Essential University Physics: Volume 1 (4th Edition) Clone

Published by Pearson
ISBN 10: 0-134-98855-8
ISBN 13: 978-0-13498-855-9

Chapter 18 - Exercises and Problems - Page 341: 16

Answer

$W = 1.5 p_1V_1$

Work Step by Step

The work done by the gas is given by: $ W = \int p dV$ Thus, the work is given by the area under the given graph. We will break it up into a triangle and a rectangle to find: $W=p_1(2v_1-v_1)+(.5)(2p_1-p_1)(2v_1-v_1)$ This gives: $W = 1.5 p_1V_1$
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