Fundamentals of Physics Extended (10th Edition)

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 607: 57

Answer

$\Delta S=3.59\frac{J}{K}$

Work Step by Step

We know that: $dQ=dE_{int}=\frac{3}{2}nRdT$ Now, $\Delta S=\int \frac{dQ}{T}$ Therefore; $\Delta S=\int _{T_i}^{T_f}\frac{\frac{3}{2}nRdT}{T} $ $\implies \Delta S=\frac{3}{2}nR\ln(\frac{T_f}{T_i})$ We plug in the known values to obtain: $\Delta S=\frac{3 }{2}(1.00)(8.31)\ln(\frac{400}{300})=3.59\frac{J}{K}$
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