College Physics (4th Edition)

Published by McGraw-Hill Education
ISBN 10: 0073512141
ISBN 13: 978-0-07351-214-3

Chapter 10 - Problems - Page 399: 30


The volume decreases by $6.71\times 10^{-6}~m^3$

Work Step by Step

We can find the the change in volume: $\frac{\Delta V}{V_0} = -\frac{1}{B}~\Delta P$ $\Delta V = -\frac{1}{B}~\Delta P~V_0$ $\Delta V = -\left(\frac{1}{60\times 10^9~Pa}\right)~(1.75\times 10^6~Pa)(0.230~m^3)$ $\Delta V = -6.71\times 10^{-6}~m^3$ The volume decreases by $6.71\times 10^{-6}~m^3$.
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