## Chemistry 10th Edition

Published by Brooks/Cole Publishing Co.

# Chapter 12 - Gases and the Kinetic-Molecular Theory - Exercises - The Combined Gas Law - Page 441: 34

#### Answer

337$^{\circ}$C

#### Work Step by Step

$V_{1}$ = 385 mL $T_{1}$ = (26+273) K = 299 K $P_{1}$ = 670. torr $P_{2}$ = 940. torr $V_{2}$ = 560. mL Using combined gas law, $T_{2}$ =$\frac{P_{2}\times V_{2}\times T_{1}}{P_{1}\times V_{1}}$= $\frac{940. torr\times560. mL\times299K}{670.torr\times385 mL}$= 610. K or (610.-273)$^{\circ}$C

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