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.
CDS 14 Physical Properties of Gases III
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From the course by Rice University
General Chemistry: Concept Development and Application
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From the lesson
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.
Meet the Instructors
John Steven HutchinsonDean of Undergraduates and Professor of Chemistry
In this lecture, we're going to wrap up
our study of the physical properties of gasses.
This will conclude the material from Concept Development Study number 14.
So you may want to review that reading as well.
In this lecture, we're going to discuss a few applications of the
Ideal Gas Law which we developed in the previous two lectures.
And we're also going to extend the Ideal Gas
Law to a discussion of mixtures of gasses.
To review very quickly, you'll recall that the Ideal
Gas Law tells us that for any gas regardless
of it's type, type of molecules, that the pressure times the volume is equal to
the number of moles of the gas multiplied
by the temperature multiplied by a constant R.
That constant is given here on the screen.
It is in fact just experimentally measured
to be 0.08206 liter atmospheres per mole Kelvin.
The Ideal Gas Law is a summary of the three experimental
laws that we've produced. Boyle's Law, relating pressure and volume.
Charles's Law, relating volume and temperature.
And Avogardo's Law, relating the volume and the numbers of moles.
Remember that, ideal means that it is independent of the type of gas molecule.
What that means is, we can apply it to just
about any kind of gas that we are interested in.
Here is a particular example. What volume of gas is occupied
by one mole of the gas if we are at room temperature and one atmosphere pressure?
Again the constant is given here on the screen.
Well, let's write this out.
We know that PV is equal to nRT and we are now asking for what
the volume of the gas is, so we can solve this equation for the volume.
It's just the number of moles times RT divided by the pressure.
And we have given here in this equation, or
in this problem, the temperature which is in degrees centigrade 25.
the pressure on atmosphere and the number of moles here.
So you can tell we have n and temperature and pressure.
And it's just a matter of then plugging
those numbers in to get the calculation correct.
So what do we have?
1.00 moles multiplied by R which is 0.08206
liter atmospheres per mole Kelvin
multiplied by the temperature which is 25 degrees centigrade which is 298.15 Kelvin.
We always have to use absolute temperatures there.
And we divide that by the pressure which is given to
be 1.00 atmospheres and this now just becomes a calculator problem.
You can just plug these in. We'll check the units here.
Here is moles per mole.
Here is Kelvin per Kelvin. Here is atmosphere per atmosphere.
The only unit left is liters.
So in fact our answer's going to be in liters.
Plugging the numbers in and calculating them out.
We wind up with a value 24.5 litres is the volume of a single mole of gas.
And it is interesting that because this is the Ideal Gas Law,
whatever gas this is, regardless of the identity of that gas, a single mole of
gas under normal conditions, 25 degrees centigrade and
one atmosphere of pressure will occupy 24.5 liters.
Let's consider then, another application sort of stepping it up one level farther.
In this case, we're actually going to use the Ideal
Gas Law to determine the molar mass of a gas.
So we're going to study now a
hydrocarbon gas, that'll be a gas consisting
entirely of hydrogen and carbon or molecules.
And consists entirely of hydrogen and carbon.
And what we are given here is the density
of the gas at 25 degrees centigrade and one atmosphere.
But that density is not moles per litre but rather grams per litre.
But from the Ideal Gas Law, we can write
that the number of moles divided by the volume
is equal to the pressure divided by R times T.
That's just a rewriting of PV is equal to nRT.
If we then calculate the pressure divided by the temperature and with R,
you'll notice back over here that we do in fact have the temperature.
And we have the pressure.
So we have everything that we need to know to be able
to plug in this equation and find the number of moles per volume.
Let's do that.
The pressure is, what did we say, 1.00 atmospheres.
R is 0.8206 liter atmosphere per mole Kelvin.
And the temperature is, let's see, 298.15K.
And if we plug all those numbers in, let's check the units first.
Atmospheres per atmospheres,
Kelvin per Kelvin and the denominator, we still have units of litre
per mole which will ultimately then give us units of moles per litre.
And if we just plug the numbers into the calculator, we discover
that the number of moles per litre is 0.0409 moles per litre.
But of course that's not the question we were asked.
We were asked what is the molar mass of
this hydrocarbon?
And right now what we have is, instead
the number of moles per litre of the hydrocarbon.
But what we also know of course is that the
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number of moles per litre or the number of moles is equal to
the mass divided by the molar mass. So, if
we simply write this as the mass per
volume times molar mass. Just substituting n
into the equation up above, is equal
to 0.0409 moles per litre. Then let's
see, we have mass per volume is here. That's the number that we've been given.
