# Unit L Reading Notespage 1 of 11

Unit L Reading NotesPage 1 of 11

Our goal for this unit is to understand the behavior of gases so that we can predict how the pressure, volume, and temperature of gases vary with respect to each other. We will next use this knowledge to help us understand and predict volumes of gases reacting at non-STP (remember our old friends Standard Temperature [273 K] and Pressure [101.3 kPa]?) conditions.

Scientists define pressure as the “force per unit of area:” , where force is often measured in Newtons, and area is often measured in square meters (m2). We define 1 as the SI unit of the pressure, the pascal, Pa. Most of the pressure units you will see will be in kilopascals (how many pascals is this?), kPa. We will discuss other common units of pressure in a bit. The text gives an example in Figure 10-6 (page 310) of an “empty can” which has no liquid contents. Instead, it is filled with air molecules that are at the same temperature as the air outside the can. That means that these particles will have the same average KE and density (think of density here as the number of particles per unit of volume) as the air particles outside the can, so the pressure exerted by the air molecules inside the can must be the same as the pressure being exerted by the air molecules on the outside surface of the can. We conclude that this must be true because when we remove gas particles from the inside of the can with a vacuum pump, the can will collapse, or implode, as seen in Figure 10-6 (b).

Even though we really don’t sense it, gas particles have both weight (a force to due gravity acting on a mass) and velocity. Because velocity is related to kinetic energy, KE, these are both forces that the gas particles exert on any surface that they contact (or collide with). This is the source of what we call atmospheric pressure, air pressure, or barometric pressure of a gas. There are different ways in which we can measure pressure.

A barometer is a glass tube filled with mercury (Hg) which stands in a reservoir of Hg (see Figure 6, page 419). A scientist named Torricelli first recognized that this tube of Hg almost always had a height of 760 mm. He concluded that, because of the way he set up this system, the space above the column of Hg in the tube was essentially a vacuum with no gas particles in it. Because the height of the mercury in the tube stayed at the same height and didn’t “fall out” of the tube and overflow the Hg reservoir, he recognized that the molecules of gases must be pushing back with the same force as the weight of the Hg was pushing down inside the tube. Torricelli’s experiments were conducted at sea level, and we have since found that the height of Hg will vary depending on the altitude where a barometer is set up. It is also dependent on temperature—the greater the temperature, the higher the Hg column will be. Changes in weather also affect the height of the Hg column—bad weather causes it to drop, so we say that bad weather is characterized by low air pressure. High air pressure values (above 760 mm Hg) occur when the weather is good (clear skies, no precipitation).

We have since established that Torricelli’s measurement of 760 mm Hg at sea level at 0 C (273 Kelvins, as we will find out later) is considered to be standard pressure. (0 C [273 Kelvins] is called standard temperature. Sometimes mm Hg is called a torr, in honor of Torricelli. This same height of mercury (760 mm) is equivalent to 101.3 kPa, or another unit of air pressure called the atmosphere (atm) We can convert between different units of pressure using the following equivalency:

760 mm Hg (torr) = 1 atm = 101.3 kPa (or 101,300 Pa)

This information is summarized in Table 10-1 on page 311. Let us now practice converting between the different units of pressure. We will look at practice problem #1 on page 312:

Convert a pressure of 1.75 atm to kPa and to mm Hg.

A)1.75 atm  ? kPa
177 kPa

B)1.75 atm  ? mm Hg
1330 mm Hg

The term kinetic applies to anything that is in motion, and the kinetic-molecular theory of matter will help to explain gas behavior. We define an ideal gas as an imaginary gas that fits all 5 of the assumptions of the kinetic theory of ideal gases outlined below (please forgive the fact that I go in a different order from the textbook):

1. Gas particle are so far apart (on average) that the total volume of the particles is extremely small compared to the volume of the container that they occupy. This means that, for ideal gases, the actual particle volumes are considered unimportant compared to the space they occupy. THIS MEANS THAT GASES ARE EASILY SMUSHED TOGETHER, OR COMPRESSED (remember the syringe demo we did in class?)
2. Gas particles are in constant, rapid motion in all directions—they have kinetic energy (the energy associated with motion), KE, and they constantly collide with each other and with the walls of their container. THIS CAUSES PRESSURE TO BE EXERTED BY THE GAS PARTICLES ON THE WALLS OF THEIR CONTAINER OR ANY OTHER SURFACE WITH WHICH THEY ARE IN CONTACT.
3. No energy is lost during collisions with other gas particles or surfaces (e.g., container walls). THIS MEANS THAT THE TOTAL KE OF A GAS SYSTEM IS CONSTANT IF THE VOLUME, TEMPERATURE, AND PRESSURE OF THE SYSTEM ARE UNCHANGED. Collisions in which no energy is lost are called elastic collisions.
4. Ideal gas particles don’t want to stick together because they aren’t attracted to, or repelled by, other gas particles or surfaces. Collisions that involve “sticking” are called inelastic collisions, so we say that ideal gas particles do not have inelastic collisions—all of them are elastic.
5. Temperature is a measure of the average KE of any sample of matter. If we define KE as:
KE = , where m = particle mass, and v = particle velocity, then it is clear that KE of a gas system will depend directly on its mass and on the square of velocity. What I want you to take away from this, is that the higher the KE for a given gas sample, the faster the gas particles will be moving. This will be used to explain a lot of gas behaviors that we will study, so please make sure you understand it.

