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23. The Second Law of Thermodynamics

1

concept

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Hey everyone. So in earlier videos we saw how to solve PV diagram problems. And in more recent videos, we've looked at heat engine problems. Well, some problems like the one that we're going to work out below. We'll give you a PV diagram for a heat engine and then ask you to calculate something like the total work done by the engine. So we're gonna put those ideas together in this video and I'm gonna show you how to calculate the work in heat engines by using the PV diagram. And what we're gonna see is we're gonna see the relationship between the PV diagram steps and the quantities that we use in our energy flow diagrams. So let's go ahead and get started. We're actually just gonna jump straight into our problem because there's a lot of stuff that we already know how to do. So let's check it out in this problem. We have two moles of a gas and it's basically undergoing a cyclic process. We were told here, is that heat is removed from this gas at constant pressure from A to B. Remember that just means that it's an ice a barrick process. Then from B to C, we're adding heat at constant volume. So this is ice, a volumetric and then finally, the gas is going to expand ice a thermally. So that just means it's going to sort of ride along this curve right here, go back to a and then the whole thing is going to repeat over again. So that's actually the first point I want to make here, remember that heat engines are cyclic processes. They start and end at the same point at the same initial state. But what you need to know here is that on PV diagrams, they will actually always run in clockwise loops. Notice how we've gone through a cycle like this, but the overall pattern of this loop is in the clockwise direction. Alright, so that's always going to be true for heat engines. Alright, so let's take a look at this problem here, because the first thing we want to calculate in part A is the heat transfer of each process. So we have three processes. We want to calculate Q four. So we have Q from A to B, and then we have que from B to C and then we have Q from C. Back to A again. Alright, so we spend a lot of time developing our equations in this table here for special processes. That's the that's basically we're going to do here. We're just gonna look at each process and then figure out which equation of Q that we use from this row here. So let's get started. So from A to B. We know this is an ice a barrick process, constant pressure. So we looked through the table and this is the equation we're gonna use we're just gonna plug and chug, we're gonna fly through this real quick, this is gonna be N C, p times delta T. We have two moles. Cp just comes from this table here, it's a mono atomic gas. So we're gonna use five halves are 5/2 times 8.314. And now, what's the change in temperature? We're going between two ice. Affirms 200 back to 100. So this is basically just going to be delta T. Equals negative 100. So you go ahead and work this out. What you're gonna get is negative 41, jewels. Now, for the second one, we're gonna use a very similar process. Q. B. C. Is equal to uh this is an ice a volumetric process. So we're gonna use this equation over here very much the same process. We're just gonna use a slightly different moller specific heat. This is gonna be too. Now we're not gonna use five halves, we're gonna use three halves for mono atomic. Right? So there's gonna be three halves times 8.314. And now, what's the delta T. Well here we're going from 100 back to 200 again, so this is going to be delta T equals 100. So you can use 100 here. And what you're gonna get is 20 for 90 for jules. Alright, now, finally these last process here, cue from C to a is a nicer thermal one. So we look through equations and Q equals w whatever the work done in this process is, that's also the heat transfer. Now do we use this equation then. Well actually we don't need to because in this