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

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

## Discussion

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## Table of Contents

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### Energy, Heat, and Work

- Temperature and Heat are NOT the same thing.
- Thermodynamic quantities always have two parts: a numerical value and a sign. The sign indicates the direction of flow. We always take the System’s point of view.
- Heat is the thing that “flows” due to a temperature difference: heat flows from a hotter object to colder object.
- There are two ways to transfer energy: Heat and Work.
- The change in energy of a system is the sum of the Heat transferred to/from the system + the Work done to/by the system. In equation form,

ΔE = q + w

### Energy, Heat, and Work

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 0:00
- Thermochemistry 0:25
- Temperature and Heat
- Work
- System, Surroundings, Exothermic Process, and Endothermic Process
- Work & Gas: Expansion and Compression
- Example 1
- Example 2
- Example 3

### AP Chemistry Online Prep Course

### Transcription: Energy, Heat, and Work

*Hello, and welcome back to Educator.com.*0000

*Welcome back to AP Chemistry.*0002

*We're going to start, today, on our unit discussing thermochemistry, and there are going to be about three or four lessons discussing thermochemistry--chemistry related to heat.*0004

*Today, we're going to talk about energy, heat, and work--some of the basic, fundamental concepts--and eventually, we will get more into the chemistry of it.*0015

*Let's just jump in and get started.*0023

*The first thing that we want to definitely make sure you know is that temperature and heat are not the same thing.*0026

*So, temperature and heat are not the same thing; this is probably the single biggest misconception.*0032

*OK, so temperature is a measure of the average kinetic energy of the particles in the sample that you are taking the temperature of.*0048

*Heat is the thing that actually flows when there is a temperature differential.*0078

*In other words, heat is the thing that flows from a hot object to something that is colder.*0085

*Or, it doesn't necessarily need to be an object; any time there is...let's say you have some hot thing, or something that is at a higher temperature; it doesn't have to be hot; there just has to be a difference in temperature.*0089

*So, if something is 30 degrees Celsius, and let's say the air surrounding it is 10 degrees Celsius, heat is that energy, if you will, that flows from the higher-temperature object into the lower-temperature object.*0099

*Heat is the energy that actually flows; what temperature does is actually measure the random motions of the molecules.*0113

*There is a difference between the two; they are related, but they are not the same thing.*0121

*Heat is the thing that flows from hotter to colder--we'll say objects (objects could also just be the air)--due to the temperature differential.*0126

*Differential, as you know, just means the difference.*0162

*Heat is a form of energy; it is a very disordered form of energy.*0166

*OK, so heat--profoundly important; thermochemistry is going to be all about the heat.*0179

*Work is another form of energy.*0187

*Clearly, energy can come in different ways; you can have light energy; it could be heat energy; work is another form of energy; all of these things are energy.*0199

*They are represented by the same unit, and the idea, essentially, of major areas of science, is: How do we convert one form of energy to another form of energy?*0208

*That is really what is going on.*0217

*Energy...work is another form of energy, and we define it as Work=Fd, which is force times distance.*0221

*So, the best way to think about that is: if I apply a certain force to an object, and if I move it a distance d, well, the work is going to be the force that I apply, times the distance that I actually move it.*0232

*Now, notice: work, the way we discuss it in science, is not the same sort of way of everyday usage of the word "work."*0243

*I can stand and I can push against a wall, and I'll definitely start sweating, and I'll definitely start getting tired, eventually.*0252

*But, the wall isn't moving; I'm applying a force, but the wall isn't moving.*0258

*Force is a positive value, but distance is 0; the wall isn't moving.*0261

*Scientifically, there is no work actually being done; yes, I am doing work, in the sense that I am actually expending energy, but no actual work is being done by the force.*0267

*So, there is a difference: when we say "work," in chemistry, in physics, in engineering, we're talking about a force that causes something to move.*0277

*OK, so let's look at some units.*0287

*Force equals mass times acceleration; that is Newton's second law.*0290

*So, if you want to give a certain body of a certain mass a given acceleration, you have to apply a certain amount of force to it, and the relation between these three variables is F=ma.*0297

*Well, mass is in kilograms; that is the standard unit of mass.*0306

*Acceleration is in meters per square second.*0313

*Well, this unit right here--the kilogram-meter per square second--this is called a newton.*0317

