Essential University Physics: Volume 1 (3rd Edition)

Published by Pearson
ISBN 10: 0321993721
ISBN 13: 978-0-32199-372-4

Chapter 18 - Exercises and Problems: 28

Answer

$C_v = 16.25 $ $C_p = 24.56$

Work Step by Step

We know that there are 2.5 moles of $O_2$, which has an average kinetic energy of $2.5kT$. Thus, we find that its volume is: $ V_{O_2}=2.5\times2.5\times RT=6.25RT$ We know that there are 3 moles of $Ar$, which has an average kinetic energy of $1.5kT$. Thus, we find that its volume is: $ V_{Ar}=3\times1.5\times RT=4.5RT$ Thus, we find: $C_v = \frac{V_t}{nT} = \frac{6.25RT+4.5RT}{5.5T} = 16.25 $ $C_p = 16.25 + 8.314 = 24.56$
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