College Physics (4th Edition)

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

Chapter 10 - Problems - Page 399: 27


The volume would increase by $1.44\times 10^{-6}~cm^3$

Work Step by Step

We can find the pressure difference $\Delta P$: $\Delta P = (10^{-9}~Pa)-(1.01\times 10^5~Pa)$ $\Delta P = -1.01\times 10^5~Pa$ We can find the the change in volume: $\frac{\Delta V}{V_0} = -\frac{1}{B}~\Delta P$ $\Delta V = -\frac{1}{B}~\Delta P~V_0$ $\Delta V = -\left(\frac{1}{70\times 10^9~Pa}\right)~(-1.01\times 10^5~Pa)(1.00~cm^3)$ $\Delta V = 1.44\times 10^{-6}~cm^3$ The volume would increase by $1.44\times 10^{-6}~cm^3$.
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