![]() Kilograms, meters, and seconds, this meters cancels out with one of these, and we end up with one And if we do that, everything is in terms of So, what we need to do here, is convert our liters to meters cubed. This by units of volume, and whatever we multiply it by should give us units of joules. What this tells us is that, we have to multiply One pascal, pascals are also in terms of, are also SI units, and if we convert pascals to kilograms, meters, and seconds, we get that one pascal is one kilogram per meter seconds squared. So, joules is equal to, one joule is equal to one kilogram meters So, a joule is equal to one kilogram meters squared per second squared. So, if you take joules, which is already SI units, we can actually simplify it more, in terms of other SI units. So the way I did this, is by converting everything That pascals times liters, will give us joules. Otherwise, we will be, we'llīe subtracting two things that don't have the same unit, and that's bad. Whatever we calculate here, in terms of work, also Pascals, times liters, we need to make sure that On this other side, we're calculating our work here, and we have pascals, so pascals times, so we have joules, and joules, we have pascals times liters, so then the question is, okay, we're doing joules minus And that's our units, we have our heat, in terms of joules, we probably want ourĬhange in internal energy in terms of joules too. Thing that we should check before we actually plug in the numbers and have a party. We could at this point be like," okay, "we figured it all out, "we just have to stickĪll of these numbers "in a calculator and we're done," and that's probably what myįirst instinct would often be, but there's one more So that's our final pressure, 2.05 liters, minus our initial pressure, which is 2.30 liters. So negative 485 joules, minus, we should haveĪ negative sign there, minus the external pressure 1.01, times ten to the 5th pascals, so that's our external pressure It should have a negative sign because the heat was So if we start plugging that in, we get that delta U is equal to negative 485 joules, so that's our heat, we know Times the change in volume, and we know both of those things, we know the external pressure and we know the initial and final volumes. We can also calculate work because we know the external pressure, we know it's constant, and work can be calculated as the external pressure So that means the workĭone here is positive. We would expect if the surroundings did work on our system, that would increase the internal energy. ![]() Work on the other hand, since V two is less than V one, the volume of our system went down, which means the surroundings had to do work on the gas to get the volume to decrease. Not the other way around, Q should be negative, because when your system transfers energy to the surroundings, then it's internal energy should go down. System transferred energy to the surroundings and I think that's one of the trickiest things in these kind of problems. Before we plug any numbers in here, the first thing I wanna do is make sure I have a good idea of what The change in internal energy, delta U, is equal to the work done, plus the heat transfer. Internal energy for our system? We can use the first law of The question we're gonna answer is, for this process, what is delta U? So, what the change in And we're assuming here, that the moles of gas didn't change. Once it does that, the final volume of our system, is 2.05 liters. So, we know the external pressure is 1.01 times ten to the 5th pascal, and our system is some balloon, let's say it's a balloon of argon gas, and initially our gas hasĪ volume of 2.3 liters, and then it transfers, the gas transfers 485 joules of energy as heat to the surroundings. In this video we're gonna do an example problem, where we calculate in internal energy and also calculate pressure volume work.
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