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: 54

Answer

$\Delta S=4.46\frac{J}{K}$

Work Step by Step

We know that: $\Delta S=nR\ln(\frac{V_f}{V_i})+nC_V\ln(\frac{T_f}{T_i})$ As volume is constant, $V_f=V_i=V$ Hence: $nR\ln(\frac{V}{V})=nR\ln(1)=nR(0)=0$ $\implies \Delta S=0+nC_V\ln(\frac{T_f}{T_i})$ We plug in the known values to obtain: $\Delta S=3.20(\frac{3}{2}\times 8.314)\ln(\frac{425}{380})=4.46\frac{J}{K}$
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