*A newton is a unit of force, symbolized with an N.*0323

*Now, we just said that work equals a force times a distance; well, force, we just said, is a newton; and distance is in meters; so now, we have this unit of a newton-meter.*0327

*This unit, which is a (well, let me go ahead and actually write out everything so you see it)...*0346

*We said that a newton is a kilogram-meter per square second, and then we multiply it by a meter, which is there; so this is the force; this is the distance; and we end up with kilogram-meter squared per second squared, and this is called a Joule.*0355

*This is our standard unit of energy, the Joule.*0377

*If you want, you can think of it as kilogram-meter squared per second squared, or you can think of it as kilogram-meter per square second, times meter--a newton-meter.*0381

*So, a newton-meter is a Joule, or if you want to break down the newton and sort of put some things together, you can get kilogram-meter squared per second squared.*0390

*It depends on the problem you are working on; sometimes you will just deal with newtons; sometimes, you will need to break it up.*0399

*As you go on in science, you will realize that you will need to sort of break up more of these things, in terms of units, to deal with certain constants.*0404

*Constants--depending on the problem, you might have to use different constants, which are the same constant, but in different units, they have different values.*0412

*OK, so the biggest piece of advice I can give is: watch your units.*0421

*If we're talking about a mass, make sure that kilogram matches with gram; or, if something is in two different units, make sure that kilograms, meters, seconds, Joules, newtons...that everything matches; otherwise, your values will be wrong.*0426

*OK, so let's talk about the two ways to actually transfer energy.*0442

*The two ways to transfer energy--to transfer energy from one object to another, from one thing to another, from one environment to another, from the outside to an inside, to transfer energy: we have two ways of doing it.*0448

*We can transfer it as heat or work, or both.*0465

*That is it; so, in some sense, this is really simple.*0473

*We're going to have some sort of a system, and we want to either give energy to it or take energy away from it; well, there are only a couple of ways we can do that.*0477

*We can either put heat into it or take heat out of it, or we can do work on it, or it can do work on us.*0485

*So, let's be a little more specific here; we'll draw a picture, and we're going to start talking about systems and surroundings.*0491

*OK, so I have a system; a system is defined as the thing that we are interested in.*0500

*I know that that is kind of a vague definition--"thing that we are interested in"--it will make more sense as we do some problems and as we talk about it more.*0513

*Well, for chemistry, most systems are going to be the reaction: the reaction, and maybe the vessel that is containing it...a solution of hydrochloric acid plus sodium hydroxide--that is your system...the beaker plus the water; that is the system.*0525

*The surroundings--that is everything else.*0540

*That is everything else; so, very, very intuitive--nothing strange is going on here.*0545

*We said that we can transfer energy in two ways: we can do it via heat, or we can do it via work.*0550

*This is our system; this is (I'm sorry, I should put the system in here)...I'll call this sys, and out here is the surroundings.*0557

*So now, this is the boundary, and it's always going to be something; there is going to be some sort of a boundary that separates your system from your surroundings.*0575

*Now, we're going to transfer heat in and out; we're going to transfer work in and out.*0583

*We have to pick a point of view.*0588

*In chemistry, we always look at the system's point of view; it's not always the case; there might be other circumstances, maybe in engineering, where you need to look at the surroundings' point of view; but in chemistry, we always look at it from the system's point of view.*0591

*From the system's point of view, when heat is added to a system (OK, so we say heat in), that heat is positive.*0602

*When heat is given off of a system, that heat is negative; in other words, the system is losing heat--it's going from a certain value of heat, it's losing heat, and its final heat is actually less than that.*0617

*So, energy and heat is flowing out; so heat is negative.*0631

*That is the sign convention; if heat is put into the system, that means more energy; so it has a certain amount of energy; we put more heat in; it goes up; that is positive.*0635

*We are looking at it from the system's point of view; notice, it would be the reverse if it were the surroundings' point of view.*0643

*The other way to transfer energy is work.*0648

*Now, the system could do work on the surroundings; the surroundings could do work on the system.*0651

*We will be a little more specific about what that means, but the same sign convention applies.*0657

*If I do work on the system--work on the system--that is positive work.*0662

*That means I have put energy into the system via the work that I have done to it.*0676

