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 8.5 - The Combined Gas Law - Questions and Problems - Page 269: 8.34b

Answer

The final volume of the gas is: $1450 \space mL$.

Work Step by Step

1. The temperatures used in gas law calculations must be converted to Kelvin values. $C^o + 273 = K$ $112 + 273 = K$ $K = 385$ Therefore: $T_1 = 385 \space K$ $C^o + 273 = K$ $75 + 273 = K$ $K = 348 $ Therefore: $T_1 = 348 \space K$ 2. Write the combined gas law, and rearrange it to solve for $V_2$, which is the final volume. $\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$ - Divide both sides by $P_2$: $\frac{P_1V_1}{T_1P_2} = \frac{V_2}{T_2}$ - Multiply both sides by $T_2$: $\frac{P_1V_1T_2}{T_1P_2} = V_2$ 3. Substitute the values and find the $V_2$ value: $\frac{1.20 \space atm \times 735 \space mL \times 348 \space K}{385 \space K \times 0.55 \space atm } = V_2$ $V_2 = 1450 \space mL$
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