The outline also asked you to make sure you understood the connection between the kinetic theory of matter and the following gas properties:

Expansion: You probably remember from your physical science courses in middle school that gases don’t have definite shape or volume—they adopt both the shape and volume of their container. The Kinetic Theory explains this via assumptions #2 and #4 above. If they are moving really quickly, and aren’t attracted to, or repelled by, other gas particles, there is absolutely nothing to keep them from expanding to fill any container.

Fluidity:Because gas particles aren’t attracted to each other (assumption #4), they are able to flow past each other like liquids. They are therefore referred to as fluids.

Density:Gases have very low density compared to solids and liquids because the particles are, on average, so far apart (assumption # 1).

Compressibility:The actual volume of gas particles is small compared to the container volume (assumption # 1), so that means that they are easily smushed together, or compressed.

Diffusion:Because gas particles are in constant, rapid motion (assumption # 2) they readily mix with each other. This spontaneous mixing of gases is called diffusion. The rate of diffusion will depend on particle speed, diameter, and any attractions between particles (see Figure 17, p. 436)

Effusion:Gases under pressure will pass through a small opening in the container wall. The rate at which this occurs is related only to the relative velocities of the gas particles. The faster a particle is moving (the velocity), the greater its chances are for hitting the walls of a container. This higher rate of wall collisions increases the probability that a gas particle will find the hole in the wall and pass through it.

There are simple mathematical relationships between pressure and volume @ constant temperature, and between volume and temperature when pressure is held constant, and we can use the kinetic molecular theory of gases to explain them.

Boyle’s Law (Pressure-volume relationships at constant temperature)

If you look carefully at Figure 9 on page 424, you will clearlysee that when the piston on a car cylinder is pushed in, the gases inside the barrel of the piston are compressed. The volume holding the gas particles has gotten smaller (V), and if you do this yourself with a syringe, you will realize that the smaller the volume gets, the harder it is to push the plunger any further. Essentially the gas is pushing back harder, or exerting greater pressure than before you started pushing in the plunger (P). In the meantime, you have not changed the temperature of the gas or the number of particles—these are the same as they were before you started compressing the gas with the plunger, so the individual force of a particle hitting the wall of the container is the same. What is then causing this increase in pressure? The answer lies in the fact that the volume is decreasing—with less space to move around in, the gas particles are likely to hit the wall more frequently than they did before the volume was decreased. This increases the net force exerted over the inner surface area of the syringe plunger, so the pressure increases. Because the pressure and volume seem to go in different directions (V P) we say that at constant temperatures and number of particles, pressure is inversely proportional to volume. We see the reverse situation (V P) to be true as well. We can therefore express Boyle’s Law in this way—“at constant temperature, whatever direction the volume goes in, the pressure goes in the other direction.” This is represented graphically in Figure 10 on page 424 in your text.Now lets see how we can use this to predict changes in pressure and volume at constant temperature:

“A sample of gas @ 822 kPa has a volume of 312 cm3. The pressure is increased to 948 kPa. What volume will the gas occupy at the new pressure if the temperature remains unchanged?”

PROBLEM-SOLVING STRATEGY:

• Write the given and unknown information:

P1 = 822 kPaP2 = 948 kPa

V1 = 312 cm3V2 = ?

• Is the pressure increasing or decreasing (your answer should be increasing)
• What is going to happen to the volume (according to Boyle’s Law, the volume should decrease)? We now write the given info again with out predictions:

P1 = 822 kPaP2 = 948 kPaP

V1 = 312 cm3V2 = ?V

We can now do a kind of conversion factor set-up:

312 cm3 ? cm3

You will notice this is different from what we did with factor label—we will have a pressure conversion factor that we will use to either reduce or increase the volume according to our prediction that looks something like this:

? cm3

Pressure conversion factor

The last question we now have to ask, is:

• Where do I put the pressure values to make sure the volume decreases? From math we know that if we multiply any number by something bigger than 1, it will get bigger. Similarly, if we multiply any number by something smaller than 1, it will get smaller. Because our prediction was that volume would decrease, we are looking to make a pressure conversion factor that is less than 1. If we put the smaller pressure value on the top, and the larger pressure value on the bottom of the conversion factor, we will have a fraction that is a number that is less than 1, and we can solve for the predicted new volume:

271 cm3 (rounded to SF = 3)

We can also go in the opposite direction:

“The volume of a gas is 204 cm3 when the pressure is 925 kPa. If the volume of the gas is changed at constant temperature to 306 cm3, what must be the new gas pressure?

