liquid-filled brake lines?
If you had a $250 \;\mathrm{mL}$ container full of helium gas and you transferred all of the gas into an empty $1.00\;\mathrm{liter}$ container, would the $1.00\;\mathrm{liter}$ container be full or only $1/4$ full? If something seems odd about this question, what do you think it is?

Vocabulary

phase: Any of the forms or states, solid, liquid, gas, or plasma, in which matter can exist, depending on temperature and pressure.

kinetic: The term "kinetic" refers to the motion of material bodies and the forces associated with them.

molecule: In the kinetic theory of gases, any gaseous particle regardless of composition

Labs and Demonstrations for Kinetic Molecular Theory

Brownian Motion Demonstration

Investigation and Experimentation Objectives

In this activity, the student will observe the movement of small particles due to collisions with molecules.

Brief description of demonstration

The Brownian Motion Apparatus consists of a metal chamber with a glass viewing window on top and a lens on one side to allow a light source to shine in the chamber. Smoke from a smoldering rope or an extinguished match is drawn into the chamber through an inlet tube by squeezing the rubber bulb. The chamber is illuminated by light shining through the side lens (laser light is best, if you have a laser pointer). The smoke cell sits on a microscope stage and by focusing through the top viewing window, smoke particles are visible (like shiny stars) against a dark background. The smoke particles jiggle about as they are bombarded by air molecules. There is also some sideways drift due to convection currents. If you have a projecting microscope, the image can be projected on a screen so everyone can see at the same time. Sometimes the smoke particles seem to explode as they rise or sink out of focus. The smoke activity in the cell will die down after a while so a new puff of smoke must be drawn in now and then.

Materials

Microscope or projecting microscope
Source of smoke (smoldering rope or an extinguished match)
Brownian Motion Apparatus

Procedure

Place the Brownian Motion Apparatus on the microscope stage.
Adjust the side light into the lens.
Squeeze the rubber bulb and hold it empty.
Light and blow out a match. Hold smoking match near the inlet aperture and release the squeeze bulb.
Have the students take turn viewing unless you have a projecting microscope or a TV camera.

Hazards

None.

Disposal

Dispose of cooled matches in waste basket.

Discussion

Make sure the students understand that the smoke particles are much too large to exhibit molecular motion. The movement of the smoke particles is due to collisions from air molecules that do have molecular motion.

Gases

Lesson Objectives

The student will describe the relationship between molecular motion and Kelvin temperature.
The student will describe random motion of gaseous molecules and explain how their collisions with surfaces cause pressure on the surface.
The student will state that zero kinetic energy of molecules corresponds to $0\;\mathrm{K}$ and that there is no lower temperature.

Introduction

Gases are tremendously compressible, can exert massive pressures, expand nearly instantaneously into a vacuum, and fill every container they are placed in regardless of size. All of these properties of gases are due to their molecular arrangement.

Gases Readily Change Volume

In dealing with gases, we lose the meaning of the word “full.” A glass of water may be $1/4$ full or $1/2$ full or full, but a container containing a gaseous substance is always full. The same amount of gas will fill a quart jar, or a gallon jug, a barrel, or a house. The gas molecules separate farther from each other and spread out uniformly until they fill whatever container they are in. Gases can be compressed to small fractions of their original volume and expand to fill virtually any volume. If gas molecules are pushed together to the point that they touch, the substance would then be in the liquid form. One method of converting a gas to a liquid is to cool it and another method is to compress it.

Gases Exert Pressure

The constant random motion of the gas molecules causes them to collide with each other and with the walls of their container. These collisions of gas molecules with their surroundings exert a pressure on the surroundings. When you blow up a balloon, the air particles inside the balloon push against the elastic sides, the walls of the balloon are pushed outward and kept firm. This pressure is produced by air molecules pounding on the inside walls of the balloon.

When you look at the side of a blown-up balloon, the balloon wall is held out firmly and with no visible vibration or movement in its position. It is not apparent that the wall is being held in position by billions of collisions with tiny particles – but that is exactly what is happening. If you place a book on its edge and tilt it over slightly so that without support, it would fall, you can keep the book from falling by tapping it very rapidly with your finger on the underside. The book doesn’t stay steady because you can’t tap it fast enough to keep it exactly in one position. If you can imagine being able to tap it millions of times per second, you can see how a steady position could be maintained with many small collisions.

Gas Temperature and Kinetic Energy

Kinetic energy is the energy of motion and therefore, all moving objects contain kinetic energy. The mathematical formula for calculating the kinetic energy of an object is $KE = 1/2\;\mathrm{mv}^2$ . This physics formula applies to all objects in exactly the same way whether we are talking about the moon moving in its orbit, a baseball flying toward home plate, or a gas molecule banging around in a bottle. All of these objects have kinetic energy and their kinetic energies can all be calculated with the same formula. As you can see from the formula, the kinetic energy is dependent on both the mass of the object and the velocity of the object. The kinetic energy of a $0.20 \;\mathrm{kg}$ ball moving at $20. \;\mathrm{m/s}$ would be $KE = 1/2 (0.20 \;\mathrm{kg})(20. \;\mathrm{m/s})^2 = 40.\;\mathrm{Joules}.$ The kinetic energy of a molecule would be calculated in exactly this same way. You should note that if the mass of an object is doubled while its velocity remains the same, the kinetic energy of the object would also be doubled. If, on the other hand, the velocity is doubled while the mass remains the same, the kinetic energy would be quadrupled because of the square in the formula.

It was mentioned at the beginning of this lesson that the molecular motion of molecules is related to their temperature. If you think of the average kinetic energy of a group of molecules and temperature measured in degrees Kelvin, the relationship is a direct proportion. When you measure the temperature of a group of molecules, what you are actually measuring is their average kinetic energy. They are the same thing but expressed in different units. The formula for this relationship is $KE_{ave} = 3/2 \;\mathrm{RT}$ where $R$ is a constant of proportionality and $T$ is the absolute temperature (Kelvin). When a substance is heated, the average kinetic energy of the molecules increases. Since the mass of the molecules cannot be increased by heating, it is clear that the velocity of the molecules is increasing. The relationship between the kinetic energy of an object and its velocity, however, is not linear. Because of the exponential dependence of the velocity in the formula for kinetic energy, when the absolute temperature is doubled, the average kinetic energy is also doubled but the velocity is increased only by a factor of $1.4$ .

It is absolutely vital that you keep in mind that the mathematical relationship between the temperature and the average kinetic energy of molecules only exists when the temperature is expressed in the Kelvin scale. In order for the direct proportion to exist, the molecules must have zero kinetic energy when the temperature is zero. It is surely apparent to you that molecules do NOT have zero kinetic energy at $0^\circ C$ . Balloons and automobile tires do not go flat when the outside temperature reaches $0^\circ C$ . The temperature at which molecular motion stops is $0 \;\mathrm{K} \ (-273^\circ C)$ . If temperature is measured in Kelvin degrees, then the average kinetic energy of a substance at $100 \;\mathrm{K}$ is exactly double the average kinetic energy of a substance at $50 \;\mathrm{K}$ . Make sure all the work you do dealing with the kinetic energy of molecules is done with Kelvin temperatures.

Lesson Summary

Gases are composed of tiny particles called molecules.
Molecules of a gas are so far apart, on average, that the volume of the molecules themselves in negligible compared to the volume of the gas.
Gas molecules are in constant, random, straight-line motion that is constantly interrupted by collisions with other molecules or with container walls.
Molecular collisions with container walls cause the gas to exert pressure.

Review Questions

Ball $A$ has a mass of $4\;\mathrm{daltons}$ and a speed of $16\;\mathrm{meters}$ per second. Ball $B$ has a mass of $16\;\mathrm{daltons.}$ What velocity is necessary for ball $B$ to have the same kinetic energy as ball $A$ ?
Suppose you blow up a balloon, tie off the opening, and place the balloon in a freezer for one hour. When you take the balloon out of the freezer, what will the most significant difference be in its appearance from when you put it in the freezer? What do you think will happen to the this difference as the balloon sits out in the room for a while?
Suppose you drive home from school on a hot day and when you get home, you check the pressure in your automobile tires. You find the tire pressure is over the manufacturer's recommended pressure and so you let some air out of the tires until the pressure is appropriate. What will the tire pressure be in the morning when you go out to go to school?
Weather balloons are large balloons that are used to carry meteorological instruments up through the atmosphere and radio back measurements on temperature, pressure, humidity, etc. as it passes through many different altitudes. When these balloons are filled with helium before they are released from earth, they are only filled a little more than $10\%$ full. This provides enough lift to carry the instruments but the balloon would have more lift if it were filled completely. Why don't they fill the weather balloons to maximum capacity?

Vocabulary

kinetic energy: Kinetic energy is the energy a body possesses due to it motion, $KE = 1/2\;\mathrm{mv}^2$ .

Kelvin temperature: The absolute temperature scale where $0\;\mathrm{K}$ is the theoretical absence of all thermal energy (no molecular motion).

Labs and Demonstrations for Gases

Molecular Motion/Kinetic Energy Demo

Investigation and Experimentation Objectives

In this activity, the student will observe evidence that molecular motion increases with increasing temperature.

Brief description of demonstration

The apparatus is a glass tube with a small amount of mercury in the bottom and some glass chips floating on the mercury. (The apparatus is sometimes referred to as Stoekle’s Molecular Apparatus) At room temperature, the molecular motion of the mercury molecules is not forceful enough to move the glass chips.

When the tube is held over a Bunsen burner, however, the molecular motion of the mercury molecules increases, and the collisions of the mercury molecules with the bottom of the glass chips sends them flying to the top of the tube.

Materials

Molecular motion demonstration tube
Utility clamp
Bunsen burner
Pot holder (something to lay the hot tube on while cooling)
Matches

Procedure

Clamp the demonstration tube in a utility clamp so you can hold it without burning your fingers. Show the tube to the students at room temperature. Hold it over a Bunsen burner flame until the glass chips begin flying to the top of the tube. Allow the tube to cool before returning to storage.

Hazards

The tube gets very hot so be careful not to burn fingers. Mercury vapors are hazardous so it is important to be very careful not to break the tube. If you see any cracks in the tube, do not use it. It should be discarded according to disposal rules for mercury. It’s best to store the tube from year to year in bubble wrap.

Discussion

As the mercury molecules leave the liquid due to heating, they collide with the glass chips and knock them high into the tube. When the tube is removed from the heat, tiny droplets of mercury can be seen to condense on the walls of the tube.

Gases and Pressure

Lesson Objectives

The student will define pressure.
The student will convert requested pressure units.
The student will read barometers and open and closed-end manometers.
The student will apply the gas laws to relationships between the pressure, temperature and volume of a gas.
The student will state standard conditions for gases.

Introduction

The tremendous gas pressures that are possible are due to the pounding of trillions of tiny particles on a surface.

