CDS 14 Physical Properties of Gases IV

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Del curso dictado por Rice University

General Chemistry: Concept Development and Application

228 calificaciones

Rice University

General Chemistry: Concept Development and Application

228 calificaciones

This course will cover the topics of a full year, two semester General Chemistry course. We will use a free on-line textbook, Concept Development Studies in Chemistry, available via Rice’s Connexions project. The fundamental concepts in the course will be introduced via the Concept Development Approach developed at Rice University. In this approach, we will develop the concepts you need to know from experimental observations and scientific reasoning rather than simply telling you the concepts and then asking you to simply memorize or apply them. So why use this approach? One reason is that most of us are inductive learners, meaning that we like to make specific observations and then generalize from there. Many of the most significant concepts in Chemistry are counter-intuitive. When we see where those concepts come from, we can more readily accept them, explain them, and apply them. A second reason is that scientific reasoning in general and Chemistry reasoning in particular are inductive processes. This Concept Development approach illustrates those reasoning processes. A third reason is that this is simply more interesting! The structure and reactions of matter are fascinating puzzles to be solved by observation and reasoning. It is more fun intellectually when we can solve those puzzles together, rather than simply have the answers to the riddles revealed at the outset. Recommended Background: The class can be taken by someone with no prior experience in chemistry. However, some prior familiarity with the basics of chemistry is desirable as we will cover some elements only briefly. For example, a prior high school chemistry class would be helpful. Suggested Readings: Readings will be assigned from the on-line textbook “Concept Development Studies in Chemistry”, available via Rice’s Connexions project. In addition, we will suggest readings from any of the standard textbooks in General Chemistry. A particularly good free on-line resource is Dickerson, Gray, and Haight, "Chemical Principles, 3rd Edition". Links to these two texts will be available in the Introduction module.

De la lección

Ideal Gas Law and the Kinetic Molecular Theory

One of the powers of chemistry is the ability to relate the properties of individual molecules to the physical and chemical properties of the compounds of these molecules. In other words, we want to relate the atomic molecular world to the macroscopic world of materials. We begin this study by observing the physical properties of gases and deriving an equation which relates these properties. From this law, we can devise a model which describes how these physical properties result from the properties and motions of individual molecules. Understanding the significance of temperature is a critical part of this study.

CDS 14 Physical Properties of Gases I12:15

CDS 14 Physical Properties of Gases II13:17

CDS 14 Physical Properties of Gases III12:24

CDS 14 Physical Properties of Gases IV6:11

Conoce a los instructores

John Steven Hutchinson
Dean of Undergraduates and Professor of Chemistry

At the beginning of the previous lecture, I said we were going to wrap up

our study of the physical properties of

gases including the discussion of mixture of gases.

But on the fly I decided to cut that last part out and

save it for a second lecture, so that the lecture wouldn't get too long.

So this lecture will be the last lecture in our study

of the physical properties of gases and concept development study number 14.

Once again, recall that we are studying the Ideal Gas Law.

That tells us for any gas,

that's a pressure times the volume is the number of moles times R times T.

In this case what we want to actually consider is, what happens if we

are measuring the pressure of the gas, when we mix two gasses together?

So, for example, let's consider taking two flasks here.

The flask on the left contains only Xenon.

We captured one atmosphere pressure of Xenon at 298K.

Not really need to know what the volume of the flask

is, other than the fact that it is the same size as the flask on the right.

The flask on the right contains only Argon, in this

case at a pressure of 1.5 atmosphere is at 298 Kelvin.

And we've separated these two by a a stop cock here in middle of

valve, which has cut them off from each other, so that they can't intermingle.

We have pure Xenon on the left and pure Argon on the right.

If we in fact now open the stop cock and ask, what happens?

We know at least in part what will happen.

We know that the gases are going to mix together.

We know that we are going to wind up with Argon

on the left and as well as Xenon on the right.

In fact, we are going to wind up with a mixture of both gases on both sides.

Fully intermixed and furthermore the pressures will

of course, can become exactly the same

on the two different sides.

And notice that the pressure is 1.25 atmosphere.

Whereas if we go back and take a look, the pressure of

the Xenon in the, in, before we open the flask, was 1 atmosphere.

And the pressure of the Argon was 1.5 atmospheres.

If you've got a quick eye for number, you'll notice that the

final pressure is the average of the pressures that we started off with.

But there's actually

something a little bit more interesting that we can examine going beyond that.

What we can examine is a calculation of, what would the

pressure of the Xenon be, if it were the only gas present?

If there were no Argon in the right flask before we open the stopcock.

Well the answer is pretty simple.

We've doubled the volume of the, that the Xenon can move

around in.

We've held the temperature on the number of moles constant.

So if we double the volume, we have to half, ha, half the pressure.

So the pressure of the Xenon, if it were the only

gas in that flask, would be half of the original one atmosphere.

Since we started off with one, one atmosphere.

So would be 0.50 atmospheres of pressure.

Furthermore, what if we calculated the pressure as

of the Argon, if we had only Argon.

In other words, what if we had started this experiment with

only Argon in the right side and nothing in the left side.

And we open the stop clock, then the Argon will flow through there.

What would the pressure then be of the Argon?

Well, we started off with 1.50 atmospheres, we've doubled the volume.

So the pressure of the Argon

must become one half, of the original pressure,

since we've doubled the volume and Boyle's Law applies.

And the pressure would be 0.75 atmospheres.

Notice actually something rather interesting here.

The final pressure is actually equal to the sum of the two pressures here.

That the observed pressure of 1.25 atmospheres,

is equal to the sum of the pressure of the Xenon if it were the only gas present.

Plus the pressure of the Argon if it were the only gas present.

Within defined something we will call the partial pressure of the gas.

The partial pressure is the pressure the gas

would exhibit if it were the only gas present.

We've actually then calculated the partial pressures of each of these,

using the ideal gas law for each of the gas components.

So the partial pressure is simply defined as the pressure

calculated by the ideal gas law for each component alone.

And notice experimentally what we have observed is the total

pressure is equal to the sum of the partial pressures.

That actually is something called Dalton's Law of partial pressures of gases.

Dalton's Law says,

that the total pressure of the gas is equal to the sum of the partial pressures.

That actually makes a lot of sense, because each gas is

not the gas pressure is independent to the type of gas.

So it shouldn't matter whether we are dealing

with a mixture of gases or an individual gas.

But Dalton's Law then is an extension of the ideal gas law to mixtures of gases.

Let's actually extend that slightly. If we define the partial pressure to

be the pressure defined by the ideal gas law as if it were the only gas present.

Then the total pressure if we had two partial pressures.

If we had two gasses present the total

pressure is just the sum of the partial pressures.

The sum of the partial pressures is then,

the ideal gas law applied to each pressure.

Well, we can factor that out.

Notice when we factor out the RT over V, we just

have the total number of moles multiplied by RT over V.

What that tells us then, is that the ideal gas law gives us the total pressure.

Calculated from the number of moles in total, multiplied by RT over V.

The total pressure is the sum of the pressure of the two gases,

as if each gas were the only gas which is present.

We'll actually use Dalton's Law in a variety of different contexts.

Not the least of which is the development of the Kinetic

Molecular Theory, which we will take up in the next lecture.

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