Answer
$Q_{12}=3.74kJ$
Work Step by Step
We can find heat for $1 \rightarrow 2$ as:
$Q_{12}=nC_v(T_1-T_2)$
We plug in the known values to obtain:
$Q_{12}=(1)(\frac{3}{2})(8.31)(600-300)=3739.5J=3.74kJ$
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: 63a
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