Pressure

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

$\text{Pressure} = \frac {\text{Force}} {\text{Area}}$

You must be careful when thinking about force and pressure. The concept of force is quite straightforward but the concept of pressure is a little trickier. Consider a large man and a smaller woman dressed to go out for an evening (Figure below). The man weighs $200\;\mathrm{pounds}$ and is wearing normal man’s shoes so that when he walks, at times, all his weight is exerted on the heel of one shoe. If that heel is $2\;\mathrm{inches}$ long and $2\;\mathrm{inches}$ wide, then his weight is exerted over an area of $4\;\mathrm{inches}^2$ and the pressure exerted by the heel of his shoe would be

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

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

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

This huge pressure has little to do with her weight but has a great deal to do with the area of the heels of her shoes. If these two people attempted to walk across the lawn, the $200\;\mathrm{pound}$ man would likely have no problem whereas the $100\;\mathrm{pound}$ woman might well have her heels sink into the grass. It is not uncommon for women wearing spike heels to put dents in hardwood flooring. 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 and such things 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 (because there are so many of them) was once demonstrated by a German physicist named Otto von Guericke who invented the air pump. Von Guericke placed two hemispheres about the size of dinner plates together and pumped the air out from between them. When both the inside and the outside of the hemispheres have air molecules, the pounding of the molecules on the plates is balanced on the two sides but when the air is removed from between the plates, there is no equalizing pounding of air molecules on the inside. Only the outside air molecules push on the plates and hold them together with amazing force. Teams of horses were unable to pull the hemispheres apart. The air molecules pounding on the outside of the hemispheres were exerting a greater pressure than the horses could produce trying to pull the hemispheres apart. When von Guericke opened a valve and allowed air inside, he could easily separate the hemispheres by hand.

The air molecules in our atmosphere exert pressure on every surface that is in contact with air. The air pressure of our atmosphere at sea level is approximately $15\;\mathrm{pounds/inch}^2$ . This pressure is unnoticed because the air is not only outside the surfaces but also inside allowing the atmospheric air pressure to be balanced. The pressure exerted by our atmosphere will become quickly noticed, however, if the air is removed or reduced inside an object. A common demonstration of air pressure makes use of a one-gallon metal can. The can has a few drops of water placed inside and is then heated to boiling. The water inside the can vaporizes and expands to fill the can pushing the air out. The lid is then tightly sealed on the can. As the can cools, the water vapor inside condenses back to liquid water leaving the inside of the can with a lack of air molecules. As the water vapor condenses to liquid water, the air pressure outside the can slowly crushes the can flat. People, of course, also have atmospheric pressure pressing on them. An averaged sized person probably has a total force exerted on them from the atmosphere in excess of $25,000\;\mathrm{pounds.}$ Fortunately, people also have air inside them to balance the force. A device to measure atmospheric pressure, 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 was constructed by filling a glass tube, open at one end and closed at the other, with liquid mercury and then inverting the tube in a dish of mercury as shown below.

Barometer. (Source: CK-12 Foundation. CC-BY-SA)

The empty tube was first turned so that the closed end was down and the tube was filled with liquid mercury. The dish was also filled two-thirds full of mercury. The open end of the mercury-filled tube was then covered and the tube was inverted. No mercury could flow out. The open end covered by the finger was then submerged completely beneath the mercury in the dish and the finger removed. Since the open end was under the surface, 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 $760\;\mathrm{millimeters.}$ The volume of empty space above the mercury in the tube was a vacuum.

The explanation for why the mercury stays in the tube is that there are air molecules pounding on the surface of the mercury in the dish and there are no air molecules pounding on the top of the mercury in the tube. 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. If you are quick with math, you will recognize that the diameter of the tube makes no difference. This is because doubling the diameter of the tube doubles the volume of mercury in the tube and doubles the weight of the mercury. It also doubles the area over which the force is exerted and therefore, the pressure will be the same for any size tube. No matter what diameter 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 $760.\;\mathrm{millimeters}$ when air pressure is normal and at sea level. The atmospheric pressure changes due to weather conditions and the height of the mercury in the barometer will change with it. 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 $9600\;\mathrm{feet}$ , the normal atmospheric pressure will only support a mercury column of $520\;\mathrm{millimeters}$ .

Measuring Gas Pressure

For various reasons, chemistry has many different units for measuring and expressing gas pressure. You will need to be familiar with most of them so you can convert them into preferred units. 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. The original unit in chemistry for gas pressure was $\;\mathrm{mm}\;\mathrm{of}\;\mathrm{Hg}$ (millimeters of mercury). Standard atmospheric pressure at sea level is $760.\;\mathrm{mm}$ of $\;\mathrm{Hg}$ . This unit is something of a problem because while it is a pressure unit, it looks a lot like a length unit. Students, in particular, occasionally leave off the “of $\;\mathrm{Hg}$ ” and then it definitely appears to be a length unit. To eliminate this problem, the unit was given another name. It was called the torr in honor of Torricelli. $760\;\mathrm{torr}$ is exactly the same as $760\;\mathrm{mm}$ of $\;\mathrm{Hg}$ . For certain work, it became convenient to express gas pressure in terms of multiples of normal atmospheric pressure at sea level and so the unit atmosphere (atm) was introduced.

$1.00\ \text{atm} = 760.\ \text{mm}\ \text{of}\ \text{Hg} = 760.\ \text{torr}$

Pressures expressed in $\;\mathrm{mm}$ of $\;\mathrm{Hg}$ can be converted to atmospheres using the conversion factor $1.00$ $\;\mathrm{atmosphere} = 760. \;\mathrm{mm}$ of $\;\mathrm{Hg}$ . In recent times, scientists decided that all branches of science should have the same units for measurements rather than each science using different units for the same measurement. The purest units are those from physics and therefore chemists have now added physics units for pressure. In physics, force is expressed in a unit called the Newton $(N)$ and area is expressed in $\;\mathrm{meters}^2 (m^2)$ . Therefore, pressure in physics is expressed in $\;\mathrm{Newtons/meter}^2$ $(N/m^2)$ . This has also been renamed the Pascal $(Pa)$ .

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

As it happens, one Pascal is an extremely small pressure, so it is convenient to use kilopascals or $kPa$ when expressing gas pressure in that unit. $1.00\;\mathrm{atm}= 101.325\;\mathrm{kPa}$ .

Pressure Unit Conversion Example 1

Convert $425\;\mathrm{torr}$ to $\;\mathrm{atm}$ .

Solution

The conversion factor is $760.\;\mathrm{torr} = 1.00\;\mathrm{atm}$ .

$(425\ \text{torr}) \left (\frac {1.00\ \text{atm}} {760.\ \text{torr}}\right ) = 0.559\ \text{atm}$

Pressure Unit Conversion Example 2

Convert $425\;\mathrm{torr}$ to $\;\mathrm{kPa}$ .

Solution

The conversion factor is $760.\;\mathrm{torr} = 101.325\;\mathrm{kPa}$ .

$(425\ \text{torr}) \left (\frac {101.325\ \text{kPa}} {760.\ \text{torr}}\right ) = 56.7\ \text{kPa}$

Pressure Unit Conversion Example 3

Convert $0.500\;\mathrm{atm}$ to $\;\mathrm{mm}$ of $\;\mathrm{Hg}$ .

Solution

The conversion factor is $1.00\;\mathrm{atm}= 760.\;\mathrm{mm}$ of $\mathrm{Hg}$ .

$(0.500\ \text{atm}) \left (\frac {760.\ \text{mm of Hg}} {1.00\ \text{atm}}\right ) = 380.\ \text{mm of Hg}$

Pressure Unit Conversion Example 4

Convert $0.500\;\mathrm{atm}$ to $\;\mathrm{kPa}$ .

Solution

The conversion factor is $1.00\;\mathrm{atm}= 101.325\;\mathrm{kPa}$ .

$(0.500\ \text{atm}) \left (\frac {101.325\ \text{kPa}} {1.00\ \text{atm}}\right ) = 50.7\ \text{kPa}$

You might notice if you wanted to measure a gas pressure that was around $2.0\;\mathrm{atm}$ with a barometer, you would need a glass column filled with mercury that was over $1.5\;\mathrm{meters}$ high. That would be a fragile and dangerous instrument since mercury fumes are toxic. If we used water (which is one-thirteenth as dense as mercury) in the tube to be free of toxicity, the column would have to be $50\;\mathrm{feet}$ high. Other instruments, called manometers, have been designed to measure gas pressure in flasks. There are two kinds of manometers in use, open-end manometers and closed-end manometers.

The closed-end manometer is easier to read so we will begin with it (see Figure below). The empty tube above the mercury level for the outside tube contains a vacuum. Therefore, there are no molecules pounding on the surface of the mercury in the outside tube. In the manometer in Figure A below, we represent a flask that contains no gas so that there are no molecules in the flask to exert pressure, $P_{gas} = 0$ . The two mercury levels will be exactly even so the mercury level in the outside arm balances the mercury level in the inside arm.

In Figure B below, we represent a flask that contains a gas at $1.00\;\mathrm{atm}$ pressure. The mercury level in the outside tube will rise to a height of $760\;\mathrm{mm}$ of $Hg$ so that the excess mercury in the outside tube balances the gas pressure in the flask. In the manometer in Figure C below, we would read the gas pressure in the flask as $200.\;\mathrm{mm}$ of $Hg$ . 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 open-end manometers (Figure below), the open-end of the tube allows atmospheric pressure to push down on the top of the column of mercury. In manometer $A$ in Figurebelow, the pressure inside the flask is equal to atmospheric pressure. The two columns of mercury are the same height and so they balance each other and 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 $760\;\mathrm{mm}$ of $Hg$ . In Figure Bbelow, the pressure inside the flask is balancing the atmospheric pressure PLUS the pressure of $300.\;\mathrm{mm}$ of $Hg$ . If the actual atmospheric pressure is $750.\;\mathrm{mm}$ of $Hg$ , then the pressure in the flask is $1050\;\mathrm{mm}$ of $Hg$ . In the flask in Figure C below, the pressure in the flask is less than the atmospheric pressure so the excess mercury is in the inside arm of the manometer. If the atmospheric pressure is $750.\;\mathrm{mm}$ of $Hg$ , then the pressure in the flask is $650.\;\mathrm{mm}$ of $Hg$ . For open-end manometers, when the excess mercury is in the outside arm, the height difference is added to atmospheric pressure and 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 they are more susceptible to reactive gases.

Lesson Summary

Because of the molecular motion of molecules, they possess kinetic energy at all temperatures above absolute zero.
The collisions between molecules are perfectly elastic. The phrase "perfectly elastic collision" means that kinetic energy is conserved in the collisions.
The molecules of an ideal gas have no attraction or repulsion for each other.
At any given moment, the molecules of a gas have different kinetic energies. We deal with this variation by considering the average kinetic energy of the molecules. The average kinetic energy of a group of molecules is measured by temperature.

Review Questions

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

Vocabulary

barometer: A barometer is an instrument used to measure atmospheric pressure.

manometer: A manometer is a liquid column pressure measuring device.

