Answer
$Q_{31}=-3.22kJ$
Work Step by Step
We know that:
$Q_{31}=nC_p(T_1-T_3)$
We plug in the known values to obtain:
$Q_{31}=(1)(\frac{5}{2})(8.31)(300-455)=-3220.125=-3.22\times 10^3=-3.22kJ$
Fundamentals of Physics Extended (10th Edition)
by Halliday, David; Resnick, Robert; Walker, Jearl
Published by
Wiley
ISBN 10:
1-11823-072-8
ISBN 13:
978-1-11823-072-5
Chapter 19 - The Kinetic Theory of Gases - Problems - Page 581: 63g
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