For more information, please see full course syllabus of General Chemistry

For more information, please see full course syllabus of General Chemistry

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### Gases

- The Kinetic Molecular Theory of gases are (5) postulates which describe gas behavior. Any gas that is assumed to follow this theory is called an ideal gas.
- A series of gas laws relates the gas parameters pressure, temperature, volume and moles to each other holding all else constant.
- The ideal gas law can be used to interpret gas density and its relationship with temperature.
- We can apply stoichiometry techniques to reactions involving gases.

### Gases

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

- Intro
- Lesson Overview
- The Kinetic Molecular Theory of Gases
- Parameters To Characterize Gases
- Parameters Cont'd
- Units For Expressing Pressure: Psi, Pascal
- Units For Expressing Pressure: mm Hg
- Units For Expressing Pressure: atm
- Units For Expressing Pressure: torr
- Parameters Cont'd
- Parameters Cont'd
- Parameters To Characterize Gases: Temperature
- Particulate Level
- Parameters To Characterize Gases: Moles
- The Simple Gas Laws
- The Simple Gas Laws
- The Simple Gas Laws
- The Simple Gas Laws
- The Ideal Gas Law
- Applications of the Ideal Gas Law
- Applications of the Ideal Gas Law
- Gas Pressures and Partial Pressures
- Gas Stoichiometry
- Stoichiometry Problems Involving Gases
- Using The Ideal Gas Law to Get to Moles
- Using Molar Volume to Get to Moles
- Gas Stoichiometry Cont'd
- Summary
- Sample Problem 1: Calculate the Molar Mass of the Gas
- Sample Problem 2: What Mass of Ag₂O is Required to Form 3888 mL of O₂ Gas When Measured at 734 mm Hg and 25°C?

- Intro 0:00
- Lesson Overview 0:07
- The Kinetic Molecular Theory of Gases 1:23
- The Kinetic Molecular Theory of Gases
- Parameters To Characterize Gases 3:35
- Parameters To Characterize Gases: Pressure
- Interpreting Pressure On a Particulate Level
- Parameters Cont'd 6:08
- Units For Expressing Pressure: Psi, Pascal
- Units For Expressing Pressure: mm Hg
- Units For Expressing Pressure: atm
- Units For Expressing Pressure: torr
- Parameters Cont'd 8:09
- Parameters To Characterize Gases: Volume
- Common Units of Volume
- Parameters Cont'd 9:11
- Parameters To Characterize Gases: Temperature
- Particulate Level
- Parameters To Characterize Gases: Moles
- The Simple Gas Laws 10:43
- Gas Laws Are Only Valid For…
- Charles' Law
- The Simple Gas Laws 13:13
- Boyle's Law
- The Simple Gas Laws 15:28
- Gay-Lussac's Law
- The Simple Gas Laws 17:11
- Avogadro's Law
- The Ideal Gas Law 18:43
- The Ideal Gas Law: PV = nRT
- Applications of the Ideal Gas Law 20:12
- Standard Temperature and Pressure for Gases
- Applications of the Ideal Gas Law 21:43
- Ideal Gas Law & Gas Density
- Gas Pressures and Partial Pressures 23:18
- Dalton's Law of Partial Pressures
- Gas Stoichiometry 24:15
- Stoichiometry Problems Involving Gases
- Using The Ideal Gas Law to Get to Moles
- Using Molar Volume to Get to Moles
- Gas Stoichiometry Cont'd 26:03
- Example 1: How Many Liters of O₂ at STP are Needed to Form 10.5 g of Water Vapor?
- Summary 28:33
- Sample Problem 1: Calculate the Molar Mass of the Gas 29:28
- Sample Problem 2: What Mass of Ag₂O is Required to Form 3888 mL of O₂ Gas When Measured at 734 mm Hg and 25°C? 31:59