Labs and Demonstrations for Gases and Pressure

Magdeburg Hemispheres Demonstration

Investigation and Experimentation Objectives

In this activity, the student will observe the force exerted by air pressure (the very large force generated by many tiny particles).

Brief description of demonstration

Two five inch diameter hemispheres are placed together and the air is removed from inside the resulting sphere with a vacuum pump. A strong student is asked to pull the two hemispheres apart (unsuccessfully).

Purpose

To demonstrate the powerful force of air pressure.

Materials

Magdeburg Hemispheres
Vacuum grease
Vacuum pump (a hand pump will do)

Procedure

When the valve is opened and air allowed in, the hemispheres come apart easily.

If you have a bell jar large enough to enclose the hemispheres, you can put the hemispheres inside the bell jar and remove the air from the bell jar (surrounding the hemispheres) and the hemispheres will fall apart.

Hazards

An air leak may allow the hemispheres to come apart suddenly.

Discussion

The actual Magdeburg hemispheres were around $20 \ inches$ in diameter and were designed to demonstrate a vacuum pump that von Guericke had invented. When the air was removed from inside the hemispheres, and the valve closed, the hemispheres were held together by the air pressure of the surrounding atmosphere.

It is not known how good a vacuum von Guericke’s pump was able to produce. If all the air were removed from the inside, the hemispheres would have been held together with a force of around 20,000 Newtons, equivalent to lifting a car or a small elephant.

Von Guericke’s demonstration was performed on May 8, 1654. Thirty horses, in two teams of 15, could not separate the hemispheres until the vacuum was released.

Gas Laws

Lesson Objectives

The student will state Boyle's' Law, Charles' Law, and Gay-Lussac's Law.
The student will solve problems using Boyle's' Law, Charles' Law, and Gay-Lussac's Law.
The student will state the combined gas law.
Using the combined gas law, $\frac {P_1V_1} {T_1}$ = $\frac {P_2V_2} {T_2}$ , and given any five of the six variables, the student will solve for the sixth variable.

Introduction

The gas laws are mathematical relationships that exist for gases between the volume, pressure, temperature, and quantity of gas present. They were determined experimentally over a period of 100 years. They are logically derivable from our present day definitions of pressure, volume, and temperature.

Boyle’s Law

Gases are often characterized by their volume, temperature, and pressure. These characteristics, however, are not independent of each other. Gas pressure is dependent on the force exerted by the molecular collisions and the area over which the force is exerted. The force exerted by the molecular collisions is dependent on the absolute temperature and so forth. The relationships between these characteristics can be determined both experimentally and logically from their mathematical definitions.

The relationship between the pressure and volume of a gas was first determined experimentally by an Irish chemist named Robert Boyle (1627-1691). The relationship between the pressure and volume of a gas is commonly referred to as Boyle’s Law.

When we wish to observe the relationship between two variables, it is absolutely necessary to keep all other variables constant so that the change in one variable can be directly related to the change in the other. Therefore, when the relationship between gas volume and gas pressure is investigated, the quantity of gas and its temperature must be held constant so these factors do not contribute to any observed changes.

You may have noticed that when you try to squeeze a balloon, the resistance to squeezing is greater as the balloon becomes smaller. That is, the pressure inside the balloon becomes greater when the volume is reduced. This phenomenon can be studied more carefully with an apparatus like that in Figure 9. This is a cylinder tightly fitted with a piston that can be raised or lowered. There is also a pressure gauge fitted to the cylinder so that the gas pressure inside the cylinder can be measured. The amount of gas inside the cylinder cannot change and the temperature of the gas is not allowed to change.

In the picture on the left in Figure below, the volume of the gas is $4.0\;\mathrm{liters}$ and the pressure exerted by the gas is $2.0\;\mathrm{atm}$ . If the piston is pushed down to decrease the volume of the gas to $2.0\;\mathrm{liters}$ , the pressure of the gas is found to be $4.0\;\mathrm{atm}$ . The piston can be moved up and down to positions for several different volumes and the pressure of the gas read at each of the volumes. Several trials would generate a data set like that shown in Table below.

PV Data
Trial	Volume	Pressure
1	$8.0\;\mathrm{liters}$	$1.0\;\mathrm{atm}$
2	$4.0\;\mathrm{liters}$	$2.0\;\mathrm{atm}$
3	$2.0\;\mathrm{liters}$	$4.0\;\mathrm{atm}$
4	$1.0\;\mathrm{liters}$	$8.0\;\mathrm{atm}$

We might note from casual observation of the data that doubling volume is associated with the pressure being reduced to half and if we move the piston to cause the pressure to double, the volume is halved. We can analyze this data mathematically by adding a fourth column to our table – namely, a column showing the product of multiplying pressure times volume for each trial (see Table below).

PV Data
Trial	Volume	Pressure	Pressure $\times$ Volume
1	$8.0\;\mathrm{liters}$	$1.0\;\mathrm{atm}$	$8.0\;\mathrm{liters-atm}$
2	$4.0\;\mathrm{liters}$	$2.0\;\mathrm{atm}$	$8.0\;\mathrm{liters-atm}$
3	$2.0\;\mathrm{liters}$	$4.0\;\mathrm{atm}$	$8.0\;\mathrm{liters-atm}$
4	$1.0\;\mathrm{liters}$	$8.0\;\mathrm{atm}$	$8.0\;\mathrm{liters-atm}$

The data in the last column shows that with constant temperature and quantity of gas, the pressure times the volume of this sample of gas yields a constant. A mathematical constant (often represented by $k$ ) is a number that does not change even when other quantities in the formula do change. The value of $k$ will change if a different quantity of gas is used or if the trials are carried out at a different temperature, but for a particular mass of a particular gas at a particular temperature, the value of $k$ will always be the same. This relationship can be shown in a mathematical equation.

$PV = K$

This equation is a mathematical statement of Boyle’s Law. This particular equation is what is called an inverse proportionality. When one of the variables is increased, the other variable will decrease by exactly the same factor (see Figure below).

This result is the same as we would draw from logic. If the pressure a gas exerts is equal to force divided by the area over which it is exerted and we decrease the area but keep the force constant, we would expect the pressure to increase. Similarly, if we maintain the same number of molecules of gas and we keep the same temperature, the total force exerted by the molecules will be the same, and if we then expand the volume of the gas so that the area over which the force is exerted increased, we would expect the pressure to decrease.

Charles’s Law

The relationship between the volume and temperature of a gas was investigated by a French physicist, Jacques Charles (1746-1823). (As a piece of trivia, Charles was also the first person to fill a large balloon with hydrogen gas and take a solo balloon flight.) The relationship between the volume and temperature of a gas is often referred to as Charles’s Law.

An apparatus that can be used to study the relationship between the temperature and volume of a gas is shown in Figure below. Once again, we have a sample of gas trapped inside a cylinder so no gas can get in or out. Thus we have a constant mass of gas. We also have a mass set on top of a moveable piston to keep a constant force pushing against the gas. This guarantees that the gas pressure in the cylinder will be constant because if the pressure inside increases, the piston will be pushed up expanding inside volume until the inside pressure becomes equal to outside pressure again. Similarly, if the inside pressure decreases, the outside pressure will push the cylinder down, decreasing volume, until the two pressures again become the same. This system guarantees constant gas pressure inside the cylinder.

With this set up, we can adjust the temperature to various temperatures and measure the volume at each temperature to produce a data table in the same way we did for pressure versus volume. The picture on the left in Figure 11 shows the volume of a sample of gas at $250\;\mathrm{K}$ and the picture on the right shows the volume when the temperature has been raised to $500\;\mathrm{K}$ . After two more trials, the data is shown in Table below.

Charles's Law Data
Trial	volume	Temperature	Volume/Temp
1	$1000.\;\mathrm{mL}$	$250.\;\mathrm{K}$	$4.00\;\mathrm{mL/K}$
2	$1200.\;\mathrm{mL}$	$300.\;\mathrm{K}$	$4.00\;\mathrm{mL/K}$
3	$2000.\;\mathrm{mL}$	$500.\;\mathrm{K}$	$4.00\;\mathrm{mL/K}$
4	$2400.\;\mathrm{mL}$	$600.\;\mathrm{K}$	$4.00\;\mathrm{mL/K}$

In order to find a constant form this data, it was necessary to divide volumes by corresponding Kelvin temperatures. The mathematical expression for Charles’s Law is

$\frac {V} {T} = K$

This relationship would also be expected when we recognize that we are increasing the total force of molecular collisions with the walls by raising the temperature and the only way to keep the pressure from increasing is to increase the area over which that larger force is exerted. This mathematical relationship is known as a direct proportionality. When one variable is increased, the other variable also increases by exactly the same factor.

Gay-Lussac’s Law

The relationship between temperature and pressure was investigated by the French chemist, Joseph Gay-Lussac (1778-1850). An apparatus that could be used for this investigation is shown in Figure below. In this case, the cylinder does not have a moveable piston because it is necessary to hold the volume constant as well as the quantity of gas. This apparatus allows us to alter the temperature of a gas and measure the pressure exerted by the gas at each temperature.

After a series of temperatures and pressures have been measured, a data table like the others can be produced.

Pressure vs. Temperature Data
Trial	Temperature	Pressure	Pressure/Temp
1	$200.\;\mathrm{K}$	$600.\;\mathrm{torr}$	$3.00\;\mathrm{torr/K}$
2	$300.\;\mathrm{K}$	$900.\;\mathrm{torr}$	$3.00\;\mathrm{torr/K}$
3	$400.\;\mathrm{K}$	$1200.\;\mathrm{torr}$	$3.00\;\mathrm{torr/K}$
4	$500.\;\mathrm{K}$	$1500.\;\mathrm{torr}$	$3.00\;\mathrm{torr/K}$

Like Charles’s Law, in order to produce a constant when operating on the data in Table above, we must divide pressure by temperature. The relationship, again like Charles’s Law is a direct proportionality.

The mathematical form of Gay-Lussac’s Law is:

$\frac {P} {T} = K$

This relationship is also logical since by increasing temperature, we are increasing the force of molecular collision and keeping the area over which the force is exerted constant requires that the pressure increases.

Standard Temperature and Pressure (STP)

It should be apparent by now that expressing a quantity of gas simply by stating its volume is totally inadequate. Ten liters of oxygen gas could contain any mass of oxygen from $4,000\;\mathrm{grams}$ to $0.50\;\mathrm{grams}$ depending on the temperature and pressure of the gas. Chemists have found it useful to have a standard temperature and pressure with which to express gas volume. The standard conditions of temperature and pressure (STP) were chosen to be $0^\circ C (273\;\mathrm{K})$ and $1.00\;\mathrm{atm}$ ( $760\;\mathrm{mm}$ of $Hg$ ). You will commonly see gas volumes expressed as $1.5\;\mathrm{liters}$ of gas under standard conditions or $1.5\;\mathrm{liters}$ of gas at STP. Once you know the temperature and pressure conditions of a volume of gas, you can calculate the volume at other conditions and you can also calculate the mass of the gas if you know the formula.

