Introductory Chemistry (5th Edition)

Published by Pearson
ISBN 10: 032191029X
ISBN 13: 978-0-32191-029-5

Chapter 11 - Gases - Exercises - Problems - Page 402: 53


$758 mL$ is the volume of the balloon at a depth of 25 m underwater.

Work Step by Step

Using the combined gas law with the involved quantities we can see that $\frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2}}$ Since the question asks for the final volume, we must solve for $V_{2}$. For this question we must also convert the two temperatures into the standard units of Kelvin by simply adding 273.15 to the Celsius values. $18^{\circ}C + 273.15 = 291.15 K$ $34^{\circ}C + 273.15 = 307.15 K$ Rearranging for $V_{2}$, we get: $\frac{P_{1}V_{1}T_{2}}{T_{1}P_{2}}=V_{2}.$ Therefore, $\frac{1.0 atm\times 2.8 L \times 291.15 K}{307.15 K\times 3.5 atm}=V_{2}.$ $V_{2} = 0.758 L$ $V_{2} = 0.758 L \times 1000 = 758 mL$
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.