Introductory Chemistry (5th Edition)

Published by Pearson
ISBN 10: 032191029X
ISBN 13: 978-0-32191-029-5

Chapter 11 - Gases - Exercises - Problems - Page 402: 54

Answer

The volume will be $0.945 L$ or $945 mL$

Work Step by Step

Using the combined gas law with the involved quantities we can see that $\frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2}}$ Since the question asks for the final volume, we must solve for $V_{2}$. For this question we must also convert the two temperatures into the standard units of Kelvin by simply adding 273.15 to the Celsius values. $5^{\circ}C + 273.15 = 278.15 K$ $25^{\circ}C + 273.15 = 298.15 K$ Rearranging for $V_{2}$, we get: $\frac{P_{1}V_{1}T_{2}}{T_{1}P_{2}}=V_{2}.$ Therefore, $\frac{765 mm Hg \times 0.585 L \times 278.15 K}{298.15 K\times 442 mm Hg}=V_{2}.$ $V_{2} = 0.945 L$ $V_{2} = 0.945 L \times 1000 = 945 mL$
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