problem we're told that this process here or this gas does 2300 joules of work. So this is Q. Which equals W. Which equals 2300 jewels. And there's no calculation necessary. Alright, so that's all there is to it. Right. That's just all the heat transfer is using a bunch of equations we've seen before. So let's now go on to part B. We want to calculate the work done by the heat engine. So that's going to be the work done by the engine. Now, how do we do this? Well, you may remember from our video on cyclic processes that whenever you have a cycle, the work that's done is going to be the area that is enclosed within the loop. So, in other words, the work done is gonna be the area. So how do we do that? Well, unfortunately, there's a couple of problems here because one we have this sort of curved ice a thermal process. So we can't use something like a triangle or anything like that. And we also don't have any of the values for pressure or volume or anything that we need to calculate the area. So, while the work done by the engine is actually the area that's in this shape, we just can't calculate it. So we're gonna have to use a different equation now in more recent videos, remember we've talked a lot about heat engines and we've been using these energy flow diagrams. And remember for heat engines, the work done is just equal to Q H minus Q. C. Right in the heat engines, some heat comes in from the hot reservoir. The engine produces some usable energy work and then spits out the waste heat to the cold reservoir and the work done is just the Q N minus the queue out. We just have to find those. So how do we do that? Which numbers are going to be? Are they gonna be? It's gonna be this one? Is it gonna be this one or this one? The answer is we're gonna actually have to use all of them. So remember that we're gonna use Q H minus Q. C. And Q. H. Is really just the heat that's added into the system. Remember we have, we have heat added into the system from the hot reservoir but in this problem here, we noticed and that in part a we actually calculated to cues that were positive numbers 24 94 2300 in both of these processes. What happens is that Q is flowing into the system right as the heat is flowing into the system here. So we have multiple heats that are going into the system. What's happening here is that qh is actually just the sum of all of the positive cues over the cycle, that's all there is to it. So this process adds some heat and this process also adds some heat. Both of them, make up both of them combined to make up the heat flowing in from the hot reservoir. So this is gonna be 24 94 plus 2300. And this is just gonna be 47 90 for jules. Alright, that's the heat that's flowing in now. We do the same exact thing for the Q. C. In this problem, we calculated one value of Q. That was negative in this. In this process. What happens is heat is flowing out of the system and it's flowing to the cold reservoir. So Q. C. Is the heat removed out of the system To the cold reservoir and it's going to be the sum of all negative cues over the cycle. So this is just gonna be negative 41, Now, just be very careful here because whenever we use these energy flow diagrams, we always want these to be positive numbers. So you may see a bunch of absolute value signs. So, basically what's happening is that this negative sign? When you calculate it just means the heat is being removed. But when we use it in our energy flow diagrams, we just use it as a positive number. So you may just see these written with a bunch of absolute value signs. All right. So that just means that the work done by the engine is just 94 minus 41 57 because we're using the absolute value here. All right. Just be really careful with the signs here. You don't want to subtract this because this already has a negative sign, and if you mess this up, you're gonna get the wrong answer, right? So just be very careful, make sure you're subtracting them and you're gonna get 637 jewels. Alright, So that's it for this one. Guys, let me know if you have any questions.

