General Chemistry Gases

Hi, welcome back to Educator.com.0000

Today's lecture from general chemistry is on gases.0002

Going to start off with a brief introduction followed by the following series of topics.0009

The first is what we call the kinetic molecular theory of gases0014

which is basically a bunch of postulates that describe gas behavior.0018

We are then going to go over the parameters that are used to characterize gases--namely pressure, volume, temperature, and moles.0024

When we combine those, we are going to get a series of gas laws which relates all four of those parameters.0035

All of these simple gas laws are then going to culminate into what we call the ideal gas law.0042

After we study the ideal gas law, we are then going to go over some applications of the ideal gas law0048

because we can come up with several additional parameters straight from that law.0054

A unique topic then is going to be gas mixtures and what we call partial pressures.0061

Finally the last topic is going to be stoichiometry and applying it to reactions that involve a gas.0067

Then of course as always, we will finish off with a summary followed by some sample problems.0077

Basically there are five postulates to the kinetic molecular theory of gases.0085

The first one is the following.0092

It deals with gas motion and basically tells us that gases travel in straight lines obeying Newton's laws.0095

They have straight trajectories; they are in constant motion.0106

Number two, the molecules in the gas occupy no volume.0110

That is we treat them as individual points.0114

In other words, if you look at a gas sample, most of it is actually empty air.0119

Number three, when gas molecules collide, we say that they follow elastic collisions.0124

That is upon collision, there is no loss of energy.0131

There is no transfer of energy.0136

You can imagine a bunch of billiard balls colliding with each other.0138

After they collide, they bounce off each other and go their separate ways.0143

Imagine that but pretty much going on infinitely with no loss of energy.0147

Number four, there are no attractive or repulsive forces between gas molecules0153

which explains why gases are so diffuse if you will.0159

Finally when we talk about the kinetic energy of a gas,0166

the kinetic energy of a molecule is really related to its kelvin temperature.0171

Any gas that follows these five postulates of the kinetic molecular theory, we call that gas an ideal gas.0178

In reality, there is no such thing as an ideal gas.0190

But by applying this model, it allows us to make a lot of simplifications and0195

a lot of assumptions which allows us to further study the gases and0200

use models which help us to describe gas behavior pretty well actually.0206

That is the kinetic molecular theory of gases.0215

We now next turn to the parameters that are used to characterize gases.0219

The parameters are basically pressure, volume, temperature, and moles.0224

From physics, pressure is formally defined as the amount of force per unit area.0231

Pressure is equal to force per unit area.0237

If we had for example a flat surface and we put a box on it,0246

that box is applying a downward force on the surface.0252

In other words, the box is applying a certain pressure on the surface.0256

If we take the same box but we stand it upright this time, same box,0260

my area of contact is now a lot smaller than in the first picture.0266

In this case, because the area is smaller, the pressure is going to be larger.0273

That is our formal definition of pressure as defined from physics.0284

But in terms of gases, we are going to describe pressure from a particulate level diagram.0290

Imagine a container.0297

Basically we have gas particles that are moving in random directions once again following0301

the kinetic molecular theory of gases and moving in straight lines, constant motion.0307

Not only do the gas particles collide with each other but the gas particles0314

also collide with the container wall; collision with container wall.0319

When that collision occurs with the wall of the container, that itself0331

generates a force just like billiard balls colliding with each other.0334

It is this force of impact that we tend to relate to gas pressure.0339

Force of impact is proportional to gas pressure.0348

For chemistry, for gases, this is our interpretation of pressure.0362

Now that we have defined pressure, let's go ahead and talk about0370

the common units of pressure that are used to make measurements.0373

From physics, the typical units of pressure are going to be psi which is pounds per square inch and pascal.0380

All of us see psi in tires; we also see psi for water pressure too.0388

We are not going to use psi and pascal too much.0400

Really in chemistry, we are going to use these three--millimeters of mercury, atm, and torr.0402

Millimeters of mercury is what we see on a barometer for the weather report.0407

We also see that whenever you take your blood pressure reading.0414

Atm stands for atmospheres or one atmosphere of pressure.0419

Basically one atmosphere of pressure is going to be the pressure roughly at sea level.0430

