Chemistry: A Molecular Approach (3rd Edition)

Published by Prentice Hall
ISBN 10: 0321809246
ISBN 13: 978-0-32180-924-7

Chapter 5 - Sections 5.1-5.10 - Exercises - Problems by Topic - Page 239: 34

Answer

$V_2=1.15mL$

Work Step by Step

According to Charles law $\frac{V_1}{T_1}=\frac{V_2}{T_2}$ This can be rearranged as $V_2=\frac{V_1T_2}{T_1}$.............eq(1) $T_1=95.3+273.15=368.15 K$ $T_2=0.0+273.15=273.15K$ We plug in the known values in eq(1) to obtain: $V_2=\frac{1.55\times 273.15}{368.15}=1.15mL$
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