A thermodynamic process is shown in figure. In process ab, 600 J of heat is added, and in process bd 200 J of heat is added. The total heat added in process acd is

A thermodynamic process is shown in figure. In process ab, 600 J of he

1.	A thermodynamic process is shown in figure. In process ab, 600 J of heat is added, and in process bd 200 J of heat is added. The total heat added in process acd is
Answer» 550 J 650 J 750 J 850 J Solution :There is no volume change in process ab, so W = 0Process BD occurs at constant pressure, so the work done by the system is `W = P(V_(2) - V_(1)) = (8 xx 10^(4) Pa)(5 xx 10^(-3)m^(3) - 2 xx 10^(-3) m^(3)) = 240 J` Thus the total work for abd is W = 240 J and the total heat for abd is Q = 800 J. `:. Delta U = Q - W = 800 J - 240 J = 560 J` Because `Delta U` is independent of path, the INTERNAL energy change is the same for the path acd as for abd, that is, 560 J. The total work for path acd is `W = (3 xx 10^(4) Pa)(5 xx 10^(-3)m^(3) - 2 xx 10^(-3) m^(3)) = 90 J` `:.` Total heat added in the path acd is `Q = Delta U + W = 560 J + 90 J = 650 J`