00:01
Hello, here, mona -atomic ideal gaze initially fills volume of 0 .45 cubic meters and the initial pressure, p0, is 85 khalpascals.
00:18
Then the gaze undergoes isabaric extension to v1, so in the first process, pressure is constant, and the gas expands to v1 of 0 .75 cubic meters.
00:33
It undergoes is a volumetric cooling to the initial temperature t0.
00:41
So then v is constant in the second process and t2 equals to t0.
00:51
And finally, in the third step, it under goes an eithothermal compression to its initial volume and pressure.
01:03
So then p3 is the same as p0 and v3 is the same as v0.
01:09
So, first of all, we have to calculate work done by the gaze during the isobaric process.
01:19
Let's do this.
01:20
So this work, w1, equals to p0 times v1 minus v0, or 85 times 10 power by 3 pascal, multiply by the difference between volumes.
01:41
Let's calculate it.
01:56
That is 25 .5.
02:00
Times 10 power by 3 joules which is 25 .5 kilojoules.
02:06
That is solved.
02:09
Let's move on to question b.
02:11
In question b we have to calculate heat absorbed during the isobaric expansion and this heat equals to w1 work plus changing internal energy.
02:24
On the other hand that is p changing volume plus 3 over 2 and are changing temperature and as we know that is also that can be written as 1 .5 times p changing volume it means that overall u1 is 2 .5 times w1 that is 63 .75 kilojoules which is roughly 63 .8 kilojoules that is soft let's move on now question c in question c we have to calculate the heat absorbed in the stage q2 and in let's look at the process q2 in process 2 is under constant volume that's why work is zero and then change heat absorbed yeah the absorbed heat is simply changing internal energy which is negative 1 .5 and r t 2 minus t1 which is negative 1 .5 and r t0 minus t 1 which is negative 1 .5 times yeah that basically equals to negative delta e1...