The Combined Gas Law

Boyle’s Law shows how the volume of a gas changes when its pressure is changed (temperature held constant) and Charles’s Law shows how the volume of a gas changes when the temperature is changed (pressure held constant). Is there a formula we can use to calculate the change in volume of a gas if both pressure and temperature change? The answer is “yes”, we can use a formula that combines Boyle’s Law and Charles’s Law.

Boyle's Law states that for a sample of gas at constant temperature, every volume times pressure trial will yield the same constant. If we use subscripts of "1" to represent one set of conditions and subscripts of "2" to represent a second set of conditions,

$P_1 V_1= P_2 V_2\ && \text{and}&& V_2 = V_1 \times \frac {P_1} {P_2}$

Charles' Law is similar.

$\frac{V_1} {T_1} = \frac{V_2} {T_2}&& \text{and} && V_2 = V_1 \times \frac{T_2} {T_1}$

Combining the two equations yields

$V_2 = V_1 \times \frac {P_1} {P_2} \times \frac {T_2} {T_1}$

This equation is most commonly written in the from shown below and is known as the Combined Gas Law.

$\frac {P_1V_1} {T_1} = \frac {P_2V_2} {T_2}$

The majority of problems you will solve with this equation will be of the type where known values will be given to you for five of the six variables and you will be asked to solve for the sixth. When solving problems with the combined gas law, temperatures must always be in Kelvin. The units for pressure and volume may be any appropriate units but the units for each value of pressure must be the same and the units for each value of volume must be the same.

Sample Problem 1

A sample of gas has a volume of $400$ . liters when its temperature is $20.^\circ C$ and its pressure is $300.\;\mathrm{mm}$ of $Hg$ . What volume will the gas occupy at STP?

Solution:

STEP 1: Assign the known values for each of the variables representing them.

$& P_1 = 300.\ \text{mm\ of\ Hg} \ & V_1 = 400.\ \text{liters} \ & T_1 = 20.^\circ C + 273 = 293\ \text{K} \ & P_2 = 760.\ \text{mm of Hg (standard pressure)}\ & V_2 = x\ \text{liters (the unknown)}\ & T_2 = 0^\circ C + 273 = 273\ \text{K}$

STEP 2: Solve the combined gas law for the unknown variable.

$\frac {P_1V_1} {T_1} = \frac {P_2V_2} {T_2}&& \text{so} && V_2 = \frac {P_1V_1T_2} {P_2T_1}$

STEP 3: Substitute the known values into the formula and solve for the unknown.

$V_2 = \frac {(300.\ \text{mm of Hg})(400.\ \text{L})(273\ \text{K})} {(760.\ \text{mm of Hg})(293\ \text{K})} = 147 \ \text{liters}$

Sample Problem 2

A sample of gas occupies $1.00\;\mathrm{liter}$ under standard conditions. What temperature would be required for this sample of gas to occupy $1.50\;\mathrm{liters}$ and exert a pressure of $2.00\;\mathrm{atm}$ ?

STEP 1

$& P_1 = 1.00\ \text{atm (standard pressure in atmospheres)}\ & V_1 = 1.00\ \text{liter} \ & T_1 = 273\ \text{K (standard temperature)}\ & P_2 = 2.0\ \text{atm} \ & V_2 = 1.50\ \text{liters} \ & T_2 = x \ \text{(the unknown)}$

STEP 2

$\frac {P_1V_1} {T_1} = \frac {P_2V_2} {T_2} && \text{so} && T_2 = \frac {P_2V_2T_1} {P_1V_1}$

STEP 3

$T_2 = \frac {(2.00\ \text{atm})(1.50\ \text{L})(273\ \text{K})} {(1.00\ \text{atm})(1.00\ \text{L})} = 819\ \text{K}$

Sample Problem 3

A $1.00\;\mathrm{liter}$ sample of oxygen gas under standard conditions has a density of $1.43\;\mathrm{g/L}$ . What is the density of oxygen gas at $500.\;\mathrm{K}$ and $760.\;\mathrm{torr}$ ?

Solution:

You can find the mass of oxygen in the $1.00\;\mathrm{liter}$ sample by multiplying volume times density which yields a mass of $1.43\;\mathrm{grams.}$ Changing the temperature and/or pressure of a sample of gas changes its volume and therefore density BUT it does NOT change the mass. Therefore, when the new volume of the gas is found, the mass of oxygen gas in it will still be $1.43\;\mathrm{grams.}$ The density under the new conditions can be found by dividing the mass by the volume the gas now occupies.

$P_1 = 760.\ \text{torr} && P_2 = 760.\ \text{torr} \ V_1 = 1.00\ \text{L} && V_2 = \text{unknown} \ T_1 = 273\ \text{K}&& T_2 = 500.\ \text{K}$

$V_2 = \frac {P_1V_1T_2} {P_2T_1} = \frac {(760.\ \text{torr})(1.00\ \text{L})(500.\ \text{K})} {(760.\ \text{torr})(273\ \text{K})} = 1.83\ \text{liters}$

$D_{O2@500\ \text{K}} = \frac {\text{mass}} {\text{volume}} = \frac {1.43\ \text{g}} {1.83\ \text{L}} = 0.781\ \text{g/L}$

Lesson Summary

Because of the molecular motion of molecules, they possess kinetic energy at all temperatures above absolute zero.
The collisions between molecules are perfectly elastic. The phrase “perfectly elastic collision” comes from physics and means that kinetic energy is conserved in collisions.
The molecules of an ideal gas have no attraction or repulsion for each other.
At any given moment, the molecules of a gas have different kinetic energies. We deal with this variation by considering the average kinetic energy of the molecules. The average kinetic energy of a group of molecules is measured by temperature.
For a fixed sample of ideal gas at constant temperature, volume is inversely proportional to pressure.
For a fixed sample of ideal gas at constant pressure, volume in directly proportional to temperature.
For a fixed sample of ideal gas at constant volume, pressure is directly proportional to temperature.
The volume of a mass of gas is dependent on the temperature and pressure. Therefore, these conditions must be given along with the volume of a gas.
Standard conditions of temperature and pressure are $0^\circ C$ and $1.0\;\mathrm{atm}$ .

Review Questions

When a sample of gas is placed in a larger container at the same temperature, what happens to the total FORCE of the molecules hitting the walls?
When a sample of gas is placed in a larger container at the same temperature, what happens to the pressure exerted by the gas?
If molecules of $H_2$ ( $\;\mathrm{mol. mass} = 2)$ , $O_2$ $(\;\mathrm{mol. mass} = 32)$ , and $N_2$ $(\;\mathrm{mol. mass} = 28)$ are all placed in the same container at the same temperature, which molecules will have the greatest average kinetic energy?
If molecules of $H_2$ $(\;\mathrm{mol. mass} = 2)$ , $O_2$ $(\;\mathrm{mol. mass} = 32)$ , and $N_2$ $(\;\mathrm{mol. mass} = 28)$ are all placed in the same container at the same temperature, which molecules will have the greatest velocity?
If $X$ and $Y$ are quantities that are related to each other by inverse proportion, what will the value of $Y$ become when the value of $X$ is increased by a factor of five?
Under what conditions will the value for the constant, $K$ , change in the equation for Boyle’s Law, $PV =\;\mathrm{K}$ .
A sample of gas has a volume of $500.\;\mathrm{mL}$ under a pressure of $500.\;\mathrm{mm}$ of $Hg$ . What will be the new volume of the gas if the pressure is reduced to $300.\;\mathrm{mm}$ of $Hg$ at constant temperature?
A graph is made illustrating Charles’s Law. Which line in the picture at right would be appropriate assuming temperature is measured in Kelvin?
At constant pressure, the temperature of a sample of gas is decreased. Will the volume of the sample
1. increase
2. decrease
3. remain the same?
A sample of gas has its temperature increased from $-43^\circ C$ to $47^\circ C$ at constant pressure. If its volume at $-43^\circ C$ was $500.\;\mathrm{mL}$ , what is its volume at $47^\circ C$ ?
A gas is confined in a rigid container and exerts a pressure of $250.\;\mathrm{mm}$ of $Hg$ at a temperature of $17^\circ C$ . To what temperature must this gas be cooled in order for its pressure to become $216\;\mathrm{mm}$ of $Hg$ ? Express this temperature in $^\circ C$ .
What is the abbreviation used to indicate standard conditions of temperature and pressure?
A sample of gas has a volume of $500.\;\mathrm{mL}$ at standard conditions. Find its volume at $47^\circ C$ and $800.\;\mathrm{torr}$ .
A sample of gas has a volume of $100.\;\mathrm{L}$ at $17^\circ C$ and $800.\;\mathrm{torr}$ . To what temperature must the gas be cooled in order for its volume to become $50.0\;\mathrm{L}$ at a pressure of $600. \;\mathrm{torr}$ ?

Vocabulary

barometer: An instrument used to measure atmospheric pressure.

dalton: The unified atomic mass unit, or Dalton, is a unit of mass used to express atomic and molecular masses. It is the approximate mass of a hydrogen atom, a proton, or a neutron. The precise definition is that it is one-twelfth of the mass of an unbound carbon-12 atom at rest.

manometer: A liquid column pressure measuring device.

Labs and Demonstrations for Gas Laws

Charles Law with a Balloon and a Bunsen Burner

Investigation and Experimentation Objectives

In this activity, the student will observe the expansion of volume due to increased temperature.

Brief description of demonstration

A Bunsen burner is lit and adjusted to a cool flame. Students then hold a partially inflated balloon over the flame. The balloon inflates rapidly, which they feel.

Materials

Bunsen burner
Balloons, several
Matches

Procedure

Light the Bunsen burner, and adjust to a cool flame. Inflate the balloons by blowing into them until they reach a diameter of $20$ to $25 \ cm$ . Tie the end off. Have the students hold the inflated balloon about $30 \ cm$ over the flame. The balloon will rapidly inflate.

Hazards

The Bunsen burner can cause burns or ignite clothing if the student gets too close. Demonstrate how to hold the balloon before allowing the students to do it. The balloon inflates so rapidly and is such a strange tactile feeling that the students often think it’s alive, which can be startling to some. Be aware of over-reaction, and clear away any breakable items that may come into harms’ way.

Disposal

Throw the used balloons away.

Discussion

For a balloon that is $20 \ cm$ in diameter, the circumference of the balloon will change about $7 \ mm$ for every applied $10^\circ C$ temperature change. This is a relatively small change, but it does happen almost instantaneously, which surprises students.

Universal Gas Law

Lesson Objectives

The student will solve problems using the Universal Gas Law, $PV = nRT$ .
The student will state Avogadro’s Law of equal molecules in equal volumes.
The student will calculate molar mass from $\;\mathrm{mm} = gRT/PV$ given mass, temperature, pressure, and volume.

