Heat Engines & PV Diagrams - Online Tutor, Practice Problems & Exam Prep

In a cyclic heat engine process, work is calculated using the area enclosed in the PV diagram. For isobaric, isovolumetric, and isothermal processes, specific heat equations are applied: q = ncpΔT for isobaric, q = ncvΔT for isovolumetric, and q=w for isothermal. The total work done is W = q_h-q_c, emphasizing the importance of understanding energy flow in thermodynamics.

Calculating Work in Heat Engines Using PV Diagrams

Everyone, 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 will give you a PV diagram for a heat engine and then ask you to calculate something, like the total work done by the engines. We're going to put those ideas together in this video, and I'm going to show you how to calculate the work in heat engines by using the PV diagram. And what we're going to see is we're going to 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 going to 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 2 moles of a gas, and it's basically undergoing a cyclic process. What we're 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 isobaric process. Then from b to c, we're adding heat at constant volume. So this is isovolumetric. And then finally, the gas is going to expand isothermally. 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 3 processes we want to calculate q for. So we have q from a to b, and then we have q from b to c, and then we have q from c back to a again. Alright? So we spent a lot of time developing our equations in this table here for special processes. That's the that's basically what we're going to do here. We're just going to 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 isobaric process, constant pressure. So we look th...

We have 2 moles. CP just comes from this table here. It's a monoatomic gas, so we're going to use \( \frac{5}{2} R = \frac{5}{2} \times 8.314 \). Now what's the change in temperature? We're going between 2 isotherms, 200 back to 100. So this is basically just going to be \( \Delta T = -100 \). So you go ahead and work this out. What you're going to get is -4157 joules. Now for the second one, we're going to use a similar process. \( Q_{bc} \) is equal to, this is an isovolumetric process. So we're going to use this equation over here. Very much the same process. We're just going to use a slightly different molar specific heat. This is going to be 2. Now we're not going to use \( \frac{5}{2} \). We're going to use \( \frac{3}{2} \) for monoatomic. Right? So this is going to be \( \frac{3}{2} \times 8.314 \). And now what's the delta t? Well, here we're going from 100 back to 200 again. So this is going to be \( \Delta T = 100 \). So we're going to use 100 here, and what you're going to get is 2494 joules.

Finally, this last process here, \( q \) from c to a, is an isothermal one. So we look through our equations and \( q = w \). 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 23100 joules of work. So this is \( q = w = 23100 \) joules, and there's no calculation necessary. Alright? So that's all there is to it. Right? That's just all the heat transfers 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 going to be the area. So how do we do that? Well, unfortunately, there's a couple of problems here because, 1, we have this sort of curved isothermal 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 going to 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 \( q_h - q_c \). Right? In a heat engine, 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 in minus the q out. So we just have to find those. So how do we do that? Which numbers are going to be are they going to be? Or is it going to be this 1? Is it going to be this 1 or this 1? The answer is we're going to actually have to use all of them. So remember that we're going to use \( q_h - q_c \), 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

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 key 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 \( q \)'s over the cycle. So this is just going to be -4157. Now just be very careful here because whenever you 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. Alright? So that just means that the work done by the engine is just, 4794 minus 4157 because we're using the absolute value here. Alright? 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 going to get the wrong answer. Right? So just be very careful. Make sure you're subtracting them, and you're going to get 637 joules. Alright? So that's it for this one, guys. Let me know if you have any questions.

