Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 17 - Work, Heat, and the First Law of Thermodynamics - Exercises and Problems - Page 497: 3


$V_1 = 200~cm^3$

Work Step by Step

We can find $V_1$ as: $W = -P\Delta V$ $W = -P ~(V_f-V_i)$ $W = -P ~(V_1-3V_1)$ $W = 2P ~V_1$ $V_1 = \frac{W}{2P}$ $V_1 = \frac{80~J}{(2)(2.00\times 10^5~Pa)}$ $V_1 = 2.0\times 10^{-4}~m^3$ $V_1 = (2.0\times 10^{-4}~m^3)(\frac{10^6~cm^3}{1~m^3})$ $V_1 = 200~cm^3$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.