Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 13 - Temperature and Kinetic Theory - Search and Learn - Page 389: 2

Answer

a) $2034\;\rm L$ b) $85\;\rm min$ c) $26\;\rm min$

Work Step by Step

a) We can assume that air is an ideal gas. The ideal gas law is given by $$PV=nRT$$ The author here asks about the change of air volume when we change pressure while the temperature remains constant. So, $$P_1V_1=P_2 V_2$$ Solving for $V_2$; $$V_2=\dfrac{P_1V_1}{P_2 }$$ Plugging the given; $$V_2=\dfrac{180\cdot 11.3}{1}=\boxed{\bf 2034\;\rm L}$$ b) The volume of air consumes in one minute $$V_{min}=2 \cdot 12 =24\;\rm L$$ The time he takes him to finish the tank is $$\Delta t=\dfrac{V_2}{V_{min}}=\dfrac{2034}{24}$$ $$\Delta t\approx \boxed{\bf 85\;\rm min}$$ c) We need to find the volume of air in this new situation. The tank is 23 m below the water surface which means that the pressure increased and the temperature decreased to 10$^\circ\;\rm C$. Using the same ideal gas law we used above, first, we need to find the new pressure. $$P_2=P_a+h\rho_w g\tag 1$$ whereas $P_a$ is atmospheric pressure and $\rho_w$ is the density of the sea water. From the ideal gas law; $$\dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}$$ Solving for $V_2$; $$V_2=\dfrac{P_1V_1T_2}{P_2T_1} $$ Plugging $P_2$ from (1) and then plugging the given and recall to convert pressures and temperatures to SI units. $$V_2=\dfrac{180\cdot 1.013\times10^5\cdot11.3\cdot 291 }{\left([1.013\times10^5]+[23\cdot 1030\cdot 9.8]\right)\cdot 283} =\;\rm \bf 635.4\;L$$ The time he takes him to finish the tank is $$\Delta t=\dfrac{V_2}{V_{min}}=\dfrac{635.4}{24}$$ $$\Delta t\approx \boxed{\bf 26\;\rm min}$$
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