Once again at sea level, we are at roughly one atm of pressure.0440

Finally torr, torr is named after Torricelli who invented the barometer.0445

Given at the bottom, we have the relationships for each of these units.0462

One atm is equal to 14.7 pounds per square inch which is equal to 760 millimeters of mercury0467

which is equal to 760 Torr which is equal to 101.325 kilopascals.0475

Of course, you see all of these equivalent statements.0481

From equivalent statements, we can then use them as conversion factors.0483

Now that we have talked about pressure, let's continue on.0490

From the kinetic molecular theory, we are told that gases travel in straight paths.0494

This implies the following.0500

That gases are going to travel in straight paths until they collide with something,0502

either with each other or the wall of the container.0505

What that says is that gases are going to expand to fill their entire container.0508

Hence the volume of a gas is strictly determined by the container that it is placed into.0514

The reason why we can do this is because gases are compressible.0522

Remember that most of a gas is empty air, that they have negligible volume.0526

I could take the same amount of gas in a ten liter bottle0533

and compress it easily to a smaller bottle, no problem.0536

The common units of volume are going to be liters of course and milliliters.0541

Sometimes milliliters, you also see cc or cubic centimeters.0545

Another postulate from the kinetic molecular theory is the kelvin temperature.0553

We say that the kinetic energy of a gas is proportional to its kelvin temperature.0559

In other words, temperature of a gas is directly related to how fast these particles are moving.0563

Because it is directly related, this says the following.0570

The faster a gas is, the hotter its kelvin temperature.0573

We can also interpret this at the particulate level.0577

We are saying that kinetic energy is going to be proportional to the temperature in kelvin.0583

If you consider a gas sample, if we apply heat to this, of course that is going to result in0593

much faster motion because the gases will get more kinetic energy; faster motion, higher kinetic energy.0606

Finally the last parameter is the mole amount of a gas.0625

For gases, the amount of a gas is going to be related to moles of course.0632

Those are the four parameters--pressure, volume, temperature in kelvin, and moles.0637

Now that we have introduced the four parameters, we now get into what is called the simple gas laws.0645

The simple gas laws basically do the following.0650

They are a series of equations that relate the four parameters of a gas that we just covered.0653

There are restrictions for these gas laws to work.0660

First we assume ideal behavior.0663

That is the gas is going to follow all five postulates of the kinetic molecular theory.0665

All other parameters are held constant.0671

If I compare pressure and volume for example, that means temperature and moles are being held constant.0675

All else is held constant.0680

Let's go ahead and tackle each of these gas laws now.0682

The first gas law is called Charles's law.0685

Charles's law states that volume and temperature are directly related, holding pressure and moles constant.0688

Volume is directly related to temperature; let's take a look at this.0696

If I have this container and these gas particles are moving in random directions,0702

if I heat it, the gas particles are going to have more kinetic energy.0712

We are going to result in an expansion of the container if the container is flexible.0719

This is going to be larger volume.0728

That is why if you try heating a balloon up, you see the balloon expanding.0733

It is because the temperature is going to increase the kinetic energy of the molecules.0739

They are going to push outward on the container, on the wall of the balloon.0746

The equation to do this, to quantify Charles's law, is V₁ over T₁ is equal to V₂ over T₂.0751

That is if we know the initial volume and initial temperature, we can get0762

either the final volume or the final temperature, whichever is not given.0765

The restriction for this equation is that this must be in kelvin.0771

Temperature must be in kelvin.0775

Volume, it can be in any units as long as they are identical units.0777

That is a pretty straightforward equation to use.0787

Volume and temperature are directly proportional to each other.0790

The next gas law is what we call Boyle's law.0794

Boyle's law states the following.0796

That pressure and volume of a gas are inversely related when temperature and moles are held constant.0799

That is pressure is inversely related to volume.0805

Let's think about this; let's say I had two gas particles in a small container.0810

All of a sudden, let's say the size of the container has increased.0819

I am holding everything else constant.0824

When this happens, I am going to have a smaller rate of collision with the container wall.0828

Because my rate of collisions with the container wall is going to be smaller, my force drops off.0844

Therefore my pressure is going to drop off.0853

Once again pressure is inversely related to volume.0859

If you ever go to the higher elevation, you notice that a potato chip bag or a snack bag is always larger.0862

It is because at the higher elevation, the outside pressure is much smaller.0870

To compensate, the air inside the bag is going to expand.0875

This is why balloons also tend to pop the higher they go.0880

Because as the elevation increases, the air pressure gets lower.0884

The air molecules inside the balloon expand against the walls of the balloon.0887