### General Chemistry Online Course

I. Basic Concepts & Measurement of Chemistry | ||
---|---|---|

Basic Concepts of Chemistry | 16:26 | |

Tools in Quantitative Chemistry | 29:22 | |

II. Atoms, Molecules, and Ions | ||

Atoms, Molecules, and Ions | 52:18 | |

III. Chemical Reactions | ||

Chemical Reactions | 43:24 | |

Chemical Reactions II | 55:40 | |

IV. Stoichiometry | ||

Stoichiometry I | 42:10 | |

Stoichiometry II | 42:38 | |

V. Thermochemistry | ||

Energy & Chemical Reactions | 55:28 | |

VI. Quantum Theory of Atoms | ||

Structure of Atoms | 42:33 | |

VII. Electron Configurations and Periodicity | ||

Periodic Trends | 38:50 | |

VIII. Molecular Geometry & Bonding Theory | ||

Bonding & Molecular Structure | 52:39 | |

Advanced Bonding Theories | 1:11:41 | |

IX. Gases, Solids, & Liquids | ||

Gases | 35:06 | |

Intermolecular Forces & Liquids | 33:47 | |

The Chemistry of Solids | 25:13 | |

X. Solutions, Rates of Reaction, & Equilibrium | ||

Solutions & Their Behavior | 38:06 | |

Chemical Kinetics | 37:45 | |

Principles of Chemical Equilibrium | 34:09 | |

XI. Acids & Bases Chemistry | ||

Acid-Base Chemistry | 43:44 | |

Applications of Aqueous Equilibria | 55:26 | |

XII. Thermodynamics & Electrochemistry | ||

Entropy & Free Energy | 36:13 | |

Electrochemistry | 41:16 | |

XIII. Transition Elements & Coordination Compounds | ||

The Chemistry of The Transition Metals | 39:03 | |

XIV. Nuclear Chemistry | ||

Nuclear Chemistry | 16:39 |

### Transcription: 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 gases*0014

*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 law*0048

*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 molecules*0153

*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 and*0195

*a lot of assumptions which allows us to further study the gases and*0200

*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 following*0301

*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 particles*0314

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

*When that collision occurs with the wall of the container, that itself*0331

*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 about*0370

*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 mercury*0467

*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 bottle*0533

*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 in*0593

*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 V _{1} over T_{1} is equal to V_{2} over T_{2}.*0751

*That is if we know the initial volume and initial temperature, we can get*0762

*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 P _{1}V_{1} is equal to P_{2}V_{2}.*0894

*Once again for this equation to work, P _{1} and P_{2} must be identical units.*0902

*V _{1} and V_{2} 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 to*0960

*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

*P _{1} over T_{1} is equal to P_{2} over T_{2}.*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 V _{1} over n_{1} is equal to V_{2} over n_{2}.*1064

*Once again the volume units must be identical.*1073

*n _{1} and n_{2} 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 times*1322

*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 of*1431

*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 O _{2} gas.*1613

*Because I am at STP, the liters of O _{2} gas is going to*1622

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

*But in order to get the moles of O _{2}, I first need the moles of H_{2}O.*1641

*Before moles of H _{2}O, we then have our mass of H_{2}O 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 ratio*1681

*which is 1 mole of O _{2} over 2 moles of H_{2}O.*1684

*Finally now that I am at moles of O _{2}, I can go ahead and*1690

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

*22.4 liters for every one mole of O _{2}.*1697

*When all is said and done, we get a volume of 6.5 liters that are*1703

*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 the*1728

*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 O _{2} gas*1923

*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 O _{2}.*1967

*That is three out of four parameters.*1971

*We can go ahead and get the moles of O _{2}.*1973

*n of O _{2} 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 O _{2}, we get 0.015 moles of oxygen gas.*2015

*We want to go from moles of O _{2} 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 O _{2} 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 O _{2}.*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

1 answer

Last reply by: Professor Franklin Ow

Thu Aug 4, 2016 5:14 PM

Post by Parth Shorey on August 2, 2016

I still don't understand why gas density is inversely proportional to temperature?

0 answers

Post by Peter Ke on October 13, 2015

For the first example at the last lecture, you put .8205 L X ATM/ K X MOL.

Shouldn't it be .08206 L X ATM / K X MOL.

1 answer

Last reply by: Professor Franklin Ow

Mon Mar 30, 2015 11:52 AM

Post by Muhammad Ziad on March 29, 2015

Hello Professor Ow, what does it mean for some parameters to be held constant? Thanks!

1 answer

Last reply by: Okwudili Ezeh

Tue Oct 28, 2014 10:59 PM

Post by Okwudili Ezeh on October 28, 2014

Do you mind checking example 1 again. I got 13.06 liters.

0 answers

Post by Saadman Elman on June 14, 2014

It was really helpful! Thanks a lot!

1 answer

Last reply by: Professor Franklin Ow

Tue Jan 14, 2014 11:14 PM

Post by felicia ekeson on January 14, 2014

hello, I 've tried 734mmHg X 1 atm /0.388L /0.08206 X 25+273.15 but unable to get 0.015 mol

1 answer

Last reply by: Professor Franklin Ow

Thu Nov 7, 2013 5:14 PM

Post by Mark Medina on September 29, 2013

ive been having trouble determine which unit of the universal gas constant to use during a problem. During some i know im suppose to use 0.0821 and during others 8.314. how do i determine which unit of measure to use?