Introduction

The individual gas laws and the combined gas law all require that the quantity of gas remain constant. The Universal Gas Law (also sometimes called the Ideal Gas Law) allows us to make calculations on different quantities of gas as well.

Avogadro’s Law

Avogadro’s Law was known as Avogadro’s hypothesis for the first century of its existence. Since Avogadro's hypothesis can now be demonstrated mathematically, it was decided that it should be called a law instead of a hypothesis. Avogadro’s Law postulates that equal volumes of gas under the same conditions of temperature and pressure contain the same number of molecules.

If we think about the definitions of gas volumes, pressures, and temperatures, we can develop Avogadro’s conclusion. Suppose we have a group of toy robots that are all identical in strength. They don’t all look the same and they are not all the same size but they all have exactly the same strength at an assigned task. We arrange a tug-of-war between groups of our robots. We arrange the rope they will be pulling on so that we can see one end but the other end disappears behind a wall. On the visible end, we place eight robots to pull. On the other end, an unknown number of robots will pull. When we say “go,” both sides pull with maximum strength but the rope does NOT move. How many robots are pulling on the hidden end of the rope? Since each robot has exactly the same strength to pull, to balance eight robots on one end, there MUST be eight robots on the other end. If we can think of molecules as robots and we recognize that molecules at the same temperature have exactly the same striking power when they collide, then equal volumes of gas with equal pressures and equal temperatures must contain an equal number of molecules.

In the early 1800s, the first attempts to assign relative atomic weights to the atoms were accomplished by decomposing compounds to determine the mass ratios in the compounds and assigning hydrogen to have an atomic mass of $1.0.$ Some atomic weights found in this way were accurate but many were not. In the 1860s, Stanislao Cannizzaro refined relative atomic weights using Avogadro's Law. If gas $X$ and gas $Y$ were heated to the same temperature, placed in equal volume containers, and some gas was released from one of the containers until the two containers had the same pressure, Avogadro’s conditions would be present. Cannizzaro could conclude that the two containers had exactly the same number of molecules. The mass of each gas was then determined with a balance (subtracting the masses of the containers) and the relationship between the masses would be the same as the mass relationship of one molecule of $X$ to one molecule of $Y$ . That is, if the total mass of gas $X$ was $10.\;\mathrm{grams}$ and the total mass of gas $Y$ was $40.\;\mathrm{grams}$ , then Cannizzaro knew that one molecule of $Y$ must have four times as much mass as one molecule of $X$ . If an arbitrary value such as $1.0$ Dalton was assigned as the mass of gas $X$ , then $Y’s$ mass on that same scale would be $4.0$ Daltons.

The Universal Gas Law Constant

We have considered three laws that describe the behavior of gases.

Boyle’s Law, $V = \frac {k_1} {P}$ at constant temperature and constant moles of gas, where $k_1$ is a constant.

Charles’s Law, $V = k_2T$ at constant pressure and constant moles of gas, where $k_2$ is a constant.

Avogadro’s Law, $V = k_3n$ at constant temperature and pressure, where $n$ is the number of moles of gas and $k_3$ is a constant.

These three relationships, which show how the volume of a gas depends on pressure, temperature, and the number of moles of gas, can be combined to form $PV = nRT$ , where $R$ is the combination of the three constants. The equation is called the Universal Gas Law (or the Ideal Gas Law) and $R$ is called the universal gas law constant. When the pressure is expressed in atmospheres and the volume in liters, $R$ has the value $0.08206\;\mathrm{L} \cdot \;\mathrm{atm/mol} \cdot K.$ You can convert the value of $R$ into values for any set of units for pressure and volume. Moles, of course, always have the unit moles and temperature must always be Kelvin.

In our analysis of gas behavior, we have used a pair of assumptions that are usually true but not always true. We have assumed that the volume of the actual molecules in a gas is insignificant compared to the volume of the empty space between molecules. We have operated as if the molecules were geometric points and took up no space. For most gases, for most of the time, this assumption is true and the gas laws work well. If, however, a gas is highly compressed (at very high pressure), the molecules will be pushed very close together and much of the empty space between molecules will have been removed. Under such circumstances, the volume of the molecules themselves becomes significant and some calculations with the gas laws will be slightly off.

Another assumption we use is that the molecules are not attracted to each other so that every collision is a perfectly elastic collision, i.e. no energy is lost during the collision. This assumption also works out well most of the time. Even though the molecules do have some attraction for each other, usually the temperature is high enough that the molecular motion readily overcomes any attraction and the molecules move around as if there were no attraction. If, however, we operate with gases at low temperatures (temperatures near the phase change temperature of the gas), the molecular attractions have enough effect to put our calculations slightly off. If a gas follows the ideal gas laws, we say that the gas behaves ideally. Gases behave ideally at low pressure and high temperature. At low temperatures or high pressures, gas behavior may become non-ideal. As a point of interest, if you continue your study of chemistry, you will discover at the next level, there is yet another gas law equation for these non-ideal situations.

Sample Problem 4

A sample of nitrogen gas $(N_2)$ has a volume of $5.56\;\mathrm{liters}$ at $0^\circ C$ and $1.50\;\mathrm{atm}$ pressure. How many moles of nitrogen are present in this sample?

Solution:

STEP 1: Assign known values to variable.

$& P = 1.50\;\mathrm{atm} \ & V = 5.56\;\mathrm{L} \ & n = \mathrm{unknown} \ & T = 0^\circ C + 273 = 273\;\mathrm{K} \ & R = 0.08206\;\mathrm{L} \cdot \;\mathrm{atm/mol} \cdot \;\mathrm{K}$

STEP 2: Solve the universal gas law for the unknown variable.

$PV = nRT && \text{so} && n = \frac {PV} {RT}$

STEP 3: Substitute the know values into the equation and solve.

$n = \frac {\text{PV}} {\text{RT}} = \frac {(1.50\ \text{atm})(5.56 \ \text{L})} {(0.08206 \ \text{L} \cdot \ \text{atm/mol} \cdot \ \text{K})(273\ \text{K})} = 0.372\ \text{mol}$

Sample Problem 5

$2.00\;\mathrm{moles}$ of methane gas $(CH_4)$ are placed in a rigid $5.00\;\mathrm{liter}$ container and heated to $100.^\circ C$ . What pressure will be exerted by the methane?

Solution:

$P & = \text{unknown} \ V & = 5.00\ \text{liters} \ N & = 2.00\ \text{moles} \ T & = 100.^\circ C + 273 = 373\ \text{K} \ R & = 0.08206\ \text{L} \cdot \ \text{atm/mol} \cdot K \ P & = \frac {nRT} {V} = \frac {(2.00\ \text{mols})(0.08206 \ \text{L}\ \cdot \ \text{atm/mol} \cdot K)(373\ \text{K})} {(5.00\ \text{L})} = 12.2\ \text{atm}$

Sample Problem 6

A sample gas containing $0.300\;\mathrm{moles}$ of helium at a pressure of $900.\;\mathrm{torr}$ is cooled to $15^\circ C$ . What volume will the gas occupy under these conditions?

Note: If we are to use $0.08206\;\mathrm{L} \cdot \;\mathrm{atm/mol} \cdot \;\mathrm{K}$ for the value of $R$ , then the pressure must be converted from torr to atm.

$P & = (900.\ \text{torr})(1.00 \ \text{atm}/760.\ \text{torr}) = 1.18\ \text{atm} \ V & = \ \text{unknown} \ N & = 0.300\ \text{mole} \ T & = 15^\circ C + 273 = 288\ \text{K} \ R & = 0.08206\ \text{L} \cdot \ \text{atm/mol} \cdot \ \text{K} \ V & = \frac {\text{nRT}} {\text{P}} = \frac {(0.300\ \text{mols})(0.08206\ \text{L} \cdot \text{atm/mol} \cdot \ \text{K})(373\ \text{K})} {(1.18\ \text{atm})}= 6.01\ \text{liters}$

Molar Mass and the Universal Gas Law

The universal gas law can also be used to determine the molar mass of an unknown gas provided the pressure, volume, temperature and mass are known. The number of moles of a gas, $n$ , can be expressed as grams/molar mass. If we substitute $g/mm$ for $n$ in the universal gas law, we get $PV = (g/mm)RT$ which can be re-arranged to $\;\mathrm{mm} = gRT/PV$ .

Sample Problem 7

$20.0$ grams of an unknown gas occupy $2.00 L$ under standard conditions. What is the molar mass of the gas?

Solution:

$\text{molar mass} = \frac {\text{gRT}} {\text{PV}} = \frac {(20.0\ \text{g})(0.08206\ \text{L} \cdot \ \text{atm/mol} \cdot \ \text{K})(273\ \text{K})} {(1.00\ \text{atm})(2.00\ \text{L})} = 224\ \text{g/mol}$

Density and the Universal Gas Law

The density of a gas under a particular set of conditions is the mass of the sample of gas divided by the volume occupied at those conditions, $D = m/V$ . In the universal gas law, both the mass of the sample of gas and the volume it occupies are represented. We can substitute density for $g/V$ where it appears in the equation and produce an equation that contains density rather than mass and volume. For example, in the equation $\;\mathrm{mm} = gRT/PV$ , mass appears in the numerator and volume appears in the denominator so we can substitute $d$ for those two variables producing the equation $\;\mathrm{mm} = dRT/P$ .

Sample Problem 8

The density of a gas was determined to be $1.95\;\mathrm{g/mL}$ at $1.50\;\mathrm{atm}$ and $27^\circ C$ . What is the molar mass of the gas?

Solution

$\text{molar mass} = \frac {\text{dRT}} {\text{P}} = \frac {(1.95\ \text{g/L})(0.08206\ \text{L} \cdot \ \text{atm/mol} \cdot \ \text{K})(300.\ \text{K})} {(1.50\ \text{atm})} = 32.0\ \text{g/mol}$

Lesson Summary

Avogadro's Law: Equal volumes of gases under the same conditions of temperature and pressure contain equal numbers of molecules.
The Universal Gas Law: $PV = nRT.$
The universal gas law is often used along with laboratory data to find the molar mass of an unknown substance.

Review Questions

What volume will $2.00\;\mathrm{moles}$ of hydrogen gas occupy at $2.62 \;\mathrm{atm}$ of pressure and $300.^\circ C$ ?
How many moles of gas are required to fill a volume of $8.00\;\mathrm{liters}$ at $2.00\;\mathrm{atm}$ and $273\;\mathrm{K}$ ?
What is the molar mass of a gas if its density is $1.30\;\mathrm{g/L}$ at STP?