*If the system does work on the surroundings, that is going to be negative work.*0680

*This is called work done by the system.*0687

*So, in this case, these prepositions are very, very important; so, when you read your questions, makes sure you read them carefully, because it will say "work done on the system" or "work done by the system."*0693

*It won't tell you whether it is positive or negative; you have to decide.*0701

*So, work done by the system--that means the system used its energy to do something to the surroundings, and in the process of doing it, it transferred its energy to the surroundings.*0706

*Now, energy is depleted.*0715

*Work is a little more abstract than heat; with heat, you have a pretty good intuitive sense--you know when something gets colder or gets hotter.*0718

*Work is a little different, but again, we will do some problems, and we'll get used to it.*0725

*So, two ways to transfer energy: heat and work; work into the system is positive; heat into the system is positive; heat out of the system is negative; work out of the system is negative.*0730

*Again, from the system's point of view--we are always looking at it from the system's point of view.*0744

*Whenever a process actually gives off heat (the system gives off heat), we call it exothermic.*0749

*It makes sense, right?--exo, outside.*0755

*And then, heat in is called endothermic--an endothermic process; that means the system is taking in heat.*0758

*These are very, very important words, and we will use them in the context of our problems.*0767

*OK, so let's look at just a couple of quick examples.*0773

*If I take methane gas, and if I combust it (burn it in oxygen), I produce two moles of CO _{2}, 1 mole of CH_{4}; I need 2 moles of oxygen; I produce 2 moles of CO_{2} and I produce 2 moles of water, and I produce heat.*0776

*As you know, any time you burn something, it gives off heat; you feel the heat radiation; that is giving off energy as heat.*0797

*This is an exothermic process.*0804

*Burning, combustion, is an exothermic process.*0810

*If I take nitrogen gas, and if I react it with oxygen gas, in order to produce a couple of moles of nitrogen monoxide gas, as it turns out, I have to add heat to the system.*0813

*So, what happens: if I have a vessel, and if I actually ran this reaction, what would happen is: the vessel itself would get cold--I would feel the vessel get cold--and the reason is, now the reaction--in order for it to work--needs energy put into it.*0826

*So, it takes heat away from the glass (let's say it's in a glass container); it sucks heat out of the glass container, and because the heat from the surroundings is not actually going into the glass fast enough, when I touch the glass, I feel cold.*0843

*So, it isn't that the glass got cold; it is that the heat was pulled out of it, bringing it to a lower temperature.*0856

*That is what is going on.*0863

*This is an endothermic process.*0864

*The system evolves heat and gives it up.*0870

*This system pulls in heat--needs the energy in order to perform its functions (in this case, the systems are the reactions).*0873

*OK, now let's go ahead and put some mathematics to this.*0883

*We said that there are two ways to transfer energy: heat and work; well, the change in energy of a system is going to be the heat that goes in or out of the system, plus the work that goes in or out of the system.*0890

*Again, heat and work are just two different types of energy.*0909

*So, if a system starts in one state of energy, and it goes to another state of energy, a couple of things have happened.*0912

*It has either lost or gained heat, lost or gained work, or some combination of both.*0920

*This is our fundamental equation; this is actually a statement of the first law of thermodynamics--that energy can neither be created nor destroyed; that any energy change is going to take one form or the other: heat, work, or both.*0926

*But, energy is conserved; that is what this equation is saying.*0942

*OK, so these heat and work thermodynamic quantities--they have two parts.*0947

*They have a magnitude, and they have a sign.*0957

*Well, again, we dealt with the sign issue here; with thermodynamics, oftentimes we have to stop and think about the physics of what is going on, in order to decide what the sign is going to be.*0963

*Often, we know what the magnitude is, but we have to make sure to keep the sign straight; otherwise, the mathematics will not work out.*0974

*All right, now let's fiddle with this equation; let's talk a little bit more specifically about this thing called work.*0982

*OK, so a common type of work in chemistry (in fact, the most common in general) is work done by a gas.*0991

*So, it's either going to be...well, work done by a gas: notice, here is that "by" again; that is expansion; and when I draw the picture in a minute, it will show you what I mean.*1015

*Or, work done to a gas, which we call compression.*1030

*Let's say I have this container, and it has a moveable piston that can move up and down; this is a chamber that has the volume.*1044