Example 1

Guys, let's take a look at this example problem. We have a heat engine that's shown as a cycle in a PV diagram. Remember heat engines are always going to be cycles. And what we want to do with part a is calculate the work that's done over a complete cycle in this process. Alright? So let's get started. With part a, we want to calculate w. Remember, w pops up in a bunch of different equations for heat engines, but probably the most fundamental one is going to be this Q_h−Q_c. The work that's done is the Q_h, it's the heat that flows in minus the heat that comes out. That's just what comes from our energy flow diagrams. Right? So if you have a hot reservoir connected to an engine, connected to a cold reservoir here, this is your hot and cold. What happens is you have heat flowing in, that's Q_h. You have heat flowing out, that's Q_c, and then the work that's done is always going to be the difference between those two numbers. Right? So this work done is always Q_H−Q_C. Alright? So that's where that equation comes from, so let's see if we can try to use it. So what happens here is that we don't actually know from the energy flow diagram. We don't know actually what Q_h and Q_c are. All we have in this problem are just the Q_s for some of the steps that are going on in this process here. So we're probably not going to use this equation over here. The other thing is that Q_h, the heat transferred from the hot reservoir, is actually what we're going to calculate in part b. Right? So this is what we're going to calculate in part b. So we're going to have to use a different sort of method to calculate the work. And remember that for we actually do know how to do that from our discussion on cyclic processes. Remember that the work that's done over a cycle is always equal to the area that's enclosed within the loop. So the area enclosed gives you the work. So in other words, the work that's done is really just the area that is enclosed within this rectangle. And the good news is because it's a rectangle, there's a pretty easy way to calculate the area. So the work that's done here is really just going to be base times height. So in other words, this is the base over here and this is going to be the height, and I have all the numbers, you know, the pressures and volumes and things like that, so I actually can calculate this work this way. So this work that's done is going to be base times height. So in other words, we're going to use, the base, which is the difference between 0.45 and 0.60. So this base here equals 0.15. So this is going to be 0.15. And then this over here is going to be a height, which is equal to 300. So this h equals 300, and there's no need to convert or anything because all this stuff is in SI units. We have pascals and we'll have kilopascals or anything like that, so you're going to just multiply this by 300. If you work this out, what you're going to get is you're going to get 45 joules. So essentially, what happens is the area inside of this is going to be 45 joules. That's the work that's done. Alright? So that's the first part. Now let's move on to the second part here. In the second part, we want to calculate the total heat transfer from the hot reservoir. So in other words, what happens here is we've actually calculated in our energy flow diagram that this w here equals 45. So then if we have if we say, if this is 45 joules, then basically, what is Q_h? That's what we're trying to find here. Alright? So can we use this equation again now that we actually have what Q_H is? Well, let's see. If we have w=Q_h−Q_c, then what happens is we can rearrange this equation here to solve for Q_h. So what happens is I'm just going to move the Q_c over to the other side. So in other words, the work that's done and I'm going to drop all the absolute values because they can get kind of annoying here. The work plus the heat that flows to the cold reservoir is equal to the heat that flows to the hot reservoir. Alright? So we have what w is. What is Q_c? So in other words, what is this sort of heat? What is the total amount of heat that flows out of, this this sort of cycle here? And to do that, we're going to actually take a look at these two arrows over here. If you notice, what happens in this sort of cycle here is that there's 2 places where heat is flowing out of the, of the heat engine here. And that's indicated by these negative signs. This Q=−90 and Q=−25. What happens at a heat engine in general is that anytime you go up or_to...

 A heat engine follows the cycle shown below. What is the thermal efficiency (%) of this engine?

• A. 58.8%
• B. 99.9%
• C. 14.3%
• D. 12.5%

Here’s what students ask on this topic:

What is a PV diagram and how is it used in thermodynamics?

A PV diagram, or pressure-volume diagram, is a graphical representation of the changes in pressure (P) and volume (V) in a thermodynamic system. It is used to visualize the processes that a gas undergoes, such as isobaric (constant pressure), isovolumetric (constant volume), and isothermal (constant temperature) processes. In thermodynamics, PV diagrams are crucial for understanding the work done by or on the system, as the area enclosed by a cycle on the PV diagram represents the net work done during that cycle.

How do you calculate work done by a heat engine using a PV diagram?

To calculate the work done by a heat engine using a PV diagram, you need to find the area enclosed by the cycle on the diagram. For a cyclic process, the work done (W) is the area inside the loop. If the process involves simple geometric shapes, you can use basic geometry to find the area. For more complex shapes, you may need to use calculus. The work done is positive if the cycle runs clockwise and negative if it runs counterclockwise.

What are the key equations for calculating heat transfer in isobaric, isovolumetric, and isothermal processes?

For an isobaric process (constant pressure), the heat transfer (q) is given by: $q=n{c}_p∆T$. For an isovolumetric process (constant volume), the heat transfer is: $q=n{c}_v∆T$. For an isothermal process (constant temperature), the heat transfer is equal to the work done: $q=w$.

What is the relationship between heat added, heat removed, and work done in a heat engine?

In a heat engine, the relationship between the heat added (q_h), heat removed (q_c), and work done (W) is given by the equation: $W=q$_h-q_c. Here, q_h is the heat absorbed from the hot reservoir, and q_c is the heat expelled to the cold reservoir. The work done by the engine is the difference between the heat added to the system and the heat removed from it.

Why do heat engines operate in cyclic processes?

Heat engines operate in cyclic processes to continuously convert heat energy into work. In a cyclic process, the system returns to its initial state at the end of each cycle, allowing the engine to repeat the process indefinitely. This cyclic nature ensures that the engine can continuously absorb heat from a hot reservoir, perform work, and expel waste heat to a cold reservoir, maintaining a steady operation over time.