00:01
In this question we are given as delta s is equal to n cv log t2 upon t1 plus nr log v2 upon v1 so c p is equal to cv plus r so this is given and the value of row is given as a ratio of cp upon cv.
00:41
So this is the ratio of molar capacity at pressure respect to volume.
00:52
So now we know that du is equal to dq minus dw and by the second law of thermodynamics that change in entry.
01:04
Is equal to change in heat with respect to temperature so we have this d -q is equal to t -d -s now solving for ds and we have this d -s is equal to d -s is equal to where d -q is given as d -u plus d -w so we will get as d -u from this equation 1, this is d .u plus d .w.
01:41
Divided by t.
01:43
Now, delta s will be equal to integration at the initial stage of the entropy and the final stage of the entropy as d .s.
02:00
So, further, d .s will be as equal to initial and final of the entropy of.
02:09
D u plus d w upon t hence delta s is equal to u initial and u final then top initial and final d u upon t plus integration with work initial and work final as d w upon t now we know that d .u is equal to n .c .v.
02:57
D .t.
03:00
And d .w.
03:02
Is equal to p.
03:05
D .v.
03:07
Is equal to nrt upon v.
03:13
D .v.
03:13
So here we can put this equation as delta s is equal to t1 and t final n c v d t upon t plus v initial and v final of nr t upon t.
03:46
So t is canceling out...