Figure below shows two paths that may be taken by a gas to go from a state A to a state C. [figure] In process AB, 400 J of heat is added to the system and in process BC, 100 J of heat is added to the system. The heat absorbed by the system in the process AC will be
- 460 J
- 300 J
- 380 J
- 500 J
Answer
(1) 460 J
Solution
Process AB is isochoric: WAB = 0
Process BC is isobaric: WBC = P V = 6 104 2 10-3 = 120 J
Along ABC: QABC = 400 + 100 = 500 J
UAC = QABC - WABC = 500 - 120 = 380 J
Work along AC = area under the line AC:
WAC = 12(2 104 + 6 104) 2 10-3 = 80 J
Internal energy is a state function,
QAC = UAC + WAC = 380 + 80 = 460 J
Answer: (1)