Chemistry: An Introduction to General, Organic, and Biological Chemistry (12th Edition)

Published by Prentice Hall
ISBN 10: 0321908449
ISBN 13: 978-0-32190-844-5

Chapter 8 - Section 8.3 - Temperature and Volume (Charles's Law) - Questions and Problems - Page 265: 8.26a

Answer

102.375 K = -170.625 Degree Celsius

Work Step by Step

Let $T_{1} = 0^{o}C + 273 K = 273 K$, $V_{1} = 4 L$, $V_{2} = 1.5 L$, $T_{2} = ?$ Using Charles's Law, we have: $\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}$ $\frac{4 L}{273 K} = \frac{1.5 L}{T_{2}}$ $T_{2} = \frac{1.5 L}{4 L}$ x 273K = 102.375 K
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