Physics (10th Edition)

Published by Wiley
ISBN 10: 1118486897
ISBN 13: 978-1-11848-689-4

Chapter 14 - The Ideal Gas Law and Kinetic Theory - Problems - Page 385: 32

Answer

440 K

Work Step by Step

The buoyant force is equal to the total weight of the balloon, at the instant just before the balloon lifts off. Also according to Archimedes' principle, the buoyant force is the weight of the displaced outside air. So, we can write, $$\rho Vg=m_{t}g-(1)$$ We know that, Mass of the air = number of moles $\times$ Molecular mass $m_{air}=nM=(\frac{PV}{RT})M-(2)$ (2)=>(1), $$\rho V=m_{load}+m_{air}=m_{load}+(\frac{PV}{RT})M$$ $T=\frac{PVM}{(\rho V-m_{load})R}$ ; Let's plug known values into this equation. $T=\frac{(1.01\times10^{5}Pa)(650\space m^{3})(29\times10^{-3}kg/mol)}{[(1.29\space kg/m^{3})(650\space m^{3})-320\space kg](8.31\space J/mol\space K)}=440\space K$ The temperature of the air = 440 K
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