The equation for this, for Boyle's law is P₁V₁ is equal to P₂V₂.0894

Once again for this equation to work, P₁ and P₂ must be identical units.0902

V₁ and V₂ also must be identical units.0913

Once again this is a rather straightforward equation to use.0917

We can calculate any final pressure or volume given the other three parameters.0919

That is Boyle's law.0926

The next gas law is what we call Gay Lussac's law.0929

Gay Lussac's law is pressure and temperature.0933

It tells us that pressure and temperature of a gas are directly related when volume and moles are held constant.0936

That is pressure is directly related to temperature; that just makes sense.0943

When we have gas particles just like this, let's say this is colder.0950

That is going to generate some pressure.0958

But if we take the same volume, the same box, the same amount, and I heat this sample up, that is going to0960

result in obviously a higher rate of collision with the walls of the container; more collisions with container wall.0969

That is going to increase my force.0982

Because my force goes up, my pressure goes up also.0985

On extreme temperature differences, a car tire is always going to be at a lower pressure when it is colder.0992

Later in the day when it gets much hotter, the pressure will slightly increase because of this difference.1002

The equation for Gay Lussac's law is the following.1009

P₁ over T₁ is equal to P₂ over T₂.1011

Once again the units of pressure must be identical.1016

But remember that our temperature is always related in kelvin whenever discussing a gas.1021

That is Gay Lussac's law.1029

The last simple gas law is what we call Avogadro's law.1033

Avogadro's law is the following.1037

That the volume and moles of a gas are directly related when temperature and pressure are held constant.1039

V is proportional to n.1046

Just think about maybe a car tire.1049

You put more air into it; you increase the amount of air.1054

What happens?--the volume increases.1059

The equation for Avogadro's law is V₁ over n₁ is equal to V₂ over n₂.1064

Once again the volume units must be identical.1073

n₁ and n₂ will always be in moles.1080

That is Avogadro's law.1084

If you look at the four gas laws, Charles's law, Boyle's law, Gay Lussac's law,1086

and Avogadro's law, really Boyle's law is the only one that stands out.1091

It is the only where we have something times something is equal to the product of something else.1096

Every other gas law is division on the left side of the equation and on the right side of the equation.1102

Please make a note of that, Boyle's law is definitely the one that stands out.1110

In case your instructor ever requires you to memorize these gas laws.1119

When we put all of these simple gas laws together, they culminate into one equation.1124

This grand equation is what we call the ideal gas law where PV is equal to nRT.1132

When we do this, there are a couple of restrictions.1138

That pressure must be in atm; volume must be in liters; n is simply moles.1140

The temperature must be in units of kelvin as we always have said.1147

There is something we haven't introduced yet; that is what R is.1152

R is what we call the universal gas constant.1155

It is equal to 0.08206 liters atmosphere K mol.1158

Once again you may or may not have to memorize this.1164

Definitely refer to your instructor for that.1166

That is a relatively straightforward equation to use.1169

Pretty much for an ideal gas, I can determine the pressure, volume, moles,1171

or kelvin temperature given any of the other three parameters.1177

Again this is the ideal gas law.1183

Probably something you want to be comfortable with is to solve for a single variable.1187

Pressure here is going to be equal to nRT over V.1193

Volume is equal to nRT over P; n is equal to PV over RT.1198

Temperature is going to be equal to PV over nR.1206

That is again the ideal gas law.1211

The ideal gas law is relatively straightforward to use.1214

But another important aspect of it is that we can derive and come to many conclusions using this law.1217

The first relationship that we are going to get from the ideal gas law is what is called standard temperature and pressure.1226

It becomes very difficult to compare gases because there is so many parameters--pressure, volume, temperature, and moles.1235

A set of universal conditions has been defined.1242

This set of universal conditions is called standard temperature and pressure or STP for short.1246

Standard temperature is 273.15 kelvin; standard temperature is 1.0 atm.1252

When these values are plugged into the ideal gas law, we can go ahead and solve for the ratio of volume to moles.1258

When we solve for this ratio of volume to moles, we get exactly 22.4 liters per mole.1265

This is what we call molar volume; its significance is the following.1271

That at STP, one mole of any ideal gas regardless of its identity occupies a volume exactly 22.4 liters.1277

