interactive thermodynamics 3.2 show more decimal command

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.

Interactive thermodynamics 3.2 show more decimal command