In the indicator diagram, we have (a) Change in internal energy along ABC is 10 J. (b) Work done along path AB = 20 J. U_(C ) = 5 J. (d) Heat absorbed by the system along path AD is 5J. (i) Change in internal energy along the path CDA. (ii) Heat given to the system along path ABC. (iii) Value of U_(A). (iv) change in internal energy along AD.

In the indicator diagram, we have (a) Change in internal energy along

1.	In the indicator diagram, we have (a) Change in internal energy along ABC is 10 J. (b) Work done along path AB = 20 J. U_(C ) = 5 J. (d) Heat absorbed by the system along path AD is 5J. (i) Change in internal energy along the path CDA. (ii) Heat given to the system along path ABC. (iii) Value of U_(A). (iv) change in internal energy along AD.
Answer» Solution :(iv) `U_(C )- U_(A) = 10 J:> U_(A) - U_(C ) = -10 J` (II) `Delta Q = Delta U + Delta W = (U_(C ) - U_(A)) + Delta W` or `Delta Q = 10 + 20 = 30 J` (iii) `Delta Q = (U_(C ) - U_(A)) + Delta W` or `U_(A) = U_(C ) + Delta W - Delta Q = 5+20 - 30 = -5J` or `U_(A) = -5J` (d) `U_(D) - U_(A) = DeltaQ_((AD).) - 5 - 0 = 5J`.