00:01
Let's discuss this question.
00:02
So here we have a breton cycle heat engine with dynamic gas follows the cycle as follows.
00:09
So here we need to determine the engine's thermal efficiency by explicitly computing the work done per cycle.
00:18
So here we can say pressure is given as p that is equal to 1 .68 atmosphere and temperature t, 1 is equal to 290 to kelvin and volume is equal to 1 .91 meter cube.
00:41
So here we can say number of moles that is n is equal to p1 b1 divided by r t1 so this is equal to 1 .68 atmosphere multiplied by 1 .013 multiplied by 10 r to 5 pascal divided by atmosphere multiplied by 1 .91 meter cube multiplied by 1 .91 meter cube whole divided by 8 .314 jule per mole kelvin this is the gas constant and here we have 292 kelvin so here n is equal to 133 .8934 mole that is equal to 133 .89 mole that is equal to 133 .89 reaction here we can say p1 vn r that is equal to p2 v2 r so here we can say v2 is equal to v1 multiplied by p1 divided by p2 whole rise to 1 divided by r so this will be equal to 1 .91 meter cube multiplied by 1 divided by 10 1 divided by 1 .4 that is equal to 0 .91 meter cube multiplied by 1 .4 that is equal to 0 .91 meter cube multiplied by 1 .5 .1.
02:11
687 meter cube.
02:16
So here we can say from ideal gas equation p1 b1 divided by t1 is equal to p2 v2 divided by p2.
02:34
So we need to find the temperature that is t2.
02:36
So here t2 is equal to t2 multiplied by p2 divided by p1 multiplied by p1.
02:45
So here we can say this is equal to 292 kelvin multiplied by 10 atmosphere multiplied by 1 .68 divided by 1 .68 atmosphere multiplied by 0 .3687 meter cubed divided by 1 .91 meter cube.
03:08
So here we can say p2 is equal to 563 .667 kelvin and this is the final answer.
03:20
Now here we can say qh is equal to 2 .98 multi -clad by 10x to 6 joules n -cp that is specific 8 multiplied by delta t is equal to 7 divided by 2 nr delta t so here we can say 2 .98 multi -clad by 10x to 6 joul is equal to 7 divided by 2 multiplied by 111 .3 .16 jules per kelvin multiplied by delta t.
03:54
So here delta t will be equal to 764 .87 kelvin.
03:59
So here t3 is equal to 32 plus delta t.
04:03
So here we can say t3 is equal to 563 .667 kelvin plus 764 .87 kelvin.
04:15
So here we can say v3 is equal to v2 multiplied by.
04:18
T 3 divided by t 2 this is equal to 0 .3687 meter cube multiplied by 1328 .3 .3 537 kelvin divided by 563 .66 kelvin.
04:36
So here we can say v3 is equal to 0 .8690 meter cube and this will be the answer.
04:46
Now further here we can say p3 v3 r is equal to p4 v4 r that is equal to v4 that is equal to v4 that is equal to v3 multiplied by p3 divided by p4 o rest to 1 divided by r so here v4 is equal to 0 .8690 meter cube multiplied by 10 raised to 1 divided by 1 .4 is equal to 4 .50 meter cube so here therefore p4 is equal to p4 v4 v2 divided by an r which is equal to 1 .68 atmosphere multiplied by 1 .013 multiplied by tell 3 .5 pascal divided by atmosphere multiplied by 4 .50 meter cube this is divided by 133 .89 mole multiplied by 8 .314 jules per mole kelvin so here we can say therefore t4 that is temperature is equal to 687 .97 kelvin.
06:04
So this is the temperature that is t4...