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 - General Problems - Page 388: 72

Answer

$F=8200N$

Work Step by Step

$P=\frac{nRT}{V}$ $\frac{P_i}{P_f}=\frac{\frac{nRT}{V}}{\frac{nRT}{V}}=\frac{T_i}{T_f}=\frac{15^oC+273K}{165^oC+273K}=0.658$ $P_f=\frac{1 atm}{0.658}=1.52 atm$ $A=(^3\sqrt{6.15\times10^{-2}m^3})^2=0.156m^2$ $F_i=AP=(1.52 atm\frac{101325\frac{N}{m^2}}{1 atm})(0.156m^2)=24000N$ $F_t=F_i-F_o=24000N-(101325\frac{N}{m^2})(0.156m^2)=8200N$
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