General, Organic, and Biological Chemistry: Structures of Life (5th Edition)

Published by Pearson
ISBN 10: 0321967461
ISBN 13: 978-0-32196-746-6

Chapter 8 - Gases - 8.3 Temperature and Volume (Charles's Law) - Questions and Problems - Page 299: 8.25

Answer

a. $303 ^{\circ} C$ b. $-129 ^{\circ} C$ c. $591 ^{\circ} C$ d. $136 ^{\circ} C$

Work Step by Step

$$\frac{T_1}{V_1} = \frac{T_2}{V_2} \longrightarrow \frac{T_1}{V_1}V_2 = T_2$$ $$T_2 = V_2\frac{T_1}{V_1}$$ $T_1/K = 15 + 273 = 288$ a. $$T_2 = (5.00 \space L)\frac{288 \space K}{2.50 \space L} = 576 \space K$$ $T_2/^{\circ}C = 576 - 273 = 303$ $T_2 = 303 ^{\circ} C$ b. $$T_2 = (1.25 \space L)\frac{288 \space K}{2.50 \space L} = 144 \space K$$ $T_2/^{\circ}C = 144 - 273 = -129$ $T_2 = -129 ^{\circ} C$ c. $$T_2 = (7.50 \space L)\frac{288 \space K}{2.50 \space L} = 864 \space K$$ $T_2/^{\circ}C = 864 - 273 = 591$ $T_2 = 591 ^{\circ} C$ d. $$T_2 = (3.55 \space L)\frac{288 \space K}{2.50 \space L} = 409 \space K$$ $T_2/^{\circ}C = 409 - 273 = 136$ $T_2 = 136 ^{\circ} C$
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