One mole equals 22.4 liters; that is an equivalence statement.1288

From that, we can use that as a typical conversion factor.1294

Once again molar volume at STP only, 22.4 liters per mole.1298

Another application we can derive from the ideal gas law is gas density.1304

Gas density is going to be measured in grams per liter.1312

We are not going to be in its derivation.1315

But the density of a gas in grams per liter is equal to the following.1317

It is equal to the molar mass of the gas in grams per mole times1322

the pressure in atm divided by the universal gas constant times the kelvin temperature.1330

You can convince yourself that when all the units cancel, we are left with grams per liter.1337

This equation once again is relatively straightforward to use.1342

However there is an important thing that cannot be overlooked.1345

We now have a relationship between density and temperature for gases.1348

You see here that density is going to be inversely related to the kelvin temperature.1353

That means the following.1359

That as temperature of a gas goes up, the gas density decreases.1360

As temperature goes up, gases tend to become lighter.1366

Therefore they tend to rise; this explains why hot air balloons rise.1373

As you heat the gas within the walls of the balloon, the gas becomes less dense than air.1379

It results in a lower density and results in rising of the hot air, bringing the balloon upwards.1385

Once again density is inversely related to the kelvin temperature of a gas.1394

A final gas law that focuses on pressure, this is called Dalton's law of partial pressures.1401

Dalton's law of partial pressures refers to gas mixtures.1409

It tells us the following.1412

Pretty much that the whole is equal to the sum of the parts.1413

The sum of the individual pressures of each gas component is equal to the total pressure of the gas mixture.1419

These individual pressures, the technical term is called partial pressures.1426

Basically very simple--the total gas pressure of a mixture is equal to the partial pressure of1431

the first gas plus the partial pressure of the second gas, etc.1443

Once again the total pressure is equal simply to the sum of all individual pressures.1448

We now come back to stoichiometry.1457

Stoichiometry is something that we spend a great deal of time on.1463

At the basis of stoichiometry was the following.1468

We want to go from moles of A to moles of B.1472

To do this, we use the conversion factor, the mole to mole ratio.1475

From moles of B, you can go to grams using molar mass.1481

You can go to atoms and molecules using Avogadro.1489

You can go to liters if it is a solution using molarity.1501

The same thing applies on the other side to go to moles of A for example.1507

We spent a deal of time doing mole to mole conversion and also mass to mass conversions.1512

All we are going to do now, we are going to apply our knowledge of stoichiometry to gases.1517

If everything is about pretty much getting to moles first, we have an ideal gas law that helps us do that.1524

Typically for gas stoichiometry problems, we are going to use ideal gas law where n is equal to PV over RT.1532

If we are at standard temperature and pressure, we could take a shortcut.1540

We can just use the molar volume to get to moles because we know that one mole equals 22.4 liters.1544

In this case, the ideal gas law not needed.1552

But again this is only at STP.1557

Let's go ahead and do a sample problem then; the question is the following.1561

How many liters of oxygen gas at standard temperature and pressure are needed to form 10.5 grams of water vapor?1566

As soon as I see the letters STP, I know that I am dealing with 1 atm pressure and 273.15 kelvin.1573

I also know that one mole of a gas is going to be equal to exactly 22.4 liters.1582

The first thing you always do in stoichiometry is to make sure the chemical equation is balanced like we have always done.1590

Here we are going to need two hydrogens and two waters.1598

What do we have here?--we have the mass of the water vapor.1605

Somehow we want to go from mass of water vapor all the way to liters of O₂ gas.1613

Because I am at STP, the liters of O₂ gas is going to1622

come from molar volume which is one mole is equal to 22.4 liters.1627

But in order to get the moles of O₂, I first need the moles of H₂O.1641

Before moles of H₂O, we then have our mass of H₂O which is given.1651

There is our basic flow chart; it is pretty much three main steps.1656

Let's go ahead and do this.1660

10.5 grams of water vapor times 1 mole of water divided by its molar mass of 18.016 grams of water.1664