*So, I can push this piston down and squeeze this volume--make it smaller--or I can pull the piston up and make the volume bigger.*1056

*In this particular case, if there is a gas in here, and it expands, that means the gas is actually pushing against the external pressure of the atmosphere.*1064

*It is doing a certain amount of work; if I actually push the piston down, and the volume gets smaller, this pressure is higher than this pressure; I am actually pushing this down--I am doing work on the system here.*1073

*Let's take a look at what this looks like mathematically.*1088

*We just said that the change in energy is equal to q + w, the heat that goes in and out of a system plus the work that is done on or by the system.*1091

*OK, q: well, we said that work is equal to force times distance; great--simple substitution.*1101

*We'll leave q alone, and now, here is what I'm going to do: I know that there is a relationship between force and pressure and area; as it turns out, if I take the force of something, and if I divide it by the area (remember, back when we talked about gases, we said that pressure is equal to the force per unit area, so...) I have this F here; if I divide by A, in order to retain this, I have to multiply by A, so I basically have multiplied this F/d by A/A, right?*1111

*I did this in order to manipulate this F a little bit.*1144

*Now, I have F/A, and I have distance times area.*1147

*Well, this thing right here--I see it in the profile, but really, it is just a cylinder.*1153

*Let's see what we get here: we have q; well, force over area is pressure, and distance times area is...well, distance is...let's say that this force here is a pressure that is being applied to this piston.*1160

*The area is, of course, the area of this circular part; so there is a certain pressure being applied; well, the area times the distance...the distance is the distance that this piston actually goes down or up, right?*1186

*So, this is the distance; well, area times the distance, as far as the circular cylinder is concerned, is just volume.*1204

*You end up with: Change in energy equals q; P is constant, so the only thing that is changing here is, from this to this, whether we either press down or pull up.*1215

*Pressure...PΔV; this is pressure times the change in volume.*1233

*This is what we mean by pressure, volume, and work.*1238

*So, when we push down on a gas, we are actually doing work on the gas; we are doing work to the system.*1242

*When the gas expands and pushes the piston up, then the gas is actually pushing against an external pressure, and it is actually doing work on the surroundings.*1251

*This final equation...this is often how we will do it: we're going to be dealing with some gas in a container, and this is going to be the expression that we use: The change in energy of the system is going to be the heat that flows in or out of the system, plus the change in volume, based on a certain amount of pressure.*1266

*This pressure is always going to be the external pressure.*1291

*It is an external pressure that pushes down, and squeezes it, and makes the volume smaller; or, it's going to be the pressure against which the gas pushes in order to make it bigger.*1294

*In either case, it is a pressure that is pushing against the boundary here.*1306

*So, it is always an external pressure--that is what this P is, not the pressure of the gas.*1311

*OK, so now let's talk about some signs.*1318

*All right, this is going to be important; so now, we're going to actually talk about this term right here, and we're going to relate...so we know that work and pressure times volume are equivalent in terms of magnitude whenever we are doing something like this.*1322

*So now, we want to talk about what sign convention we are going to use.*1339

*All right, so let's just write "Work = PΔV"; so, work equals pressure times the change in volume.*1345

*If I start at a certain volume, and if the gas expands, that means the gas inside is pushing against an external pressure; well, the final volume is bigger than the initial volume.*1351

*Final volume, minus initial volume, is going to be bigger than 0.*1369

*Well, in this particular case, it is bigger than 0; the pressure is bigger than 0; here, this work is going to be positive; this number is going to be positive.*1375

*Pressure times the change in volume, for an expansion, is positive; but, because it is expanding, it is the system that is doing work on the surroundings; therefore, the work is negative, because again, we said we are looking from the system's point of view.*1383

*If the system is pushing against this piston and expanding, it is doing work on the surroundings.*1398

*In other words, work is leaving the system.*1404

*Therefore, work is negative; so, by putting this negative sign, we also account for the other direction.*1406

*If the outside pressure is pushing down on the piston, and making the volume smaller, that means the final volume is smaller than the initial volume.*1413

*Therefore, ΔV, which by definition is final minus initial, is less than 0.*1425

*Well, if ΔV is negative, and P is positive; this negative sign--negative times negative--makes the work positive, which works with the fact that now the surroundings are doing work on the system.*1430

