College Physics (7th Edition)

Published by Pearson
ISBN 10: 0-32160-183-1
ISBN 13: 978-0-32160-183-4

Chapter 10 - Temperature and Kinetic Theory - Learning Path Questions and Exercises - Exercises - Page 383: 32

Answer

$1.67atm$

Work Step by Step

At STP, T=273.15K, P=$1 atm = 1.01325\times10^{5}Pa$ By ideal gas law, PV=NRT PV/T is constant. So, $\frac{P1\times V1}{T1}=\frac{P2\times V2}{T2}$ or, $\frac{1\times 2.4}{273.15}=\frac{P2\times1.6}{303.15}$ P2=$1.67 atm$
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