00:01
In this problem, we are given a pv diagram for a heat engine.
00:05
And as you can see, we have temperatures given at three labeled points.
00:10
And our goal initially is to find the p and v at those labeled points also.
00:15
Then in the end, we're going to be filling in this table about the heat and the work for each of the processes in the cycle.
00:22
Well, let's first start by getting the p and the v at each of the points.
00:26
Now, we are given the number of moles, so we can use the ideal gas law.
00:29
So we can use p1, v1, nrt1.
00:36
And from that we can solve for v1 is equal to nrt1 over p1.
00:42
Now remember, t1 temperature must be in calvin pressure to get meters cubed must be in pascal.
00:49
So this becomes 0 .3, 5 moles, 8 .314, joules, mole kelvin, 300 kelvin, and one atmosphere in pascal is 1 .013 times 10 to the 5 pascal.
01:19
And this works out to be 8 .62 times 10 to the minus 3 cubic meters.
01:32
And this also is v2 because 1 to 2 is an isochoric process, constant volume process.
01:40
Now we can do the same thing to get p2 is the ideal gas law.
01:44
P2 is nr t2 over v2 and this number works out 2 .025 times 10 to the fifth pascal which effectively is two atmosphere and lastly we need v3 because we know p3 is the same as p1 one atmosphere p3 is nrt3 over p3, and this works out to be 1 .41 times 10 to the minus 2 cubic meters, and we have finished with all the entries that we need in that table.
02:42
Now, part b, part b deals with the right -hand table now, but they gave us the magnitudes of the changes in internal energy, not with the appropriate sign.
02:54
So how do we figure out that? well, for an ideal gas, u is solely dependent on the temperature.
03:01
So if the temperature increases, you increases.
03:06
Temperature decreases, u decreases.
03:08
So t1, from 1 to 2, temperature increases, the change of internal energy is positive.
03:15
2 to 3, temperature decreases, change in internal energy is negative.
03:19
3 to 1, temperature decreases, change in internal energy.
03:22
Energy is negative.
03:24
Now we can go and calculate the q and the w for each of the processes.
03:30
So 1 to 2, we use the first law, delta u, is equal to q minus w.
03:39
Now, constantifying a process.
03:43
By definition, the work is zero.
03:46
So this just is equal to, q is equal to delta u.
03:51
So that is 2180 joules.
03:56
So we have the work and the, the, heat for process for 1 to 2.
04:03
Now 2 to 3.
04:05
Start out again.
04:06
Dot to you, q minus w.
04:10
Well, that's an adiabetic process.
04:13
So q is equal to 0.
04:16
So minus w.
04:17
So that gives me the w is equal to minus of minus 785, so 785.
04:22
So 785 fuels.
04:32
And now 3 to 1.
04:36
This one's going to require a little more work.
04:38
Delta u is q minus w as always.
04:43
Now this is an isobaric process.
04:45
The work done in an isabic process is p -dil -v.
04:48
So q minus p -dil -v.
04:53
So we have everything we need...