 So we've talked about how to calculate energy and heat and work changes when we heat a system specifically at ideal gas at constant volume. Let's also consider the case where we do the heating at constant pressure. So let's consider exactly the same problem as we had before. We've got one mole of an ideal gas initially at a pressure of one atmosphere and a temperature of 298 Kelvin, which makes it occupy 24.5 liters. Once again we're going to increase the temperature by 50 Kelvin, but now we're going to do it isobarically at constant pressure. So initial pressure and final pressure are going to be the same value. We're not going to allow the pressure to change, and what that means is the volume is going to have to increase. If I have a box of gas at constant pressure and I heat it up, its volume will increase. It'll expand as I increase its temperature. So that's going to increase the volume to some new final volume. We can calculate what that final volume should be using the ideal gas law, of course. P equals nRT, so V2 is equal to nRT over P. So if I know my final temperature and final pressure, this is going to be one mole of a gas. Gas constant is 8.314 joules per mole Kelvin. I'm still multiplying in the numerator by 298 Kelvin, and I divide that by, I'm sorry, not 298 Kelvin. The final temperature is 348 Kelvin, and divide by the final pressure, which is the same as the initial pressure, which is one atmosphere. And now I realize it's a good thing that I was going to double check my units because my units are not working out. So rather than using the gas constant in units of joules per mole Kelvin, I need to cancel some units of atmospheres. So I'm going to instead use the gas constant as 0.08206 liter atmospheres per mole Kelvin. And then when I check my units, moles will cancel moles, Kelvin will cancel Kelvin, atmospheres will cancel atmospheres, and my answer will come out in units of liters. So plug these numbers into a calculator, which I've already done, and the answer works out to be 28.6 liters. So as expected, the volume of our ideal gas increases. As I heat it from room temperature of 298 Kelvin up to 50 Kelvin hotter, the volume increases by the same ratio, 24.5 up to 28.6 liters. So now we're ready to start calculating properties like the change in the internal energy and the heat and the work. Internal energy, that's exactly the same as it was before. Remember, energy is a state function, so it doesn't matter whether I took a constant pressure or constant volume path, the amount by which I changed the temperature is all I need to know to calculate the change in the energy. So 3 halves nr delta t is the same thing as it was previously. If I plug numbers into this expression with one mole and the gas constant, that's going to work out to the same 620 joules that it was for the isochoric process. So so far everything is the same for delta u, isochoric or isobaric doesn't matter for an ideal gas. If I calculate work, on the other hand, for the constant volume process, the work turned out to be zero because we weren't changing the volume, we weren't doing any PV work. In this case, we are changing the volume, so we do have to calculate something to calculate the PV work. So if we go back to our definition of PV work, negative integral of P external dv, P is constant. We're doing this under constant pressure, so the pressure we can pull out of the integral dv integrates to just delta v. So because it's a constant pressure process, I can use minus P external delta v. The pressure is 1 atmosphere. The change in volume is 24.5 up to 28.6 liters, so 28.6 minus 24.5 v2 minus v1. So that change of 4.1 liters multiplied by an atmosphere gives me a total of negative 4.1 liter atmospheres. So that's correct, it's not wrong. Liter atmospheres is kind of an inconvenient unit to express work in, so let's change liter atmospheres to joules. There's 101 joules per liter atmosphere, or if we prefer 8.314 joules for every .08206 liter atmospheres. So that unit conversion, pretty close to 100, it's a little over 100, so it works out to be negative 420 joules is the amount of PV work that gets done. So notice what's happening. I've got my box of gas by keeping the pressure constant as I increase it from T1 to T2, the volume increases from v1 to v2. In the process of changing the volume it has to do some work. The system does some work by expanding, so the work has a negative sign and it does 420 joules worth of work in pushing back the lid of the container or the walls of the container. So notice that that's pretty significant, it's almost as big as the size of delta u and that now allows us to calculate the heat. First law tells us the heat is delta u minus the PV work and we've calculated both of those. The change in the internal energy is 620 joules, the work is negative 420 joules, so I'm subtracting negative 420 joules, so the net result is over a kilojoule, 1,040 joules in this case. So let's make sure we understand what those quantities mean. I've increased the temperature of the gas, I've increased the kinetic energy of the gas, that's the form of energy that gases have, so the internal energy of the gas increased because of that temperature increase. So 620 joules worth of energy went into raising the temperature of the gas. In the process of heating it up it expanded, did some PV work, did work on the surrounding so that's a negative value of w. The reason q is this large number of 1,040 joules is if we're thinking about how much energy had to enter or leave the system, I had to put enough energy into this system to pay not only to heat up the gas to make its kinetic energy increase, but also to pay for the work of pushing back against the atmosphere and expanding the volume of the container. So the total amount of energy I had to put into the gas in the form of heat is the 620 joules it took to raise the temperature as well as the 420 additional joules it took to expand the container. So a total of 1,040 joules worth of energy in the form of heat was supplied into the system to pay for energy and work. So a few things to notice about this. Number one it's not as convenient to do heating at constant pressure because we don't have the convenience of being able to say work is zero. Isochoric processes, constant volume processes because the volume doesn't change the work calculation is very simple. So that also ends up complicating the heat. Heat has to pay for the PV work which is non-zero. So because of that inconvenience of having to deal with PV work which is really sort of a hassle, the next thing we'll do is define a new quantity that lets us not have to worry so much about PV work. So that's what we'll consider next.