Vocabulary

universal gas law constant, $R$: $R$ is a constant equal $\frac {PV} {nT}$ where the pressure, volume, moles, and temperature of the gas are $P$ , $V$ , $n$ , and $T$ , respectively. The value and units of $R$ depend on the units of $P$ and $V$ . Commonly used values of $R$ include;

$82.055\;\mathrm{mL}^3\;\mathrm{atm} \ K^{-1}\;\mathrm{mol}^{-1}, 0.082055\;\mathrm{L} \;\mathrm{atm}\ K^{-1} \;\mathrm{mol}^{-1}, 8.314 \;\mathrm{JK}^{-1} \;\mathrm{mol}^{-1}, 8.314\;\mathrm{Pa\ m}^3 \;\mathrm{K}^{-1} \;\mathrm{mol}^{-1}.$

Labs and Demonstrations for Universal Gas Law

Teacher’s Pages for Finding the Molar Mass of a Gas Experimentally Lab

Lab Notes:

The use of the butane lighters as directed in the lab produces few hazards. The use of the butane lighters in ways other than directed in the lab will produce many hazards. Alert supervision is required. The room should be well ventilated and no open flames allowed.

The most critical part of the lab, in terms of getting accurate values, is exactly filling the flask with butane to the $200. \ mL$ line while the line on the flask is held exactly at water level. Careful attention to detail is necessary.

Timing:

The calculations in this lab require use of the combined gas law, the universal gas law, and the law of partial pressures so the lab should be done after all topics have been considered in the course.

Answers to Pre-Lab Questions:

$P_{H2} = P_{TOTAL} - P_{H2O} = 755.0 \ mm \ of \ Hg - 23.7 \ mm \ of \ Hg = 731.3 \ mm \ of \ Hg$
$V_2 = \frac{P_1V_1T_2}{P_2T_1} = \frac {(730. \ mm \ of \ Hg)(137 \ mL)(273 \ K)} {(760. \ mm \ of \ Hg)(298 \ K)} = 121 \ mL$
$\frac{0.51 \ g} {0.250 \ L}=\frac {x \ g} {22.4 \ L} \qquad x = 45.7 \ g$

Answers to Post-Lab Questions:

1. Since the pressure inside the flask is determined by measuring the atmospheric pressure in the room, it is necessary to make sure the pressures are the same. If the water levels inside and outside the flask are not exactly the same, the pressures will not be the same.

2. $V_2=\frac {P_1V_1T_2} {P_2T_1} = \frac {(775 \ mm \ of \ Hg)(498 \ mL)(273 \ K)} {(760. \ mm \ of \ Hg)(373 \ K)} = 372 \ mL$

$\frac{2.31 \ g}{0.372 \ L} = \frac{x \ g}{22.4 \ L} \qquad x = 139 \ g$

Therefore, the molar mass is $139 \ g/mol$ .

3. The molar mass of $C_4H_{10}$ is $58.0 \ g/mol$ .

moles $C_4H_{10}=\frac{3.60 \ g}{58.0 \ g/mol} = 0.0621 \ moles$

4. $\# \ of \ molecules = (moles)(molecules/mole) = (0.0621)(6.02 \times 10^{23} = 3.74 \times 10^{22}) \ molecules$

5. $V = \frac{nRT}{P} = \frac {(0.0621 \ mol)(0.0821 \ L \cdot atm/mol \cdot K)(273 \ K)} {(1.00 \ atm} = 1.39 \ liters$

6. $2 \ C_4H_{10}+13 \ O_2 \rightarrow 8 \ CO_2+10 \ H_2O$

$\frac {2 \ mols \ C_4H_{10}} {13 \ mols \ O_2} = \frac {0.0621 \ mol \ C_4H_{10}} {x \ mol \ O_2} \qquad x = 0.404 \ mol \ O_2$

$grams \ O_2 = (0.404 \ mol)(32.0 \ g/mol) = 12.9 \ grams$

Finding the Molar Mass of a Gas Experimentally Lab

Investigation and Experimentation Objectives

In this activity, the student will use mathematical formulas and his/her own experimental data to calculate the molar mass of an unknown gas.

Background:

A mole of any pure gas at STP has a volume of $22.4 \ liters$ . The mass of that $22.4 \ L$ is the molar mass of the gas.

If the volume, mass, temperature and pressure of a gas is known, then, using the combined gas law, the volume can be mathematically converted to a volume at STP. The mass of $22.4 \ liters$ can be determined. This mass is the molar mass of the gas and has been determined experimentally.

When gas is collected by water displacement, you must look up the vapor pressure of water at that temperature and use Dalton's law of partial pressures to factor out the pressure of the water vapor.

Pre-Lab Questions

$200. \ mL$ of hydrogen gas was collected over water at a pressure of $755.0 \ mm \ of \ Hg$ and a temperature of $25^\circ C$ . If the vapor pressure of water at $25^\circ C$ is $23.7 \ mm \ of \ Hg$ , what was the partial pressure of hydrogen in the container?
If a volume of gas occupies $137 \ mL$ at $730. \ mm \ of \ Hg$ and $25^\circ C$ , what volume will it occupy under standard conditions?
A $250. \ mL$ sample of gas under standard conditions has a mass of $0.51 \ grams$ . What would be the mass of $22.4 \ L$ of this gas?

Purpose:

To experimentally determine the molecular mass of butane.

Safety Issues:

There should be no open flames in the room during the lab and the room should be well ventilated. Students should be closely supervised to make sure they are using the lighters only as directed.

Apparatus and Materials:

butane lighter
balance
$250-300 \ ml$ flask
china marker or water proof marker
$100 \ mL$ graduated cylinder
Water trough (bucket, dishpan)
thermometer
plastic wrap or glass plate

Procedure:

Use a graduated cylinder to place exactly $200 \ ml$ of water in the flask. Using the marker, draw a highly visible line at the $200 \ ml$ water line.
Determine and record the exact mass of the butane lighter.
Fill the trough or other large container with water.
Fill the flask completely with water and using plastic wrap, a glass plate or your hand, invert the flask into the trough without permitting any air bubbles in it.
Make sure that the lighter is turned to its highest gas flow. Hold the butane lighter under the mouth of the flask and press the release lever. Be careful that all of the gas flows into the flask. Hold the flask so that the $200 \ ml$ mark is exactly even with the water level in the trough. Fill the flask exactly to the $200 \ ml$ mark that you made on the flask.
Determine and record the temperature of the water in the trough and the barometric pressure in the lab.
Thoroughly dry the butane lighter and determine and record its exact mass.

Data

Initial mass of lighter = ____________ g
Final mass of lighter = _____________ g
Mass of butane collected = ____________ g
Volume of butane collected = $200 \ mL$
Temperature = _________ $^\circ C$
Atmospheric pressure in the lab = ____________mm of Hg
Vapor pressure of water at this temperature = __________ mm of Hg

Table of water vapor pressures at normal lab temperatures.
Temperature, $^\circ C$	Pressure, mm of Hg
$17$	$14.5$
$18$	$15.5$
$19$	$16.5$
$20$	$17.5$
$21$	$18.6$
$22$	$19.8$
$23$	$21.0$
$24$	$22.4$
$25$	$23.7$
$26$	$25.2$
$27$	$26.7$

Calculations:

Using Dalton's law of partial pressures, enter the pressure of water vapor that corresponds to the lab temperature in the data table. Then subtract this from the lab pressure to find the partial pressure of butane gas.
Convert the lab temperature from Celsius to Kelvin.
Using the combined gas law formula, calculate the volume of butane to the volume it would occupy at STP.
Knowing the mass of butane that occupied the volume found in calculation 3, use a proportion to determine the mass of this gas that would occupy $22.4 \ liters$ . This is the experimentally determined molar mass of butane.
Butane has the formula $C_4H_{10}$ . Determine its molar mass from this formula.
Calculate your percent error.

$\% \ \text{error} = \frac{actual \ value-experimental \ value}{actual \ value} \times 100 =$

Post-Lab Questions

In procedure step 5, it was required that the flask be held in the water trough so that the $200. \ mL$ line marked on the flask was exactly level with the water outside the flask. Why was this done?
If $2.31 \ grams$ of dry gas (no water vapor) occupied $498 \ mL$ at $100.^\circ C$ and $775 \ mm \ of \ Hg$ , what is the molar mass of the gas?
A disposable butane lighter contains approximately $3.60 \ grams$ of butane. How many moles of butane is this?
How many molecules of butane would be present in $3.60 \ grams$ ?
What volume would $3.60 \ grams$ of butane occupy under standard conditions?
How many grams of oxygen would be required to burn $3.60 \ grams$ of butane?

Molar Volume

Lesson Objectives

The student will apply the relationship $1.00\;\mathrm{mole}$ of any gas at standard conditions will occupy $22.4\;\mathrm{L}$ .
The student will convert gas volume at STP to moles and to molecules and vice versa.
The student will apply Dalton’s Law of Partial Pressures to describe the composition of a mixture of gases.

Introduction

When molecules are at the same temperature, they have the same kinetic energy regardless of their mass. They all exert the same force when they strike a wall. In this sense, all molecules are created equal (as long as they are at the same temperature). Since equal numbers of molecules at the same temperature will exert the same force of collision, then when such groups of molecules are trapped in identical containers, they will exert exactly the same pressure. This is the logic of Avogadro's Law. If we alter this situation a little and say we have the same number of molecules at the same temperature and insist they exert the same pressure, then similar logic allows us to conclude that these groups of molecules must be in equal size containers.

Molar Volume of Gases at STP

As you know, $1.00\;\mathrm{mole}$ of any substance contains $6.02 \times 10^{23}$ molecules. With a sort of a reverse use of Avogadro’s Law, we also know that $6.02 \times 10^{23}$ molecules of any gas will occupy the same volume under the same conditions of temperature and pressure. Therefore, we can say that $1.00\;\mathrm{mole}$ of any gas under the same conditions will occupy the same volume. You can find the volume of $1.00\;\mathrm{mole}$ of any gas under any conditions by plugging the values into the universal gas law, $PV = nRT$ , and solving for volume. We are particularly interested in the volume occupied by $1.00\;\mathrm{mole}$ of any gas under standard conditions. We can calculate that volume using $PV = nRT.$

The molar volume of any gas at STP. (Source: CK-12 Foundation. CC-BY-SA)

This is the volume that $1.00\;\mathrm{mole}$ of any gas will occupy at STP. You will be using the number to convert gas volumes at STP to moles and vice versa. You will use it often enough to make it worth memorizing.

$V = \frac {\text{nRT}} {\text{P}} = \frac {(1.00\ \text{mols})(0.08206\ \text{L} \cdot\ \text{atm/mol} \cdot \text{K})(273\ \text{K})} {(1.00\ \text{atm})} = 22.4\ \text{liters}$

Sample Problem 9

How many moles are present in $100.\;\mathrm{L}$ of hydrogen gas at STP?

Solution:

$(100.\ \text{L}) \left (\frac {1.00\ \text{mol}} {22.4\ \text{L}}\right ) = 4.46\ \text{liters}$

Sample Problem 10

What volume will $100.\;\mathrm{grams}$ of methane gas $(CH_4, \mathrm{molar\ mass} = 16.0 \;\mathrm{g/mol})$ occupy at STP?

