University Physics with Modern Physics (14th Edition)

Published by Pearson
ISBN 10: 0321973615
ISBN 13: 978-0-32197-361-0

Chapter 18 - Thermal Properties of Matter - Problems - Exercises - Page 610: 18.4

Answer

(a) $-175^{\circ}C$ (b) $1 \ L$

Work Step by Step

For (a): From the ideal gas law, $pV = nRT$. In this case $n, R,$ and $V$ are constant. So we have $\frac{p_1}{T_1} = \frac{p_2}{T_2}$. $\implies T_2 = T_1(\frac{p_2}{p_1}) = (293)(\frac{1}{3}) = 97.7 K = -175^{\circ}C$. For (b): $p_2 = 1 \ atm, V_2 = 3 \ L, p_3 = 3 \ atm$. In this case, $nRT$ is constant, so we have $p_2 V_2 = p_3 V_3$. $\implies V_3 = V_2(\frac{p_2}{p_3}) = (3)(\frac{1}{3}) =1 \ L$.
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