College Physics (4th Edition)

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

Chapter 10 - Problems - Page 399: 26


The volume decreases by $6.44\times 10^{-5}~cm^3$

Work Step by Step

We can find the pressure difference $\Delta P$: $\Delta P = (9.12\times 10^6~Pa)-(1.01\times 10^5~Pa)$ $\Delta P = 9.02\times 10^6~Pa$ We can find the fraction that the volume changes: $\frac{\Delta V}{V_0} = -\frac{1}{B}~\Delta P$ $\frac{\Delta V}{V_0} = -\frac{1}{140\times 10^9~Pa}~(9.02\times 10^6~Pa)$ $\frac{\Delta V}{V_0} = -6.44\times 10^{-5}$ We can find the decrease in volume: $\Delta V = (-6.44\times 10^{-5})(1.0~cm^3) = -6.44\times 10^{-5}~cm^3$ The volume decreases by $6.44\times 10^{-5}~cm^3$.
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