 Hi, I'm Professor Nashiva, and I want to tell you a little bit about how to calculate the work when the pressure that's been opposing the expansion of a volume is held constant. This is probably the case in most cases if the work is done against the atmosphere. So the physical idea could be something like this. We've got a beaker here again and a solution and maybe there's some reaction going on that's forming bubbles. So we just imagine that after a certain period of time enough bubbles have formed to effectively push the atmosphere out by, you know, some volume. And, you know, just for the sake of an example, we could just say that that's one liter that the that the volume has increased. So our expression for this is work is equal to the minus the integral of the external pressure dV. And in this case in which the external pressure is constant as the atmosphere would be, I could say, well, that's minus the external pressure times the integral of dV. But the integral of dV is just the volume evaluated from, you know, the ending point to the starting point. Let's just say that we started off, I have an axis here of volume. Let's suppose we started off at zero and we ended up at one liter. Okay, so the change in volume would be one liter and I can finish this up like that. The work done is minus the external pressure times change in volume. So what did we just say? I said that's one liter. But if I want to make this into an SI calculation, I would have to convert that into cubic meters, which turns out to be 10 to the minus 3 cubic meters. And what about the external pressure? Well, I said, or, you know, if it's atmospheric pressure, this would be about, let's say, one bar. But in SI units, that's, we should convert that to Pascals. So that would be 10 to the fifth Pascals. Okay, so I have 10 to the fifth Pascals times 10 to the minus 3 cubic meters. That's 100. Okay, so I get minus 100. And since this is an entirely SI calculation, I know that my answer, which has to have units of energy, it must be, it must be joules. So my answer for this expansion of going from zero to one liter of expansion against atmospheric pressure at one bar comes out to be minus 100 joules. It's a negative, of course, because now the system has done work on the surroundings. Okay.