Answer
$\Delta E_{int}=900cal$
Work Step by Step
$\Delta E_{int}=nC_V\Delta T$
We plug in the known values to obtain:
$\Delta E_{int}=3.0(6.00)(50)=900cal$
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: 75g
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