Solution:

$(100.\ \text{g}) \left (\frac {1.00\ \text{mol}} {16.0 \ \text{g}}\right ) \left (\frac {22.4\ \text{L}} {1.00\ \text{mol}}\right ) = 140. \ \text{liters}$

Dalton’s Law of Partial Pressures

The English chemist, John Dalton, in addition to giving us the atomic theory, also studied the pressures of gases when gases were mixed with each other but did not react chemically. His conclusion is Dalton’s Law of Partial Pressures.

: For a mixture of gases in a container, the total pressure exerted is the sum of the pressures that each gas would exert if it were alone.

This can be expressed as: $P_{TOTAL} = P_1 + P_2 + P_3 + \cdots$

In the last section, molecules were described as robots. The robots were identical in their ability to exert force. When a group of diverse molecules are at the same temperature, this analogy works well. In the case of molecules, they don’t all look alike but they have the same average kinetic energy and therefore they exert the same force when they collide. If we have $10,000$ gaseous molecules at $100^\circ C$ in a container and they exert a pressure of $0.10\;\mathrm{atm}$ , then adding another $10,000$ gaseous molecules of any substance at the same temperature to the container would increase the pressure to $0.20\;\mathrm{atm}$ . If there are $200\;\mathrm{molecules}$ in a container (and at the same temperature), each single molecule is responsible for $1/200^{th}$ of the total pressure. In terms of force of collision, it doesn’t make any difference if the molecules have different masses. At the same temperature, the little ones are moving faster and the big ones are moving slower so that the striking force is the same. It is this ability to exert force that is measured by temperature. We can demonstrate that different sized objects can have the same kinetic energy by calculating the kinetic energy of a golf ball and the kinetic energy of a bowling ball at appropriate velocities. Suppose a $100.\;\mathrm{gram}$ golf ball is traveling at $60.\;\mathrm{m/s}$ and a $7,200\;\mathrm{gram}$ bowling ball is traveling at $7.1\;\mathrm{m/s}$ .

$& \text{KE}_{\text{GOLF}} = \frac{1} {2} \text{mv}^2 = \frac{1} {2} (0.100\ \text{kg})(60.\ \text{m/s})^2 = 180\ \text{Joules}\ & \text{KE}_{\text{BOWL}} = \frac{1} {2} \text{mv}^2 = \frac{1} {2} (7.2\ \text{kg})(7.1\ \text{m/s})^2 = 180 \ \text{Joules}$

The kinetic energies are the same. If these balls were invisible and were to strike a pressure plate, they would exert exactly the same force and we would not be able to determine which one had struck the plate. So it is with molecules at the same temperature.

Suppose we have three one-liter containers labeled A, B, and C. Container A holds $0.20\;\mathrm{mole}$ of $O_2$ gas at $27^\circ C$ , container $B$ holds $0.50\;\mathrm{mole}$ of $N_2$ gas at $27^\circ C$ , and container $C$ holds $0.30\;\mathrm{mole}$ of $He$ gas at $27^\circ C$ . The pressure of each gas in the separate containers can be calculated with $PV = nRT$ .

$& \text{P}_{\text{O}2} = \frac {\text{nRT}} {\text{V}} = \frac {(0.20 \ \text{mols})(0.08206 \ \text{L} \cdot \text{atm/mol} \cdot \text{K})(300.\text{K})} {(1.00\text{L})} = 4.92\ \text{atm} \ & \text{P}_{\text{N}2} = \frac {\text{nRT}} {\text{V}} = \frac {(0.50\ \text{mols})(0.08206\ \text{L} \cdot \ \text{atm/mol} \cdot \text{K})(300.\text{K})} {(1.00\text{L})} = 12.3\ \text{atm} \ & \text{P}_{\text{He}} = \frac {\text{nRT}} {\text{V}} = \frac {(0.30\ \text{mols})(0.08206\ \text{L} \cdot \ \text{atm/mol} \cdot \text{K})(300.\text{K})} {(1.00\text{L})} = 7.39\ \text{atm}$

The sum of these three pressures is $24.6\;\mathrm{atm}.$

If all three gases are placed in one of the containers still at $27^\circ C$ , the pressure in the single container can also be calculated with $PV = nRT$ . (Remember that with all three gases in the same container, the number of moles is $0.20 + 0.50 + 0.30 = 1.00\;\mathrm{mole}$ .

$\text{P}_{\text{MIXTURE}} = \frac {\text{nRT}} {\text{V}} = \frac {(1.00\ \text{mols})(0.08206\ \text{L}\ \cdot \ \text{atm/mol} \cdot \text{K})(300.\ \text{K})} {(1.00 \ \text{L})} = 24.6\ \text{atm}$

You can quickly verify that the total pressure in the single container is the sum of the individual pressures when the gases were alone in their own container. The pressure exerted by each of the gases in a mixture is called the partial pressure of that gas. Hence, Dalton’s Law is the law of partial pressures.

Graham’s Law of Diffusion

In a previous section, it was mentioned that gases will spread out and occupy any and all space available. This spreading out and mixing of a substance is called diffusion. The small spaces available in liquid structure allows diffusion to occur only slowly. A drop of food coloring in a glass of water (without stirring) might take half an hour or more to spread through the water evenly. With solid structure we would expect no diffusion or mixing at all but there is a story of lead and gold bricks stacked together for many years that showed a few molecules exchanged in the surfaces. Even if the story is true, diffusion in solids is negligible. In gases, of course, diffusion is very rapid. Not even the presence of other gases offer much obstacle to gases spreading out throughout their container. If someone opens a bottle of perfume or ammonia across the room from you, it is only a matter of minutes until you smell it. The molecules evaporate from the liquid in the bottle and spread out through the room quickly.

It turns out that the diffusion of gases is not all the same. If bottles of strong smelling substances are opened across the room from you at the same time, the odors will not reach you at the same time. If we pick the strong smelling substances ammonia $(NH_3)$ and acetone $(C_3H_6O)$ and open bottles of these substances across the room from you, you will always smell the ammonia before the acetone. If we had thought about this a little, we probably would have predicted it. With the two liquids in the same room, they are probably the same temperature so the molecules will have the same kinetic energy. Remember that when molecules have the same kinetic energy, big ones move slower and little ones move faster. If we consider the molar mass of these two substances, the molar mass of ammonia is $17\;\mathrm{g/mol}$ and the molar mass of acetone is $58\;\mathrm{g/mol}$ ; so we would have realized that the ammonia molecules are traveling quite a bit faster than the acetone molecules in order for them to have the same kinetic energy. We would have expected the ammonia to diffuse through the room faster.

A demonstration commonly used to show the different rates of diffusion for gases is to dip one cotton ball in an ammonia solution and another cotton ball in a dilute $HCl$ solution and stuff the cotton balls in the opposite ends of a glass tube (Figure below). When $NH_3$ and $HCl$ react, they form $NH_4Cl$ , a white powdery substance. Molecules of $NH_3$ and $HCl$ will escape the cotton balls at opposite ends of the tube and diffuse through the tube toward each other. Since the molar mass of $HCl$ is slightly more than double that of $NH_3$ , the $NH_3$ will travel further down the tube than the $HCl$ by the time they meet.

A Diffusion Tube.

The white cloud of $NH_4Cl$ always forms nearer the $HCl$ end. If this experiment is done carefully and the distances are measured accurately, a reasonable ratio for the molar masses of these compounds can be determined.

Thomas Graham (1805-1869), an English chemist, studied the rates of diffusion of different gases and was able to describe the relationships quantitatively in what is calledGraham’s Law.

: Under the same conditions of temperature and pressure, gases diffuse at a rate inversely proportional to the square root of the molecular masses.

The velocity of particle one is to the velocity of particle two as the square root of the molecular mass of particle two is to the square root of the molecular mass of particle one.

$\frac{V_1} {V_2} = \sqrt{\frac{mm_2} {mm_1}}$

Lesson Summary

At STP, one mole of any gas occupies $22.4\;\mathrm{liters.}$
When gases are mixed and do not react chemically, the total pressure of the mixture of gases will be equal to the sum of the partial pressures of the individual gases.
Gases diffuse at rates that are inversely proportional to the square roots of their molecular masses.
Real gases tend to deviate from ideal gases at high pressures and low temperatures.

Review Questions

A $1.00\;\mathrm{L}$ container of helium gas at $1.00\;\mathrm{atm}$ pressure and a $1.00\;\mathrm{L}$ container of hydrogen gas at $2.00\;\mathrm{atm}$ are both transferred into a $1.00\;\mathrm{L}$ container containing nitrogen gas at $3.00\;\mathrm{atm}$ . What is the final pressure in the final container holding all three gases (assuming no temperature change)?
For the situation described in problem #1, what will be the partial pressure of the helium in the final container?
What conditions of temperature and pressure cause gases to deviate from ideal gas behavior?
At STP, how many molecules are in $89.6\;\mathrm{liters}$ of gas?
If $1.00\;\mathrm{liter}$ of gas A at STP and $1.00\;\mathrm{liter}$ of gas $B$ at STP are both placed into a $2.00\;\mathrm{liter}$ evacuated container at STP, what will the pressure be in the $2.00\;\mathrm{liter}$ container?
Consider the gases and . Which of the following will be nearly identical for the two gases at and ?
1. I only
2. III only
3. I and II only
4. II and III only
5. I, II, and III
  1. average molecular speed
  2. rate of effusion through a pinhole
  3. density
The density of an unknown gas at and is determined to be . Which of the following gases is the unknown most likely to be?
1. $CH_4$
2. $F_2$
3. $N_2O_4$
4. $O_2$
5. $CF_2Cl_2$

Vocabulary

diffusion: The movement of particles from areas of higher concentration to areas of lower concentration of that particle.

partial pressure: The pressure that one component of a mixture of gases would exert if it were alone in a container.

molar volume: The volume occupied by one mole of a substance in the form of a solid, liquid, or gas.

Labs and Demonstrations for Molar Volume

Rate of Diffusion at Various Temperatures Demo

Investigation and Experimentation Objectives

In this activity, the student will make measurements on the rates of diffusion and draw conclusions about rates of diffusion and their relationship to molar mass.

Brief description of demonstration

Drops of food color are added to water at three different temperatures and the time required for complete diffusion is observed.

Materials

Hot Plate
Ice
$3 - 250 \ mL$ graduated cylinders
3 – dropper pipets
Thermometer
Wall Clock with second hand

Procedure

Fill one graduated cylinder with tap water at room temperature, (this water needs to sit in the room for several hours to reach room temperature. (Fill a clean milk carton with tap water the day before the demonstration.
Heat tap water to boiling and fill the second graduated cylinder.
Make ice water and stir until no more ice melts. Fill the third graduated cylinder with ice water (but no ice cubes).
Record the temperature of the water in each cylinder.
Allow the water to settle for a couple of minutes.
Add two drops of good coloring to cylinder at the same time. Note the time.
Record how long it takes for the food coloring to be completely dispersed in each cylinder.

Hazards

Boiling water can be a burn hazard.

Disposal

All solutions can be disposed of down the sink.

Discussion

In which cylinder did the dye spread out the fastest.

In which cylinder did the dye spread out the slowest.