GIVEN INFORMATION AND PREDICTIONS:

P1 = 925 kPaP2 = ?P

V1 = 204 cm3V2 = 306cm3V

You will note that because volume increases, we must predict that the pressure must decrease. If this is the case, we must have a volume conversion factor that is <1. We can now set this up and solve:

617 kPa(rounded to SF = 3)

The above procedure for predicting gas pressures and volumes at constant temperature is a conceptual approach. We can also express Boyle’s Law as an equation:

P1V1 = P2V2

If we divide both sides of the equation by V2 in order to get P2 by itself, we end up with the following equation:

Hopefully you can recognize that we can plug in our given values to this and end up with exactly the same set-up that we got by the conceptual approach:

You can use either method in your HW, quizzes, and tests, but in either case, you must list your given information. You must show your predictions (up and down arrows) for the conceptual approach, or you must show the algebraic derivation of your final relationship before plugging in numbers.

Finally, it is important to realize that the same relationships between pressure and volume will hold true if the pressure changes occur on the outside surface of our gas container. If we can manage to reduce the pressure on the outside surface (external pressure) of a balloon, the volume of that balloon will increase until once again the force exerted over the internal area of the balloon (internal pressure) equals that of the external pressure.

Charles’ Law (volume-temperature relationships at constant pressure and number of particles)

If we do a little thought experiment with balloons, we can understand the relationship between volume and temperature at constant pressure. It’s the middle of summer and as you are hosting a 4th of July party before the fireworks, you want to make sure that you have plenty of balloons. You place your order, and go to pick them up the morning of the party. Think of yourself at the party store. The air pressure outside the store is the same as it is inside the store (or the glass windows would either implode or explode). The temperature outside is really hot, so you sigh with relief as you enter the air-conditioned store. You are very disappointed when they bring out your order, because the balloons look kind of limp—they are under-inflated, and you think the store employees are being lazy or cheap about filling up your balloons, so you tell them to go back and fill them completely. They say that they are just right for an outdoor party, but you won’t hear of it. They return a few minutes later with your perfectly inflated balloons, and you are very happy as you walk out of the store. You get into your car which, having left it sitting closed up in the hot sun, feels 10hotter than the outside air. You put the balloons in the back, start the car and pull out of the parking lot onto the road, when you hear what sounds like a series of backfires in the back of your car. You look in the rear view window, and you don’t see your balloons floating gaily behind you because they have all popped!

Believe it or not, we can explain what has happened with the kinetic molecular theory of gases. As you exited the store, the external pressure on the outside of the balloon remained the same as inside the store, and the number of gas particles inside the balloon didn’t change. The increase in temperature, however, caused the gas particles inside the balloon to move much faster (remember that temperature is directly proportional to the average KE [= ]). This means that the internal pressure of the balloon was greater than the external pressure, so the wall of the balloon was pressed outward (the volume increased) in an effort to equalize the internal pressure with the external pressure. The balloon material, however, was only able to stretch so much, and it therefore burst! You might ask what we can take away from all of this in terms of predicting volume-temperature relationships at constant pressure. We have basically seen that whatever direction temperature goes in, the volume will follow:

T V and T V (@ constant pressure)

This relationship is called Charles’s Law, and we say that volume is directly proportional to volume at constant pressure (and the same number of particles in the container). This is represented graphically in Figure 12 on page 427 in the text. As with Boyle’s Law there is also a mathematical expression that describes this realationship:

It should be clear to you that if the value for T is zero, the fraction on either side becomes mathematically “undefined” or “infinity.” We do know that at 0C, most balloons aren’t completely deflated. That is, they still have some measurable volume, and they aren’t gigantic numbers either. If you look at Figure 12 on page 427, you will notice that volume of a gas will continue to decrease at temperatures below the freezing point of water. It was for this reason that an absolute temperature scale had to be developed. This absolute temperature scale has degree intervals that are identical to the Celcius scale, but its absolute zero, which describes a point at which the particles of a substance theoretically have no kinetic energy (no motion at all), is at approximately 273C. Our ability to accurately predict temperatures will depend on which temperature scale we use—so we therefore must always make sure that our temperatures are in Kelvins before we do any calculations. We can do this by using the simple mathematical relationship to convert between the two temperature scales: