## College Physics (4th Edition)

Published by McGraw-Hill Education

# Chapter 13 - Problems - Page 498: 44

#### Answer

The volume occupied outside the tire is $0.0614~m^3$

#### Work Step by Step

We can convert the gauge pressure of 36.0 psi to units of $Pa$: $P_g = 36.0~psi \times \frac{6895~Pa}{1~psi} = 248,000~Pa$ The absolute pressure initially: $P_1 = P_{atm}+P_g$ $P_1 = (1.01\times 10^5~Pa)+(2.48\times 10^5~Pa)$ $P_1 = 3.49\times 10^5~Pa$ We can find an expression for the original volume: $P_1~V_1 = nRT$ $V_1 = \frac{nRT}{P_1}$ We can find an expression for the final volume: $P_2~V_2 = nRT$ $V_2 = \frac{nRT}{P_2}$ We can divide $V_2$ by $V_1$: $\frac{V_2}{V_1} = \frac{\frac{nRT}{P_2}}{\frac{nRT}{P_1}}$ $V_2 = \frac{P_1}{P_2}~V_1$ $V_2 = \left(\frac{3.49\times 10^5~Pa}{1.01\times 10^5~Pa}\right)~(0.0250~m^3)$ $V_2 = 0.0864~m^3$ Since the volume inside the tube is still $0.0250~m^3$ the volume occupied outside the tire is $0.0864~m^3-0.0250~m^3$ which is $0.0614~m^3$

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