Give reasons for the different times of diffusion of the three cylinders of water.

The same molecules at different temperatures are moving at different speeds. At higher temperatures, the speed of the molecules is greater. Molecular collisions occur with greater frequency and with greater force.

If you wish, you could relate what the student sees in this demo to increased gas pressure at higher temperature and increased effusion rates (escape through a pinhole) at higher temperature.

Stoichiometry Involving Gases

Lesson Objectives

The student will solve stoichiometry problems involving gas volume at STP to moles and vice versa.
The student will solve stoichiometry problems involving gas volume to gas volume.

Introduction

The knowledge of the molar volume of gases at STP and the other gas laws allows us to work stoichiometry problems from gas volumes as well as masses.

Volumes, Moles, and Molecules

Avogadro’s Law tells us that under the same conditions of temperature and pressure, equal volumes of gases will contain equal numbers of molecules. By logical extension, we can say that under the same conditions of temperature and pressure, equal numbers of molecules will occupy equal volumes. For example, $1.00\;\mathrm{mole}$ of hydrogen gas will occupy the same volume as $1.00\;\mathrm{mole}$ of oxygen gas if their temperatures and pressures are the same. With the same logical extension, we can say that at the same temperature and pressure, $2.00\;\mathrm{moles}$ of oxygen gas will occupy twice the volume of $1.00\;\mathrm{mole}$ of hydrogen gas. This logic allows us an additional way to read an all gaseous chemical equation.

$2~ H_{2(g)} + O_{2(g)} \rightarrow 2~ H_2O_{(g)}$

The first way we learned to read this equation was: $2\;\mathrm{molecules}$ of hydrogen gas react with $1\;\mathrm{molecule}$ of oxygen gas to yield $2\;\mathrm{molecules}$ of gaseous water. After we learned about moles, we could also read this equation as: $2\;\mathrm{moles}$ of hydrogen gas react with $1\;\mathrm{mole}$ of oxygen gas to yield $2\;\mathrm{moles}$ of gaseous water. Now, we have a third way we can read this all gaseous equation: $2$ volumes of hydrogen gas react with $1$ volume of oxygen gas to yield $2$ volumes of gaseous water if all substances are at the same temperature and pressure. The reacting ratio indicated by the coefficients in this equation are true for molecules, moles, and volumes for an all gaseous reaction if the gases are all under the same conditions.

Sample Problem 11

What volume of oxygen gas is necessary to react with $100.\;\mathrm{L}$ of hydrogen gas assuming all volumes are measured at the same temperature and pressure?

Solution

The reacting ratio indicated by the coefficients from the equation are true for gas volumes under equal conditions.

$2~ H_{2(g)} + O_{2(g)} \rightarrow 2~ H_2O_{(g)}$

The coefficients of the balanced equation indicate that $2\;\mathrm{molecules}$ of hydrogen will require $1$ molecule of oxygen to react completely so $100.\;\mathrm{L}$ of hydrogen will required $50.\;\mathrm{L}$ of oxygen to react completely.

$\frac {2\ \text{mols H}_2} {1\ \text{mol O}_2} = \frac {100.\ \text{L of H}_2} {\ \text{x L of O}_2} && x = 50.0\ \text{L of}\ O_2$

Volume-Volume Calculations at STP

When you first learned to solve stoichiometry problems, one of the steps was to convert known quantities to moles. In those early problems, the known quantity was also given in mass and you converted mass to moles by dividing grams by molar mass. Later, known quantities may have been given to you as the molarity and volume of a solution. To convert a volume of solution at known molarity to moles, you multiplied molarity times solution volume in liters.

From now on, you may also be given known quantities as a volume of gas either at STP or at some other conditions of temperature and pressure. You already know how to convert a volume of gas at STP to moles. At STP, one mole of any gas occupies $22.4\;\mathrm{liters.}$ If you are given a volume of gas at STP as a known, you convert it to moles by dividing by $22.4\;\mathrm{L/mol}$ . If you are given a volume of gas at conditions other than STP, you must use $PV = nRT$ to calculate the number of moles.

Sample Problem 12

$N_{2(g)} + 3~ H_{2(g)} \rightarrow 2~ NH_{3(g)}$

According to this equation, how many liters of ammonia can be formed from $50.0\;\mathrm{L}$ of nitrogen gas, $N_2$ . Assume all gases are at STP.

Solution:

Since both the given substance and the requested substance are gases under the same conditions of temperature and pressure, the reacting ratio indicated by the coefficients in the equation are true for volumes.

$\frac {1\ \text{N}_2} {2\ \text{NH}_3} = \frac {50.0\ \text{L of N}_2} {\ \text{x L of NH}_3} && x = 100.\ \text{L\ of}\ NH_3$

Sample Problem 13

$N_{2(g)} + 3~ H_{2(g)} \rightarrow 2~ NH_{3(g)}$

According to this equation, how many grams of ammonia can be formed from $30.0\;\mathrm{L}$ of hydrogen gas at STP?

Solution:

Since the hydrogen gas volume is given at STP, we can convert the volume to moles by dividing by $22.4\;\mathrm{L/mol}$ . Once we have the given quantity in moles, the remainder of the problem is the same as other stoichiometry problems.

$(30.0\ \text{L}) \left (\frac {1.00\ \text{mol}} {22.4\ \text{L}}\right ) = 1.34\ \text{mols}\ H_2$

$\frac {3\ \text{H}_2} {2\ \text{NH}_3} = \frac {1.34\ \text{mol H}_2} {x\ \text{mol NH}_3} && x = 0.893\ \text{mol}\ NH_3$

$\ \text{grams}\ NH_3 = \ \text{(moles)(molar mass)} = (0.893\ \text{mol)}(17.0\ \text{g/mol}) = 15.2\ \text{grams}$

Sample Problem 14

$N_{2(g)} + 3~ H_{2(g)} \rightarrow 2~ NH_{3(g)}$

According to this equation, how many liters of ammonia gas at STP can be formed from $25.0\;\mathrm{g}$ of hydrogen gas?

Solution:

The solution steps are to convert the grams of $H_2$ to moles, use the ratio from the equation to solve for moles of ammonia gas and then convert the moles of ammonia to liters of gas at STP. The moles of ammonia gas is converted to liters at STP by multiplying by $22.4 \;\mathrm{L/mol}$ .

$(25.0\ \text{g}) \left (\frac {1.00\ \text{mol}} {2.02\ \text{g}}\right ) & = 12.4\ \text{mols}\ H_2 \ \frac {3\ \text{H}_2} {2\ \text{NH}_3} & = \frac {12.4\ \text{mol H}_2} {\ \text{x mol NH}_3} \qquad \quad x = 8.27\ \text{mol}\ NH_3 \ \ \text{liters}\ NH_3 & = \ \text{(moles)(molar mass)} = (8.27\ \text{mol})(22.4\ \text{L/mol}) = 185\ \text{liters}$

Mole-Volume or Volume-Mole Calculations Not at STP

When volumes involved in chemical reactions are not given at standard conditions, the conversion between moles of gas and volume of gas can be determined by using the universal gas law, $PV = nRT$ .

Sample Problem 15

$1,000$ . grams of calcium carbonate are heated and reacts according to the following equation.

$CaCO_{3(s)} \rightarrow CaO_{(s)} + CO_{2(g)}$

If the carbon dioxide is collected at $500.\;\mathrm{K}$ and $2.00\;\mathrm{atm}$ , what volume will it occupy under these conditions?

Solution:

$(1000.\ \text{g}) \left (\frac {1.00\ \text{mol}} {100.\ \text{g}}\right ) = 10.0\ \text{mols}\ CaCO_3$

Since the reacting ratio between calcium carbonate and carbon dioxide in the equation is $1$ to $1$ , the $10.0\;\mathrm{mols}$ of calcium carbonate will form $10.0\;\mathrm{mols}$ of carbon dioxide. We use $PV = nRT$ to find the volume of $10.0\;\mathrm{mols}$ of gas under these conditions.

$V = \frac {nRT} {P} = \frac {(10.0\ \text{mols})(0.08206\ \text{L} \cdot \ \text{atm/mol}\ \cdot K)(500.\ \text{K})} {(2.00\ \text{atm})} = 205\ \text{liters}$

Lesson Summary

In a balanced gaseous equation, the coefficients apply to molecules, moles, and volumes of gas if the gases are under the same conditions of temperature and pressure.

Review Questions

How many liters of hydrogen gas are required to react with $25.0\;\mathrm{ L}$ of nitrogen gas according to the following equation? $N_{2(g)} + 3 H_{2(g)} \rightarrow 2 NH_{3(g)}$
How many grams of ammonia will be formed from $25.0\;\mathrm{L}$ of nitrogen gas measured at STP according to the equation in question 29?

Wednesday, September 19, 2012

Unit II Reading I

The Three States of Matter

Lesson Objectives

Introduction

The Assumptions of the Kinetic Molecular Theory

The Characteristics of Solids

The Characteristics of Liquids

The Characteristics of Gases

Lesson Summary

Review Questions

Vocabulary

Labs and Demonstrations for Kinetic Molecular Theory

Brownian Motion Demonstration

Gases

Lesson Objectives

Introduction

Gases Readily Change Volume

Gases Exert Pressure

Gas Temperature and Kinetic Energy

Lesson Summary

Review Questions

Further Reading / Supplemental Links

Vocabulary

Labs and Demonstrations for Gases

Molecular Motion/Kinetic Energy Demo

Gases and Pressure

Lesson Objectives

Introduction

Pressure

Atmospheric Pressure

Measuring Gas Pressure

Lesson Summary

Review Questions

Vocabulary

Labs and Demonstrations for Gases and Pressure

Magdeburg Hemispheres Demonstration

Gas Laws

Lesson Objectives

Introduction

Boyle’s Law

Charles’s Law

Gay-Lussac’s Law

Standard Temperature and Pressure (STP)

The Combined Gas Law

Lesson Summary

Review Questions

Further Reading / Supplemental Links

Vocabulary

Labs and Demonstrations for Gas Laws

Charles Law with a Balloon and a Bunsen Burner

Universal Gas Law

Lesson Objectives

Introduction

Avogadro’s Law

The Universal Gas Law Constant

Molar Mass and the Universal Gas Law

Density and the Universal Gas Law

Lesson Summary

Review Questions

Vocabulary

Labs and Demonstrations for Universal Gas Law

Teacher’s Pages for Finding the Molar Mass of a Gas Experimentally Lab

Finding the Molar Mass of a Gas Experimentally Lab

Molar Volume

Lesson Objectives

Introduction

Molar Volume of Gases at STP

Dalton’s Law of Partial Pressures

Graham’s Law of Diffusion

Lesson Summary

Review Questions

Vocabulary

Labs and Demonstrations for Molar Volume

Rate of Diffusion at Various Temperatures Demo

Stoichiometry Involving Gases

Lesson Objectives

Introduction

Volumes, Moles, and Molecules

Volume-Volume Calculations at STP