Answer
$\frac{Q}{p_1V_1} = -0.693$
Work Step by Step
Step 1 : Isothermal expansion to pressure $2.00p_1$
Since $ln(\frac{1}{2.00}) = -ln(2.00)$, this leads to
$Q = -p_1V_1ln(2.00)$
$\frac{Q}{p_1V_1} = -ln(2.00)$
$\frac{Q}{p_1V_1} = -0.693$
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 20 - Entropy and the Second Law of Thermodynamics - Problems - Page 605: 19g
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