Question

lem 15.39: Work done in a cyclic process. 7 of 8 I Review Figure 1), consider the closed loop 1-2-+3-+4+1. s is a cyclic process in which the initial and final states the same. Take the graduation Po-0.50 atm and graduation Vo = 0.10 L. Part A Find the total work done by the system in this process, and show that it is equal to the area enclosed by the loop. Express your answer in joules to two significant figuren. να ΑΣφ W- Previous Answers Request Answer Submit < 1 of 1 ounE X Incorrect; Try Again; 5 attempts remaining p (atm) Part B How is the work done during the process in Part Arelated to the work done if the loop is traversed in the opposite direction, 143-2-17 Express your answer in joules to two significant figures. P. V (L) ν ΑΣφ

Answer 1

Given
   P0 = 0.50 atm , V0 = 0.10 L
work done in the loop 1-->2-->3-->4--->1 is

We know that the work done W = p*dV   

Part A

from W(1-->2) W1= P(V2-V1) = 5*0.50(8*0.10-2*0.10) = 1.5 J

from 2--->3, W2 = P(V3-V2) = P(0 ) = 0 J

from W(3-->4) W3 = P(V4-V2) = 0.50(2*0.10-8*0.10) = -0.3 J

from 4--->1, W4 = P(V1-V4) = P(0 ) = 0 J

so the net work done is W = W1+W2+W3+W4 = 1.5 + O -0.3 +0 = 1.2 J
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Now area under the curve (loop 1-->2-->3-->4--->1 is )

   A1 that is from 1-->2 = (6*0.1)(5*0.5) = 1.5 J
   A2 that is from 3-->4 = (6*0.1)(1*0.5) = 0.3 J
total area is A = A1-A2 = 1.5 -0.3 = 1.2 J

so the work done = area under the cure

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part B

as we proceed in reverse direction so the work done will be negative that is -1.2 J

Part D

work done in the path 3-->4-->2-->3

From 3--> 4, W1 = Area under the curve = 6*0.1*0.5 = 0.3 J
From 4--> 2, W2 = Area under the curve= 0.5*6*0.1*(4*0.5) = 0.6 J
from 2-->3, W3 = 0 J
total work done W = W1+W2+W3 = 0.3+ 0.6+0 = 0.9 J