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 8.4 - Temperature and Pressure (Gay-Lussac's Law) - Questions and Problems - Page 267: 8.31a

Answer

The final temperature is equal to $-23 \space C^o$.

Work Step by Step

1. The temperatures used in gas law calculations must be converted to Kelvin values. $C^o + 273 = K$ $25 + 273 = K$ $K = 298$ Therefore: $T_1 = 298 \space K$ 2. Write the Gay-Lussac's law, and rearrange it to solve for $T_2$, which is the final temperature. $\frac{P_1}{T_1} = \frac{P_2}{T_2}$ - Invert both fractions: $\frac{T_1}{P_1} = \frac{T_2}{P_2}$ - Multiply both sides by $P_2$: $\frac{T_1}{P_1} \times P_2 = T_2$ 3. Substitute the values and find the $T_2$ value: $\frac{298 \space K}{740. \space mmHg} \times 620. \space mmHg = T_2$ $T_2 = 250 \space K$ 4. Conver the final temperature to degrees Celsius: $C^o + 273 = K$ $C^o = K - 273$ $C^o = 250 - 273$ $C^o = -23$ $T_2 = -23 \space C^o$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.