That gives me moles of A.1679

Now from moles of A to moles of B using the mole to mole ratio1681

which is 1 mole of O₂ over 2 moles of H₂O.1684

Finally now that I am at moles of O₂, I can go ahead and1690

use molar volume as a conversion factor to go and get volume.1693

22.4 liters for every one mole of O₂.1697

When all is said and done, we get a volume of 6.5 liters that are1703

required at STP for this reaction to make 10.5 grams of water vapor.1707

Let's now go ahead and summarize this lecture.1715

We started off today's lecture with the kinetic molecular theory of gases.1719

It is basically five postulates which describe ideal gas behavior.1723

We then proceeded to tackle the simple gas laws which basically relates the1728

four parameters used to characterize gases--pressure, volume, kelvin temperature, and moles.1732

When we culminated all of these simple gas laws, we arrived at the ideal gas law.1740

The ideal gas law allows us to come up with many applications including density and its relationship with temperature.1745

Finally all of our stoichiometry skills that we established previously can easily apply to gas problems.1754

That is our summary; let's go ahead and do a series of sample problems.1764

Here is sample problem one; you have a 827 milligram sample of a gas.1769

It occupies 0.270 liters when measured at a temperature of 88 degrees Celsius and a pressure of 975 millimeters of mercury.1775

Calculate the molar mass of the gas; let's take it one by one.1784

Here we have mass; here we have volume, temperature, and pressure.1788

The question is asking for molar mass.1797

Molar mass, we all know to be in units of grams per mole.1800

We have the grams already; that is the 827 milligrams or the 0.827 grams.1805

All we have to get then is the moles.1812

Once we have that, we can divide the two numbers to give us the molar mass.1815

We need to get the moles of this gas which is n.1821

We are given pressure, volume, and temperature; that is three out of the four parameters.1826

We can go ahead and use the ideal gas law to help us do this.1834

Moles is equal to PV over RT.1838

Pressure is 975 millimeters of mercury.1845

We have to then go ahead and convert this to atm remember.1851

That is our restriction.1853

We are going to multiply this by the volume in liters.1857

It is already in liters.1860

We are going to divide this by the universal gas constant.1863

We are going to then multiply this by the kelvin temperature.1871

88 plus 273.15; this gets us 0.012 moles.1876

Now we can go ahead and proceed to solve for the molar mass.1889

0.827 grams over 0.012 moles.1892

That is going to be equal to roughly 69 grams per mole for molar mass.1899

This is another nice application of the ideal gas law.1907

It can be used to determine the molar mass of a gas that follows ideal behavior.1911

Let's go ahead and now proceed on to sample problem two.1919

What mass of silver(I) oxide is required to form 388 milliliters of O₂ gas1923

when measured at 734 millimeters of mercury and 25 degrees Celsius?1930

Mass is what we want to get; we are given volume.1937

We are given pressure; we are given temperature; guess what?1940

You have chemical equation here which pretty much means you have a stoichiometry problem.1945

Always the first step is to balance.1951

When we go ahead and balance this, we are going to need 2 silver oxides and 4 silvers.1955

We want to go from basically the following.1962

We are given the pressure, the temperature, and the volume of O₂.1967

That is three out of four parameters.1971

We can go ahead and get the moles of O₂.1973

n of O₂ is equal to PV over RT.1976

That is going to be equal to 734 millimeters of mercury times 1 atm divided 760 millimeters of mercury.1981

Going to multiply that by the volume in liters which is 0.388 liters,1994

divided by the universal gas constant, 0.08206 liters atmosphere K mol,2000

times the temperature in kelvin, 25 plus 273.15.2009

Then the moles of O₂, we get 0.015 moles of oxygen gas.2015

We want to go from moles of O₂ which is what we have.2023

Somehow we want to go all the way to the mass of silver(I) oxide.2027

We know how to do that.2032

This is really now just a matter of doing something we have already learned.2033

We are going to go from the moles of O₂ to the moles of silver(I) oxide using the mole to mole ratio.2038

Then on from there is to the mass of silver(I) oxide using the molar mass.2048

Let's go ahead and finish this up.2055

You have 0.015 moles of O₂.2058

The mole to mole ratio is going to be 2 moles of silver(I) oxide for every 1 mole of oxygen gas.2062

Then we are going to go ahead and multiply this by the molar mass of silver(I) oxide to get to grams.2072

Its molar mass is 231.74 grams for every 1 mole of silver(I) oxide.2078

We get roughly 7.0 grams of silver(I) oxide that are required.2089

That is another stoichiometry problem that involves gases.2097

Thank you all for your attention; I will see you all next time on Educator.com.2102