00:01
This portion belongs to the loss of thermodynamics in which we have to determine for the part a, the work done in each step when the process is in the counterclockwise direction.
00:10
So the work done w, it is equals to minus p multiplied by delta v and delta v can be written as v final minus v initial.
00:17
So for the b a process, the work done will be minus and the pressure for the process a b, it is 300 kilo -pascal from the graph.
00:27
So 300 multiplied by 10 to the power 3.
00:30
V final, it will be volume at the a, which is 1 meter cube, and initial volume will be volume at the b, which is 4 meter cube.
00:37
So from here after solving, we get work done for the process b .a, it is equals to plus 9 .0 multiplied by 10 to the power 5 jule.
00:46
Now for the process b .c, so for this process, the work done will be 0 because the volume is constant.
00:55
And now for the process d .c.
00:58
So this work done will be minus p.
01:01
So for the process dc, the work done will be the pressure is 100 kilo pascal from the graph.
01:08
So 100 multiplied by 10 to the power 3.
01:11
And the final situation is c...