Chemistry 10th Edition

Published by Brooks/Cole Publishing Co.
ISBN 10: 1133610668
ISBN 13: 978-1-13361-066-3

Chapter 12 - Gases and the Kinetic-Molecular Theory - Exercises - Basic Ideas - Page 438: 5

Answer

(a) 675 mmHg (b) 0.888 atm (c) $9.00\times10^{4}\,Pa$ (d) 90.0 kPa

Work Step by Step

(a) $675\,torr= 675\,torr\times\frac{1\, mmHg}{1\,torr}$ $=675\, mmHg$ (b) $675\,torr= 675\,torr\times\frac{1\, atm}{760\,torr}$ $=0.888\, atm$ (c) $675\,torr=0.888\,atm$ $= 0.888\,atm\times\frac{1.01325\times10^{5}Pa}{1\,atm}=9.00\times10^{4}\,Pa$ (d) 675 torr= $9.00\times10^{4}\,Pa=90.0\times10^{3}Pa= 90.0\,kPa$
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