Video: Finding the Heat Exchanged by and Work Done on a Gas

An ideal gas expands isothermally along the path 𝐴𝐡, as shown in the accompanying diagram. The gas does 7.0 Γ— 10Β² J of work along this path. How much heat does the gas exchange along 𝐴𝐡? The gas then expands adiabatically along 𝐡𝐢 and does 400 J of work. When the gas returns to 𝐴 along 𝐢𝐴, it exhausts 100 J of heat to its surroundings. How much work is done on the gas along this path?

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Video Transcript

An ideal gas expands isothermally along the path 𝐴𝐡 as shown in the accompanying diagram. The gas does 7.0 times 10 to the power of two joules of work along this path. How much heat does the gas exchange along 𝐴𝐡? The gas then expands adiabatically along 𝐡𝐢 and does 400 joules of work. When the gas returns to 𝐴 along 𝐢𝐴, it exhausts 100 joules of heat to its surroundings. How much work is done on the gas along this path?

Okay, so we’ve got a multipart question here. So let’s start by looking at the very first part of the question. We’ve been told that we’ve got an ideal gas, and it expands isothermally along the path 𝐴𝐡. We’re also told that the gas does 7.0 times 10 to the power of two joules of work along this path. We’re firstly asked to find how much heat does the gas exchange along 𝐴𝐡. So for the very first part of this question, we’re just studying this part here, from 𝐴 to 𝐡.

Now, we know that we’ve got an ideal gas and we also know that the process 𝐴𝐡 is an isothermal expansion. In thermodynamics, an isothermal process is one for which the temperature remains constant. And of course we use the letter capital 𝑇 to denote temperature. Now, we also know the amount of work that the gas does along the path 𝐴𝐡. And we’re asked to find the amount of heat exchange by the gas along this path.

To do this, let’s first recall how the internal energy of an ideal gas is found. For an ideal monatomic gas, the internal energy π‘ˆ is given by three by two multiplied by 𝑛, the number of moles of gas, multiplied by 𝑅, the molar gas constant, multiplied by 𝑇, the temperature of the gas. Now of course we’ve not been told that the gas is monatomic, but this doesn’t matter. Basically if the gas is not monotonic, then the only thing that changes is this factor here, three by two.

That value is no longer three by two if the gas is not monatomic. But, as we’ll see soon, it’s not relevant to us. So, we know that we’ve got a gas undergoing a certain set of processes, 𝐴 to 𝐡, 𝐡 to 𝐢, and 𝐢 to 𝐴. When we study the process 𝐴 to 𝐡, we are told that it’s an isothermal process, so the temperature is constant. We also note that 𝑅, the molar gas constant, is a constant. The clue is in the name. Thirdly, there’s no mention of adding or taking away any particles from the gas or any change of mass for that matter.

So the number of moles of gas that we have, 𝑛, must also be constant. And of course, the value three by two or any other value for nonmonotonic gas is also a constant. Therefore, the whole of this right-hand side of the equation is a constant. Have I said constant enough times yet?! Probably not! How about one more? Constant! So here, the right-hand side is constant and so the left-hand side must be also constant.

In other words, for an isothermal process, when the temperature does not change, the internal energy does not change during this process either. And we can write this as saying Ξ”π‘ˆ because Ξ” represents a change, so a change in internal energy π‘ˆ, is equal to zero. Why is this useful? Well, it’s because we can use something known as the first law of thermodynamics. The first law of thermodynamics tells us that the change in internal energy Ξ”π‘ˆ is given by 𝑄 plus π‘Š.

𝑄 is the heat transferred to the gas, and π‘Š is the work done on the gas. So that’s what Ξ”π‘ˆ is equal to. But we know that for an isothermal process, for the process 𝐴𝐡, Ξ”π‘ˆ is equal to zero. So we can say that 𝑄 plus π‘Š is equal to zero. Now in this first part of the question, we’re trying to find out the value of 𝑄, the heat exchange by the gas along 𝐴𝐡. And we already know the value of π‘Š. It’s been given to us in the question.

