College Physics (4th Edition)

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

Chapter 13 - Problems - Page 499: 53

Answer

The balloon would have a volume of $1313~m^3$

Work Step by Step

We can find an expression for the original volume: $P_1~V_1 = nRT_1$ $V_1 = \frac{nRT_1}{P_1}$ We can find an expression for the final volume: $P_2~V_2 = nRT_2$ $V_2 = \frac{nRT_2}{P_2}$ We can divide $V_2$ by $V_1$: $\frac{V_2}{V_1} = \frac{\frac{nRT_2}{P_2}}{\frac{nRT_1}{P_1}}$ $V_2 = \frac{P_1~T_2~V_1}{P_2~T_1}$ $V_2 = \frac{(1.00\times 10^5~N/m^2)(260~K)(5.0~m^3)}{(0.33\times 10^3~N/m^2)(300~K)}$ $V_2 = 1313~m^3$ The balloon would have a volume of $1313~m^3$.
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