*The work is positive, because the system is now taking in energy in the form of work.*1444

*It is positive from the system's point of view.*1451

*The energy of the system is increasing.*1453

*So, this is the actual relationship: work is negative pressure times the change in volume (change in volume always being final volume minus initial volume--which is the definition of Δ--final minus initial).*1456

*OK, so with this little bit of a background, let's just jump into the problems, and I think we can start to make sense of some of this.*1473

*Let's do Example 1: So, we want to calculate the change in energy for an endothermic process in which 29.6 kilojoules of heat flows (and notice: we said endothermic), and where 13.7 kilojoules of energy is done on the system.*1482

*OK, so we want to calculate: we say that, for an endothermic process, 20.6 kilojoules of heat is flowing.*1540

*Endothermic means heat is coming into the system; that means heat is positive.*1545

*"Where 13.7 kilojoules of energy is done on the system": on the system means work is flowing into the system--that means work is also positive.*1551

*So, we write our equation: change in energy equals q + w; q is heat; w is work.*1559

*We said an endothermic process, so it's positive, so it's going to be positive 20.6 kilojoules; so notice, the problem will give you a magnitude; it won't necessarily tell you what direction it is going.*1569

*If it said exothermic, then I would have to use -20.6.*1582

*If this said "by the system," I would have to use -13.7.*1585

*But here, it's endothermic, so heat is positive, and then the work is also positive 13.7, because it is done on the system.*1589

*Kilojoules...this is a simple arithmetic problem.*1597

*So, we end up with 34.3 kilojoules of energy.*1600

*I put heat into it in the amount of 20.6 kilojoules; I did work on the system by 13.7 kilojoules; the total amount of energy that I imparted to the system is a positive 34.3 kilojoules of energy.*1608

*Oh, by the way, I should let you know: oftentimes, another unit of energy is something called the calorie; it's an older unit of energy, and it is defined as the amount of energy required to raise the temperature of 1 gram of water by 1 degree Celsius.*1626

*The calorie is a unit of heat; it's a unit of energy, because that is what heat is--it's just energy.*1640

*So, as a conversion factor, one calorie equals 4.184 Joules.*1646

*We are probably not going to run into calorie too much; we may or may not; we'll find out; but just so you know...*1653

*You have heard of the word "calorie"; you hear about it all the time; that is what it is: 1 calorie is that many Joules--it's a unit of heat; it's a unit of energy.*1658

*OK, so Example 2: What is the work associated with the expansion of a gas from 0.750 liters to 1.760 liters at constant external pressure of 13 atmospheres?*1667

*The gas is in a container; there is a pressure of 13 atmospheres on the outside; and again, pressure is always going to be external.*1724

*On the outside, 13 atmospheres is pressing on it.*1732

*But, the gas is actually expanding from the inside, and it goes from .750 liters to 1.760 liters.*1736

*In other words, it is pushing out against the surroundings.*1746

*It is doing work on the surroundings, which means that energy is flowing out of the system and into the surroundings, which means that it is actually going to be negative.*1750

*Well, work equals negative pressure, times change in volume, equals negative 13 atmospheres; change in volume--we have 1.760 liters, minus 0.750 liters, and when I do this, I end up with -13.13 (I don't know if you can read that) liter-atmosphere.*1761

*OK, so notice: atmosphere, liter; this is not exactly a unit of energy that you have seen, although it is a unit of energy; it just happens to be different units, and I will show you what this is.*1809

*The conversion factor here is this: 1 liter-atmosphere is equal to 101.3 Joules of energy.*1822

*I'll show you in a minute where this actually comes from.*1833

*But again, it just goes back to the point: units are very, very important; just because you end up with some unit that seems strange doesn't necessarily mean that it is strange.*1835

*So, all of the math that we have done is correct; work is equal to pressure times change in volume; pressure is in atmospheres; volume is in liters; this is actually a unit of energy--it's a unit of work.*1844

*It is actually equivalent (by a conversion factor) to a Joule.*1858

*So, don't let this throw you off.*1863

*Now, let's go ahead and do the conversion.*1865

*When we multiply this out, we end up with 1330.4 Joules, or (can't have these stray lines floating around--let's erase this) 1.33 kilojoules.*1867