We know that the amount of work done by the gas is 7.0 times 10 to the power of two joules. However, we need to be careful here. If we look more closely at the definition of Ξ”π‘ˆ from the first law of thermodynamics, we can see that π‘Š represents the work done on the gas. However, the question tells us that the gas does the work. Now this convention is very important. π‘Š must always represent the work done on the gas by the surroundings. And if the gas itself is doing the work on the surroundings, then the value of π‘Š must be negative.

And so we know that π‘Š is equal to negative 7.0 times 10 to the power of two joules. And so, 𝑄 plus negative 7.0 times 10 to the power of two joules is equal to zero, which means that we find that 𝑄, the heat exchange by the gas, is 7.0 times 10 to the power of two joules. And a slightly neater way of writing 7.0 times 10 to the power of two is 700. So, our final answer to this part of the question is that the heat exchange by the gas along 𝐴𝐡 is 700 joules.

Now let’s move on to the next bit. We’re told that the gas expands adiabatically along 𝐡𝐢 and does 400 joules of work. Once again, notice that the gas is doing the work along 𝐡𝐢. We’re also told that when the gas returns to 𝐴 along 𝐢𝐴, it exhausts 100 joules of heat to its surroundings. And notice again, the question is telling us that the gas is exhausting its heat to the surroundings. In other words, it’s giving out heat.

What we’re asked to do is to find out how much work is done on the gas along the path 𝐢𝐴. In other words, we’re given information about the path 𝐡𝐢, and we need to find out something about the path 𝐢𝐴. Specifically, we need to find out how much work is done on the gas along 𝐢𝐴. So let’s look at the two equations we’ve got at the bottom of the screen. Ξ”π‘ˆ is equal to zero. Well that cannot apply here because we saw that Ξ”π‘ˆ is equal to zero for an isothermal process.

However, process 𝐡𝐢 is not an isothermal process. It’s an adiabatic process. And we cannot assume this for process 𝐢𝐴 either. Therefore, Ξ”π‘ˆ is equal zero goes out the window. We cannot apply this to processes 𝐡𝐢 and 𝐢𝐴. However, the first law of thermodynamics, Ξ”π‘ˆ is equal to 𝑄 plus π‘Š, must always hold anyway. And we’ll need to use this equation later, so we’ll leave it on the screen for now.

Let’s also note now that we’re studying a cycle on this PV diagram. In other words, the gas went from 𝐴 to 𝐡 and then 𝐡 to 𝐢 and then back from 𝐢 to 𝐴. So the gas is going through some processes and then it is finishing at the same point that it started. It’s finishing at 𝐴, and it started at 𝐴. So there’s a useful property that we can exploit here.

Whenever we’ve got a cycle on a PV diagram, we can say that the total change in internal energy of the cycle must be zero. The reason for this is because we’re going from 𝐴 to another point, which is 𝐡, to another point, which is 𝐢, and then back to 𝐴. And of course since we’re coming back to the same point, the internal energy of the cycle must be the same.

Because it has a certain amount of internal energy here, we’ll call this π‘ˆ sub 𝐴. And regardless of whether or not the gas has undergone a cycle, the internal energy at the point 𝐴 must be the same. That’s how we know that it’s at point 𝐴. It has a certain internal energy. So regardless of what happens along the cycle, the change in internal energy therefore must be zero along a complete cycle. We can state this in symbols as saying Ξ”π‘ˆ sub total, the total internal energy change, is equal to zero.

And this is the internal energy change along 𝐴 to 𝐡 plus the internal energy change along 𝐡 to 𝐢 plus the internal energy change along 𝐢 to 𝐴. And so we can state this in the following way: Ξ”π‘ˆ sub 𝐴𝐡 plus Ξ”π‘ˆ sub 𝐡𝐢 plus Ξ”π‘ˆ sub 𝐢𝐴 is equal to zero. Now we already know at least one of these quantities. We already know what Ξ”π‘ˆ sub 𝐴𝐡 is, because we know that 𝐴𝐡 is an isothermal process.

