() 33 10Pa 40Pa Work 1.0m 1.0m 02 + =− = (a) and (b) Thus, the total work during the BC cycle is (75 – 30) J = 45 J. During the BA cycle, the “tilted” part is the same as before, and the main difference is that the horizontal portion is at higher pressure, with Work = (40 Pa)(–3.0 m 3) = –120 J. Therefore, the total work during the BAcycle is (75 – 120) J = – 45 J. 93. The net work may be computed as a sum of works (for the individual processes involved) or as the “area” (with appropriate ±sign) inside the figure (representing the cycle). In this solution, we take the former approach (sum over the processes) and will need the following fact related to processes represented in pV
This is the end of the preview. Sign up
access the rest of the document.
This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.