*There we go.*1887

*OK, so let me see what I have here.*1890

*OK, so really quickly...you know what, let's skip the unit conversion part; if you can, you are welcome to look it up, but the liter-atmosphere--one liter-atmosphere is equal to 101.3 Joules, so that conversion will always be there for you.*1898

*Let's just keep going on with these examples.*1918

*All right, another problem here: Now, a balloon is inflated from 4.00x10 ^{6} liters to 4.50x10^{6} liters by the addition of 1.4x10^{8} Joules of heat energy.*1922

*So, I add some heat to the system, and I add 1.4x10 ^{8} Joules of heat.*1972

*Now, if the balloon expands against a constant pressure of 0.95 atmospheres, what is the ΔE for the process?*1981

*OK, so let's just draw a quick picture here: we have this balloon--obviously, we are talking about a really, really large balloon; we pump in heat--pumping in heat energy--so heat is going to be positive.*2019

*This balloon is going to expand, which means that it is going to push out against the .95 atmospheres; so now, the balloon is actually doing work on the surroundings.*2034

*Work is going to be negative.*2043

*Now, let's work it out.*2049

*ΔE is equal to q plus w; we have worked out our sign convention--heat is going in; work is going out, because it is expansion.*2052

*We also know that work is equal to -PΔV, so let's go ahead and calculate our work.*2062

*The work is equal to -0.95 atmospheres (that is the external pressure; and again, pressure, volume, work; PΔV; that P is the external pressure against which you are pushing, or that is pushing on you, or pushing on the system), times 4.50x10 ^{6}, minus 4.00x10^{6}.*2076

*We end up with (when we do this mathematics) -475,000 liter-atmospheres; and then, when we multiply by 101.3 (oops, that is not kilojoules--the conversion factor is 101.3 Joules per liter-atmosphere), we end up with a total of -4.8x10 ^{7} Joules of work.*2103

*Well, we just (oh, no, we definitely don't want that...OK) said that ΔE equals q + w; we calculated w--that is -4.8x10 ^{7} Joules--so now ΔE equals q plus that.*2141

*Well, q, positive--it is 1.4x10 ^{8} Joules (you know what, I am always confused about whether I should write it out or do the letters; for years it has been like this--OK, I'll just do the letter), minus 4.8x10^{7} J, so our change in energy is 9.2x10^{7} J.*2163

*Notice what has happened here: energy flowed in as heat.*2196

*Work ended up expanding the gas; the gas ended up pushing against the external atmosphere; so energy came in as heat; energy left as work.*2202

*The total energy change for this process was 9.2x10 ^{7} Joules; this is a positive value, so there is now more energy in the system that stayed as heat than went out as work.*2214

*That is all this says--the fundamental equation; this is the first law of thermodynamics.*2231

*The change in energy is equal to the heat that goes in and out of a system, plus the work that goes in and out of a system.*2235

*Energy itself can be transformed, but it cannot be created or destroyed; it always has to be accounted for.*2241

*OK, thank you for joining us here at Educator.com.*2247

*We'll see you next time for some more thermochemistry; goodbye.*2250

2 answers

Last reply by: Professor Hovasapian

Wed Jan 18, 2017 6:15 PM

Post by Magic Fu on December 20, 2016

Hi Professor H

I think you balanced your equation wrong at 13:23. You had 2 carbons on the product side while only one on the reactant side. You had six oxygens on the product while only 4 on the reactant side.

1 answer

Last reply by: Professor Hovasapian

Sun Oct 30, 2016 5:18 PM

Post by Warwick Shaw on October 30, 2016

Hi,

If delta E equals q + w or as you put it, q + p times delta v, wouldn't that mean that delta E is equal to q + negative w ( because work is negative p times delta v and just p times delta v would be negative work)?

3 answers

Last reply by: Jason Smith

Sun Sep 27, 2015 3:54 PM

Post by Jason Smith on September 26, 2015

Hi professor. If I understand correctly, "heat" doesn't necessarily mean "hot", right? It basically depends on the difference in temperature between two objects- is this correct? Thank you in advance.

1 answer

Last reply by: Professor Hovasapian

Sun Sep 27, 2015 1:20 AM

Post by Gaurav Kumar on September 26, 2015

Hi Professor Hovasapain,

I am confused to why N2 + O2 ----> 2NO is a endothermic process. Can you please explain?

