Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 16 - A Macroscopic Description of Matter - Exercises and Problems - Page 464: 25


$p_{2}$ = 2.6 atm

Work Step by Step

We know that a rigid container holds Hydrogen gas. Therefore $V_{2}$ = $V_{1}$ . We can use the ideal gas law $\frac{p_{1}V_{1}}{T_{1}}$ =$\frac{p_{2}V_{2}}{T_{2}}$, where $p_{1}$ is the initial pressure, $p_{2}$ is the final pressure, $T_{1}$ is the initial temperature, $T_{2}$ is the final temperature, $V_{1}$ is the initial volume, and $V_{2}$ is the final volume. To find the final pressure( $p_{2}$), we are given : $p_{1}$ - 3.0 atm $T_{1}$ - 20$^{\circ}$C which must be converted to Kelvins(K) = 293K $T_{2}$ = -20$^{\circ}$C = 253K We convert to Kelvins by adding 273 to the Temperature in Celsius: $T_{K}$ = 20$^{\circ}$C + 273 = 293K. Now we are ready to solve for $p_{2}$ So $p_{2} = \frac{p_{1}V_{1}T_{2}}{V_{2}T_{1}}$ = $\frac{p_{1}V_{1}T_{2}}{V_{1}T_{1}}$ = (3 atm) $\frac{253}{293}$ = 2.6 atm
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