### Video Transcript

An ideal gas expands isothermally along the path π΄π΅ 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 π΄π΅? The gas then expands adiabatically along π΅πΆ and does 400 joules of work. When the gas returns to π΄ along πΆπ΄, it exhausts 100 joules of heat to its surroundings. How much work is done on the gas along this path?

Okay, so weβve got a multipart question here. So letβs start by looking at the very first part of the question. Weβve been told that weβve got an ideal gas, and it expands isothermally along the path π΄π΅. Weβre also told that the gas does 7.0 times 10 to the power of two joules of work along this path. Weβre firstly asked to find how much heat does the gas exchange along π΄π΅. So for the very first part of this question, weβre just studying this part here, from π΄ to π΅.

Now, we know that weβve got an ideal gas and we also know that the process π΄π΅ 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 π to denote temperature. Now, we also know the amount of work that the gas does along the path π΄π΅. And weβre asked to find the amount of heat exchange by the gas along this path.

To do this, letβs first recall how the internal energy of an ideal gas is found. For an ideal monatomic gas, the internal energy π is given by three by two multiplied by π, the number of moles of gas, multiplied by π
, the molar gas constant, multiplied by π, the temperature of the gas. Now of course weβve not been told that the gas is monatomic, but this doesnβt 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 weβll see soon, itβs not relevant to us. So, we know that weβve got a gas undergoing a certain set of processes, π΄ to π΅, π΅ to πΆ, and πΆ to π΄. When we study the process π΄ to π΅, we are told that itβs an isothermal process, so the temperature is constant. We also note that π
, the molar gas constant, is a constant. The clue is in the name. Thirdly, thereβs 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, π, 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 Ξπ because Ξ represents a change, so a change in internal energy π, is equal to zero. Why is this useful? Well, itβs 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 Ξπ is given by π plus π.

π is the heat transferred to the gas, and π is the work done on the gas. So thatβs what Ξπ is equal to. But we know that for an isothermal process, for the process π΄π΅, Ξπ is equal to zero. So we can say that π plus π is equal to zero. Now in this first part of the question, weβre trying to find out the value of π, the heat exchange by the gas along π΄π΅. And we already know the value of π. Itβs 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 Ξπ from the first law of thermodynamics, we can see that π 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 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 must be negative.

And so we know that π is equal to negative 7.0 times 10 to the power of two joules. And so, π plus negative 7.0 times 10 to the power of two joules is equal to zero, which means that we find that π, 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 π΄π΅ is 700 joules.

Now letβs move on to the next bit. Weβre told that the gas expands adiabatically along π΅πΆ and does 400 joules of work. Once again, notice that the gas is doing the work along π΅πΆ. Weβre also told that when the gas returns to π΄ along πΆπ΄, 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, itβs giving out heat.

What weβre asked to do is to find out how much work is done on the gas along the path πΆπ΄. In other words, weβre given information about the path π΅πΆ, and we need to find out something about the path πΆπ΄. Specifically, we need to find out how much work is done on the gas along πΆπ΄. So letβs look at the two equations weβve got at the bottom of the screen. Ξπ is equal to zero. Well that cannot apply here because we saw that Ξπ is equal to zero for an isothermal process.

However, process π΅πΆ is not an isothermal process. Itβs an adiabatic process. And we cannot assume this for process πΆπ΄ either. Therefore, Ξπ is equal zero goes out the window. We cannot apply this to processes π΅πΆ and πΆπ΄. However, the first law of thermodynamics, Ξπ is equal to π plus π, must always hold anyway. And weβll need to use this equation later, so weβll leave it on the screen for now.

Letβs also note now that weβre studying a cycle on this PV diagram. In other words, the gas went from π΄ to π΅ and then π΅ to πΆ and then back from πΆ to π΄. So the gas is going through some processes and then it is finishing at the same point that it started. Itβs finishing at π΄, and it started at π΄. So thereβs a useful property that we can exploit here.

Whenever weβve 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 weβre going from π΄ to another point, which is π΅, to another point, which is πΆ, and then back to π΄. And of course since weβre 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, weβll call this π sub π΄. And regardless of whether or not the gas has undergone a cycle, the internal energy at the point π΄ must be the same. Thatβs how we know that itβs at point π΄. 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 Ξπ sub total, the total internal energy change, is equal to zero.

And this is the internal energy change along π΄ to π΅ plus the internal energy change along π΅ to πΆ plus the internal energy change along πΆ to π΄. And so we can state this in the following way: Ξπ sub π΄π΅ plus Ξπ sub π΅πΆ plus Ξπ sub πΆπ΄ is equal to zero. Now we already know at least one of these quantities. We already know what Ξπ sub π΄π΅ is, because we know that π΄π΅ is an isothermal process.

And we saw earlier that the change in internal energy along an isothermal process is zero. So we can replace Ξπ sub π΄π΅ with zero. However, we cannot apply the same statement to Ξπ sub π΅πΆ or Ξπ sub πΆπ΄, because theyβre not isothermal processes. So weβll actually have to work out what Ξπ sub π΅πΆ and Ξπ sub πΆπ΄ are. We can get rid of the zero because that represented Ξπ sub π΄π΅, but we donβt need it anymore.

And we can move the equation to the left to give us more room on the screen to work out Ξπ sub π΅πΆ and Ξπ sub πΆπ΄. Letβs start with Ξπ sub π΅πΆ. We can use the second law of thermodynamics, which weβve written on the bottom right of the screen, to work out what Ξπ sub π΅πΆ is. Ξπ sub π΅πΆ ends up being π sub π΅πΆ, the heat transfer to the gas along process π΅πΆ, plus π sub π΅πΆ, the work done on the gas during π΅πΆ.

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

So we look at the question again to tell us what π sub π΅πΆ is. Weβre told that along π΅πΆ, the gas does 400 joules of work. And as we saw earlier, we need to define π as the work done on the gas by the surroundings. But here, weβve got the gas doing the work on the surroundings, so π sub π΅πΆ will be negative 400 joules. And hence, Ξπ sub π΅πΆ is negative 400 joules, so we can replace Ξπ sub π΅πΆ with negative 400 joules in the bottom left equation.

We can then go on to working out what Ξπ sub πΆπ΄ is. Once again Ξπ sub πΆπ΄ is equal to π sub πΆπ΄ plus π sub πΆπ΄. And itβs at this point that we realize that weβve actually come to a point where weβve got a quantity, which weβre trying to find in the question. After all, we are asked to find how much work is done on the gas along the path πΆπ΄. In other words, weβre trying to find out π sub πΆπ΄. Weβre also told in the question that when the gas returns to π΄ along πΆπ΄, it exhausts 100 joules of heat to the surroundings.

Now this is the value for π sub πΆπ΄. But yet again, we need to be careful, because π 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 π sub πΆπ΄ is negative 100 joules. Now of course we donβt know what π sub πΆπ΄ is because weβre actually trying to find this out. So letβs replace this expression for Ξπ sub πΆπ΄, which weβve 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 sub πΆπ΄ is. Adding 400 joules and 100 joules to both sides of the equation, we find that π sub πΆπ΄ 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 πΆπ΄ is 500 joules.