Thank you

1 answer

Last reply by: Professor Hovasapian

Sat Aug 8, 2015 9:49 PM

Post by Jim Tang on August 7, 2015

Hey!

31:15 shouldn't that be negative?

Also, in example 3, the pressure that the balloon expands AGAINST is the pressure the surroundings is pushing ON the balloon correct?

Ty!

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Last reply by: Professor Hovasapian

Sat Sep 6, 2014 11:10 PM

Post by David Gonzalez on August 31, 2014

Hi profession, great lecture. You are such an amazing teacher!

Just to help me wrap my head around this, please tell me if the analogy that I have in my head is correct: if you place a drop of food coloring into a glass of water, it will "disperse" until it reaches equilibrium. Does this mean that the heat particles/molecules are trying to do the same thing - "disperse" to achieve equilibrium?

Thanks!

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Last reply by: Professor Hovasapian

Wed Jul 9, 2014 6:41 PM

Post by ibrahim shawi on July 8, 2014

hello professor Hovasapain, my question here is about work&gas you said the work done by the gas on the system (expansion) gives us w=-p(deltaV) and i get that because earlier in the video you explained work or energy leaving the system is negative and it is positive if it is entering the system, what im confused on is in my mcat prep book it says "this equation (w=p(deltaV)) *(constant pressure) shows the work done by the gas. the work done on the gas is just the negative of the work done by the gas, so you might see tis equation written with a negative sign as (w=-p(deltaV)). why do you think this may be? and also how can pressure be constant if it is inversely proportional to V. like if volume was to increase pressure would decrease. do the mean the more gas is pumped in to keep the pressure constant?

Thank you.

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Last reply by: Professor Hovasapian

Wed May 15, 2013 2:04 AM

Post by Nawaphan Jedjomnongkit on May 14, 2013

Thank you for the lecture but I tend to have problem when try to link things up. What I've learnt from here is that when we do work ON a system , the volume will decrease and energy of the system will gain, right? But from the relationship of volume and temp from ideal gas law is in the way that when we decrease volume it will decrease temperature. And when I relate temperature to KE of the system it will be decrease, so is that mean when the system gain energy, their KE will decrease and where the energy gone?

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Last reply by: Professor Hovasapian

Fri Apr 26, 2013 8:28 PM

Post by Shaurntae Thomas on April 26, 2013

Why was the answer 1330.069 J and not -1330.069 J? This question comes from the work equals pressure and volume equation.

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Last reply by: Professor Hovasapian

Sat Apr 20, 2013 6:10 PM

Post by Antie Chen on April 20, 2013

Wï¼Pï¼ˆdeltaï¼‰V

The equation is from W=Force*Distance=P(delta)V

The unit of P is atm and the Unit of V is m3, 1Nï¼Šm=1j why 1 atm*m3 is not equal to 1 joule but need to use 1 atm*m3=101.3j to convert?

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Last reply by: Professor Hovasapian

Fri Feb 1, 2013 3:17 AM

Post by Gayatri Arumugam on January 31, 2013

Hi Professor Hovasapain,

In Example two why is the answer 1330.04 J instead of -1330.07 J?

Thanks,

Gayatri Aruumugam

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Last reply by: Professor Hovasapian

Wed Jan 16, 2013 9:02 PM

Post by Stephen Piesley on January 16, 2013

If the equation format is

(delta) E = q + w.

The Last part of the equation Example III

is set out as :

(q) 1.4 x 10 power of 8 + (w) -4.8 x 10 power of 7

Had did this equate to the answer 9.2 x 10 power of 7 ,Have I missed something in the Arithmetic.?

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Last reply by: Professor Hovasapian

Wed Nov 28, 2012 2:02 PM

Post by Julie Mohamed on November 26, 2012

How would i find the temperature of hot water with a thermometer that only reads to 50 degrees Celsius?

Supplies : styrofoam cup

50 degree thermometer

tap water

Hot water

Electronic balance

0 answers

Post by Themiya Chandraratna on May 6, 2012

v. is for volume

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Last reply by: Professor Hovasapian

Sat Jul 14, 2012 6:31 PM

Post by Miguel Ibanez on March 24, 2012

can you explain the W=P(delta)V formula? does "v" mean the change in area of the balloon?