That's 1.80 grams per mole. So, we get 1.80 grams per litre times one
over the molar mass, is equal to 0.0409 moles per liter.
If we solve
now for the molar mass, that's just taking let's see,
we'll move mass to, molar mass to the other side.
Divide both sides of the equation by 0.0409.
If we do that calculation we will wind up with the molar mass
being 44.0 grams per mole.
The units of litres cancel out.
The last part of this question then, asks us to identify
what is this hydrocarbon's molecular formula?
Well that's a trickier problem.
But, remember, this hydrocarbon consists only of hydrogen and carbon.
So we have to have a comb, combination of the mass
of carbon and the mass of hydrogen which adds to 44.
And with a little bit of work, we can see that, that is
three times 12 plus eight times one would give us in fact the gas
which is C3H8 or propane.
So in fact just by measuring the density of the gas under specified conditions,
temperature and pressure, we can figure out what the molar mass of the gas is.
Let's try another example.
Here is actually applying still at the Ideal Gas Law to do some Stoichiometry.
In this particular case, we are going to react calcium hydride, which is
a solid, with water.
When calcium hydride reacts with water, it
produces hydrogen gas as well as calcium hydroxide.
In this case, we're going to say that an experiment has been one in
which we have a five gram solid
sample that contains calcium carbon calcium hydride.
But is not exclusively calcium hydride.
It contains other impurities that don't react with water.
The question then is, if we produce a
specified amount of hydrogen gas as the product.
Here we have under given conditions.
How much calcium hydride was there in the original sample?
To do the stoichiometry, we have to know what the balanced equation is.
But we're given the information we need.
Calcium hydride reacts with water to
produce calcium hydroxide and hydrogen gas.
Let's be careful to make sure that this equation is balanced.
We have one calcium on each side of the equation.
We have an one oxygen on the left side but two on the right side.
So we'd better put a two in front of the
water so now we have two oxygens on each side.
How about hydrogen?
There are six hydrogens in the reactants.
And only two hydrogens in the products plus two more here.
We need to double the amount of hydrogen gas there.
So one calcium hydride
produces two hydrogen molecules. That's important because we can
calculate the number of moles of hydrogen which are produced from the Ideal Gas Law.
The Ideal Gas Law tells us that the number of moles is
equal to the pressure times the volume divided by R times T.
Notice if I multiplied both sides of the equation by
RT, I would have PV as equal to NRT again.
And this is useful because,
let's see, we have been given a pressure although it's in torr.
And we're given a temperature although it's in degrees centigrade.
And we are given a volume 3.90 litres.
So we have the pressure, we have the volume and we have the temperature.
Let's write it all in here then.
The pressure is 750 torr, but we know R in values of atmosphere.
So we need to convert this
to atmospheres. And there are 760 torr per atmosphere.
That's the pressures in atmospheres.
The volume is 3.90 litres. R is 0.08206
litre atmosphere per mole Kelvin.
And the temperature is let's see, 32 degrees centigrade which is 305 Kelvin.
We can check all the units here K K's cancel out.
Torrs cancel out.
Atmosphere cancels out.
Liters cancel out.
The only unit left is one over one over moles which is units of moles.
And if we simply plug all these numbers
into our calculator and do the calculation, we
discovered that the number of moles of the
hydrogen is 0.154 I, I'm sorry 0.154 moles.
That's the number of moles of hydrogen gas which are
produced by the reaction of the calcium hydride with water.
But it is not the number of moles of calcium hydride.
The number of moles of calcium hydride is going to be half that
amount because one mole of calcium hydride produces two moles of hydrogen.
That means that the number of moles of calcium hydride is 0.077.
actually zeros, yeah, 0.0770 moles.
And now we know the number of moles of calcium hydride.
But the question asks us for what mass
percentage was the sample of the calcium hydride?
So we need to know the mass of the calcium hydride.
And the mass of the calcium hydride is then given
by the number of moles of calcium hydrate multiplied by the molar
mass of the calcium hydride using the
stoichiometry we learned in the first semester.
The molar mass of the calcium hydride is
easily calculated to be 42.1 grams per mole.
Multiplying the two numbers together then, we wind up with the
mass of calcium hydride is 3.24 grams in the original sample.
But the original sample, you recall, was five grams large.
It was not 3.24 grams.
That's because apparently there was some impurity present.
What fraction of the calcium hydride was present?
The percent of the calcium hydride will be
equal to 3.24 grams divided by five grams total.
And that as a percentage works out to then be 64.7%.
So we can actually use the Ideal Gas Law to assay samples of solid samples
by reacting those solid samples out and getting a volume of a gas produced.
That's three examples now of how we can use the Ideal Gas Law.
We'll continue in the next lecture.
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