00:01
In this problem, we're going to talk about the first law of thermodynamics.
00:03
So what we need to remember is that the first law of thermodynamics tells us that delta u, the variation in internal energy of a gas, is equal to the heat provided to the gas, minus the work done by the gas.
00:21
And it's important to remember this convention because if the heat were provided from the gas to the system, then the heat would be negative.
00:28
And the work if it was done on the gas then it would be negative as well okay so what we have in a problem are several processes that a gas undergoes initially the gas has a pressure of one atm and a volume of three liters and it expands at constant pressure to five liters then its pressure increases at constant volume to 3 atm, then its pressure decreases linearly with the volume to 2 atm and 3 liters.
01:24
And finally, the gas loses pressure at constant volume to 1 atm and 3 liters.
01:36
Okay? okay, and our going question a is to draw a pv diagram of our gas, of our process actually.
01:50
So here for p we have 1 atm, 2 atm, 3 atm, and the volume is 3 liters, actually, yeah, 3 liters and 5 liters.
02:11
So initially, the initial point is this one here, okay, this point.
02:23
Then the gas expands at constant pressure to this point here.
02:36
Then it keeps the volume constant and expands to three atm, expense, actually not expends it, increases pressure to three atm.
02:50
Then the pressure decreases linearly with the volume to 2 atm and the volume to 3 liters and finally the pressure decreases to the original value of pressure and volume at constant volume so this is our graph and in question b we have to calculate how much heat flowed into the system or out of the system during this process.
03:35
We know the delta u that is a variation in internal energy is zero.
03:40
And why do we know that? the internal energy of a system depends only on its point in the pvt space, meaning that even that p, v and t are the same, then u is going to be the same...