2

example

10m

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Hey, guys, let's do an example. Ah, Proposed four step engine is produced when, in moles of a mono atomic ideal gas undergo the process shown in the following PV diagram. What would the efficiency of the engine be? Okay, The efficiency is given by the work output by the engine divided by the heat. Input it into the engine. Remember, there are two heats in a cyclic process and an engine. There's going to be amount of heat input it, which is gonna be positive heat and amount of heat output, which is gonna be negative heat. Okay, the work is easy to calculate. The work is just the area contained by the cycle with the appropriate sign. Now first, could this even be an engine? Yes, it can. It's a clockwise cycle, right? So it's absolutely an engine. Okay. The area is going to be the base times height, because it's just a rectangle. This is gonna be to you. Not this is gonna be one half peanut. So this is to V not one half peanut, right? That too, and that one have canceled. This is just peanut Vienna, and this is absolutely gonna be positive. It's gonna be positive. Right? Energy released. Bye, Engine. Okay, that's so that's just the work done by the engine. Now, in order to find how much heat is input, we have to analyze the steps individually. Okay, I'm gonna call this step one this step to this Step three and this Step four. Okay. Now, something important to know about P v diagrams. And you can show this for yourself if you want Is that when the process is going up into the right, the heat is always input into the system. So in process one and process to heat is gonna be added into the system. Okay. So cue was going to go in when the process is down or to the left. Heat is always leaving the system. All we need to know is how much he enters the system, right? This how much he enters the system. So we only have to look at steps one and two. You can use the same method, the same process that I'm using here to analyze steps three and four and show that the heats always gonna be leaving in those steps. Okay, for step one. First of all, the work is always going to be zero, because this is an isil coric process. This means that the first law of thermodynamics, which says the change in internal energy is the heat transferred. Plus, the work done just means sorry. Wrong color. It just means that the heat transferred is the change in internal energy four. Step one. Okay. Now, the internal energy off any ideal gas is f over to in our t an equation that we used a bunch of times because this is a model atomic ideal gas. The number of degrees of freedom is simply 31 for each translational directional. Now, the change in internal energy is what we're interested in. And this is gonna be free has in our delta t we're talking about in moles simply changing its state. There's no heat coming into the city. Sorry. There's no gas coming into the system. No gas leaving the system. So the only thing that changes when the internal energy changes is the temperature. So the temperature is the only thing that gets this delta. The question is, what is the change in temperature? Okay, Well, PV was nrt The ideal gas equation is gonna tell us how the temperature changes as the gas changes state. This is going to tell us because the volume is a constant. This is an I support process that V Delta P is in our delta T. The change in the left side is due entirely to a change in pressure. The change on the right side is due entirely to a change in temperature. So Delta T is v Delta p over in our now plugging that in two Delta U. Okay, okay, right. So I could say Delta t one is equal to this. This equals three halves in our times, V one Delta P one over in our okay. Notice that the n r in the numerator and then are in the denominator. Cancel. So this is three halves V one Delta P one. Okay, which is three halves. What's the volume for step one? It's just V not. What's the change in pressure and step one from peanut. It's from one half peanut too peanut. Okay, so this is just one half peanut. So this becomes 3/4 peanut Vienna. Okay, that's the change. Internal energy for step one and by the first law of thermodynamics. Therefore, the heat exchange in step one, which is Delta U one, is going to be positive. 3/4 Peanut Vienna. Okay, it's very important that the heat is positive because we're only looking for heat input into the gas. We're not looking for heat. Output it by the gas. Okay for step two, the pressure doesn't change for Step two Delta P zero. Okay, but Delta U is still equal to Q plus W. However, we do know that at constant pressure, W is equal to negative. P Delta V Okay, so this is for Step two. What's the pressure at Step two? It's just peanut, right? What's the change in volume? It's three v not minus two v. Not sorry, minus V not which is going to be positive to be not so this is negative. Two. Peanut Vienna, right? This is positive to Vienna. Okay, so that's how much work is done during step to now. What's the change in internal energy during step to? Well, it's still three halves in our delta t during step, too, right? What's Delta T during step to? Well, PV equals in our tea pressure does not change on Lee. The volume changes so we have peed Delta V equals in our delta t So p two Delta V two equals in our delta t two right in our delta T too, is this whole right portion. So Delta U two is three halves p two Delta V two, which is three halves peanut, right, The pressure for step two times once again the same change in pressure that we found here. So this is just three peanut V. Not so by the first law of thermodynamics. Hugh during Step two is just Delta U two minus w, which is three peanut V not minus negative to peanut V not which is five peanut Vienna. Okay, so for the cycle Sorry, The heat the total heat input during the cycle is just the heat input during steps one and steps to because steps three steps Step four release heat. Okay, so the amount of heat inputs just during step one Step two Step one had 3/4 peanut Vienna and Step two had five Peanut Vienna. If we find the least common denominator, this is 3/4 if you're not plus this common denominators. Four. This is 20/4 Peanut Vienna. So that's 23/4 Peanut Vienna. Okay, and what's the work done by the cycle we calculated it to be Peanut Vienna. This means that the efficiency, which is just the work over the amount of heat input, is peanut peanut over 23/4 Peanut view, not those peanut peanut's canceled. And finally, the efficiency is 4/23. Now 4 23 is an exact answer, and you can leave it like that if you want. Or you can approximate a number 0.174 which is 17.4%. So this engine actually has an incredibly low efficiency Onley 17.4%. Other engines can have much, much, much higher, um, efficiencies, right? I showed a case in an earlier problem where car, not engine, had an 80% efficiency when it was placed between a reservoir of 1000 Kelvin and a reservoir of 100 Calvin. This 17.5% efficiency sort of leaves a lot to be desired, but this is an application of the first law of thermodynamics on heat engines and how to find the efficiency using a PV diagram which could be very useful. Alright, guys, that wraps up this problem. Thanks for watching

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