 So let's explore a little more closely what happens with not just the work, but some other properties for a reversible and isothermal change in volume for a system. So just as a real quick reminder, we have this expression for what the PV work is for process minus P external dV. If we are doing a process reversibly, we can write that as minus P dV. If it's an ideal gas, pressure is nRT over V, so we can write that as minus nRT dV over V. And lastly, when we integrate dW to get W, sticking an integral sign in front of this whole thing, if it's an isothermal process, then the T can be pulled out of the integral. And the result we've seen before that I've just given a quick recap of is that the work for a reversible isothermal volume change for an ideal gas is minus nRT log V2 over V1. So I'll box that equation and we'll spend a little more time thinking about that. Notice that this is, we've only talked about expansion so far, but this could be any change from any V1 to any V2. It doesn't have to be expanding the volume of the gas. It could be also compressing the volume of gas. So just for some variety, let's work an example where we compress the gas. So let's say instead of pistons, let's say now we're talking about a balloon. I've got a balloon with a volume of one liter. Let's make that a temperature of, what temperature am I going to use? I'm going to use 298 Kelvin instead of 273. And initially the pressure is one atmosphere. So those are my initial conditions. That's enough information to use the ideal gas law to, if it's an ideal gas, to calculate the number of moles of gas inside this balloon. And that, if I use n equals PV over RT, find that there's .041 moles of gas in this balloon. And the question is, let's say I compress this gas to half of its original volume and twice its original pressure, leaving the temperature unchanged. So I've done an isothermal, left the temperature alone, volume change. I'm doing it as a compression this time instead of an expansion. And we're going to make sure and do that reversibly so that we can use this expression. So the work done, by that, involved in that process, minus nRT log V2 over V1. So minus .041 moles, 8.314 joules per mole Kelvin for the gas constant, 298 Kelvin multiplied by the log of this V2 over V1. So I'm now decreasing the volume. The final volume is smaller than the initial volume, so I've got the log of a small number over a large number. So the calculator will tell us what the value of this is, and we get a value of 70 joules. The thing that's worth pointing out at this point is the value of this number is positive. The work in this case is a positive number. As we'd expect to be true for a compression, when the gas is compressed, we're doing work on the gas. We're supplying energy into the system. So the net change in energy of the system due to this PV work is positive, so the work itself is positive. Mathematically, that shows up because when my final volume is smaller than my initial volume, this ratio is less than one. Take the log of a number less than one, and I get a negative number. When I multiply it by this negative sign, I get a positive result. So any compression where the volume is going to be smaller after the compression than before is always going to lead to a positive value for the work. So that's just a plug and chug question using the same expression we had before for the reversible isothermal change in volume for an ideal gas. But we can do a few other things as well. Let's see. If I want to calculate the change in energy of this gas, the change in the internal energy as it undergoes this same process, we can do that also. So remember for an ideal gas, in particular for a gas that we can use the 3D particle in a box model to describe, the energy of a 3D particle in a box is three-halves nRT. So if I know the temperature, I can calculate the energy. So initially, the temperature is 2 and 8 Kelvin, and I can use that temperature to calculate the energy for the final conditions because it's an isothermal compression of this gas. The final temperature is also 2 and 8 Kelvin, so the energy of the gas under these final conditions just depends on the temperature and the number of moles. If I haven't changed the number of moles, I haven't changed the temperature, I haven't changed the energy. So more formally, I could say that the change in the energy is energy of the final conditions minus the energy of the initial conditions, three-halves nRT2 minus three-halves nRT1 or three-halves nRT2 minus T1, I can write as delta T, the change in the temperature for this process. So because the temperature is not changing, delta T is zero, doesn't matter how much, but I multiply that delta T by, because the temperature changes zero, in this case, I'm going to get zero. So the internal energy change, the energy of this gas inside this balloon didn't change because its temperature didn't change. So if I go back over here and I collect my list of equations that describe properties for the reversible isothermal volume change, any of an ideal gas, we're always going to get delta U of zero because the ideal gas is isothermal. The only type of energy that ideal gas has is related to its temperature because the only type of energy we've allowed an ideal gas to have when we solve the quantum mechanical problem was kinetic energy. And if I haven't changed the temperature, I haven't changed the kinetic energy. So the energy didn't change. And of course now, since I have both delta U and W, I can also calculate the value of heat. If I rearrange this equation to say that Q is equal to delta U minus W, any time I have a reversible isothermal change in volume for a gas, so this equation is always true. If the work is minus nRT log V2 over V1 and the energy change is zero, then I'll have zero minus negative nRT log V2 over V1. So the heat is going to be positive nRT log V2 over V1. So I could plug numbers in and calculate nRT log V2 over V1. I've already done that with a negative sign. I get the same answer this time just with the opposite sign, so that's going to work out to be negative 70 joules. So let's let's do two things. Let's put an equation over here. Any time we want to calculate the heat involved in expanding or compressing an ideal gas reversibly and isothermally, we can just use this expression. We can use any of these equations without having to repeat any of the algebra over on this side of the board. Let's also take a look at this number. What does it mean to say that when I have this balloon, let's say I have an actual balloon, if I compress it to half its original volume, changing the pressure from one atmosphere to two atmospheres but not letting the temperature of the gas change in the process. If I do it isothermally and if I do it slowly and reversibly, it certainly costs me some energy to do the compression. The PV work of doing that compression, 70 joules. That's what it cost me to do compression, reversible isothermal compression of one liter of an ideal gas. Because the only type of energy that gas has is kinetic energy and I haven't allowed the temperature to change, of course its energy is not going to change. So what the heat means is that if I've supplied energy, cost me energy to do the compression, I've supplied energy in the form of work into the gas, ordinarily that would heat the gas up. That would raise its temperature. But I have not allowed the temperature to change because I did the process isothermally. So what that means is in order to keep the temperature the same, that excess energy that I put in the system in the form of work wasn't used to change the temperature. Instead, it had to leave the system in the form of heat. So some heat left the system, I supplied 70 joules of energy into the system as work. Same 70 joules, so compression supplied 70 joules into the system and then an equal 70 joules of energy left the system in the form of heat in order to keep the temperature of the balloon itself from changing. So that's the way to interpret the size and the signs on these numbers. When I compress a gas isothermally, I do work on it and it transfers some heat to the surroundings. So we have those numerical results. We also have these equations that we can use anytime we have a reversible isothermal volume change. So the next thing we'll do is look a little more closely at the fact that when I do the expansion or compression reversibly and isothermally, I get this set of equations. If you remember back a little bit, you'll remember that the work done for reversible and isothermal volume change is different, a different number than when you do an expansion or compression under different conditions. So we need to look a little more closely at why it is that we can get different values for the work under different types of conditions.