College Physics (4th Edition)

Published by McGraw-Hill Education
ISBN 10: 0073512141
ISBN 13: 978-0-07351-214-3

Chapter 9 - Problems - Page 362: 30


(a) $2.21\times 10^5~Pa$ (b) $1650~torr$ (c) $2.18~atm$

Work Step by Step

(a) We can convert $32~lb/in^2$ to units of Pa: $32~lb/in^2 \times \frac{6895~Pa}{1~lb/in^2} = 2.21\times 10^5~Pa$ (b) We can convert $32~lb/in^2$ to units of torr: $32~lb/in^2 \times \frac{51.7~torr}{1~lb/in^2} = 1650~torr$ (c) We can convert $32~lb/in^2$ to units of atm: $32~lb/in^2 \times \frac{1~atm}{14.7~lb/in^2} = 2.18~atm$
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