## College Physics (4th Edition)

Published by McGraw-Hill Education

# Chapter 13 - Problems - Page 498: 47

#### Answer

The lungs would expand by a factor of 1.50

#### Work Step by Step

We can find the gauge pressure at a depth of 5.0 meters: $P_g = \rho~g~h = (1030~kg/m^3)(9.80~m/s^2)(5.0~m) = 50,470~Pa$ We can find the absolute pressure initially: $P_1 = P_{atm}+P_g$ $P_1 = (1.01\times 10^5~Pa)+(50,470~Pa)$ $P_1 = 1.5147 \times 10^5~Pa$ 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 $V_1$: $\frac{V_2}{V_1} = \frac{\frac{nRT}{P_2}}{\frac{nRT}{P_1}}$ $V_2 = \frac{P_1}{P_2}~V_1$ $V_2 = \left(\frac{1.5147\times 10^5~Pa}{1.01\times 10^5~Pa}\right)~V_1$ $V_2 = 1.50~V_1$ The lungs would expand by a factor of 1.50

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