Chemistry 10th Edition

Published by Brooks/Cole Publishing Co.
ISBN 10: 1133610668
ISBN 13: 978-1-13361-066-3

Chapter 12 - Gases and the Kinetic-Molecular Theory - Exercises - Charles's Law: The Volume-Temperature Relationship - Page 440: 26


0.81 L

Work Step by Step

$V_{1}$ = 0.82 L $T_{1}$ = (26+273)K =299 K $T_{2}$ = (21+273)K =294 K According to Charles's Law, $V_{2}$ = $\frac{V_{1}\times T_{2}}{T_{1}}$ =$\frac{0.82L\times 294K}{299K}$ $\approx$ 0.81 L We can use 0$^{\circ}$C = 273.15 K for more precise results.
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.