Illustration Problem 2 (A B, B- C, C M). Points B, and Care both at the same pre
ID: 1606699 • Letter: I
Question
Illustration Problem 2 (A B, B- C, C M). Points B, and Care both at the same pressure of 1.5 x 10s Pa. Points ef, B, and Care all equilibrium states. a) (2 pts) During which stage is the work involved the greatest in magnitude? During which stage is the work involved the least in magnitude? b) (2 pts) If the work done by the system in going from A- B is +5.6 x 105 J, calculate the work done during a full cycle by the system. e) If the change in internal encrgy in going from A B is +3.0 x 10' J, i) (I pt) is heat added to the system or is it removed from the system in going fromExplanation / Answer
a) Work done is equal to the area under P-V diagram. So looking at the diagram the area under curve AB is largest. Thus the work done during stage A to B the work done is largest.
During stage C to A the change in volume is zero which maks work done = 0. Thus the least work is done during this stage.
b) Work done during stage A to B is WAB = 5.6X105J
Work done during stage B to C is WBC = PBC(VC - VB) = 1.5X105(1m3 - 3m3) = -3X105J
Work done during stage C to A is WCA = PCA(VA - VC) = PCA(0) = 0 (as VA = VC)
So work done during a full cycle is WAB + WBC + WCA = 5.6X105J - 3X105J + 0 = 2X105J
c) (ii) From first law of thermodynamics we have
Q = U + W
where Q = heat added to the system (+ve when added and -ve when removed)
U = change in internal energy of the system
W = work done by the system
using this formula we have for the stage A to B
QAB = UAB + WAB
or QAB = 3X105J + 5.6X105J = 8.6X105J
So heat added to the system is = 8.6X105J
d) During a full cycle the change in internal enery is zero as the system's initial and final state in a cycle is same and internal energy is a state function.
This concludes the answers. Check the answer and let me know if it's correct. If you need anymore clarification I will be happy to oblige....