And we saw earlier that the change in internal energy along an isothermal process is zero. So we can replace Ξ”π‘ˆ sub 𝐴𝐡 with zero. However, we cannot apply the same statement to Ξ”π‘ˆ sub 𝐡𝐢 or Ξ”π‘ˆ sub 𝐢𝐴, because they’re not isothermal processes. So we’ll actually have to work out what Ξ”π‘ˆ sub 𝐡𝐢 and Ξ”π‘ˆ sub 𝐢𝐴 are. We can get rid of the zero because that represented Ξ”π‘ˆ sub 𝐴𝐡, but we don’t need it anymore.

And we can move the equation to the left to give us more room on the screen to work out Ξ”π‘ˆ sub 𝐡𝐢 and Ξ”π‘ˆ sub 𝐢𝐴. Let’s start with Ξ”π‘ˆ sub 𝐡𝐢. We can use the second law of thermodynamics, which we’ve written on the bottom right of the screen, to work out what Ξ”π‘ˆ sub 𝐡𝐢 is. Ξ”π‘ˆ sub 𝐡𝐢 ends up being 𝑄 sub 𝐡𝐢, the heat transfer to the gas along process 𝐡𝐢, plus π‘Š sub 𝐡𝐢, the work done on the gas during 𝐡𝐢.

Now the question tells us that the gas expands adiabatically along 𝐡𝐢. In other words, 𝐡𝐢 is an adiabatic expansion. An adiabatic process in thermodynamics is defined as one where there is no heat exchange. In other words, 𝑄 sub 𝐡𝐢 is equal to zero, which means that Ξ”π‘ˆ sub 𝐡𝐢 the change in internal energy is simply π‘Š sub 𝐡𝐢, the work done on the gas during the process 𝐡𝐢.

So we look at the question again to tell us what π‘Š sub 𝐡𝐢 is. We’re told that along 𝐡𝐢, the gas does 400 joules of work. And as we saw earlier, we need to define π‘Š as the work done on the gas by the surroundings. But here, we’ve got the gas doing the work on the surroundings, so π‘Š sub 𝐡𝐢 will be negative 400 joules. And hence, Ξ”π‘ˆ sub 𝐡𝐢 is negative 400 joules, so we can replace Ξ”π‘ˆ sub 𝐡𝐢 with negative 400 joules in the bottom left equation.

We can then go on to working out what Ξ”π‘ˆ sub 𝐢𝐴 is. Once again Ξ”π‘ˆ sub 𝐢𝐴 is equal to 𝑄 sub 𝐢𝐴 plus π‘Š sub 𝐢𝐴. And it’s at this point that we realize that we’ve actually come to a point where we’ve got a quantity, which we’re trying to find in the question. After all, we are asked to find how much work is done on the gas along the path 𝐢𝐴. In other words, we’re trying to find out π‘Š sub 𝐢𝐴. We’re also told in the question that when the gas returns to 𝐴 along 𝐢𝐴, it exhausts 100 joules of heat to the surroundings.

Now this is the value for 𝑄 sub 𝐢𝐴. But yet again, we need to be careful, because 𝑄 has to be defined as the heat transferred to the gas. But in this case, the gas is transferring the heat to its surroundings. And so the value of 𝑄 sub 𝐢𝐴 is negative 100 joules. Now of course we don’t know what π‘Š sub 𝐢𝐴 is because we’re actually trying to find this out. So let’s replace this expression for Ξ”π‘ˆ sub 𝐢𝐴, which we’ve just found, into the equation on the bottom left of the screen.

At this point all we need to do is to rearrange to find out what π‘Š sub 𝐢𝐴 is. Adding 400 joules and 100 joules to both sides of the equation, we find that π‘Š sub 𝐢𝐴 is equal to 400 joules plus 100 joules. And this ends up being 500 joules. And so, the final answer to our question is that the work done on the gas along path 𝐢𝐴 is 500 joules.

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