## Essential University Physics: Volume 1 (3rd Edition)

Published by Pearson

# Chapter 19 - Exercises and Problems - Page 350: 45

#### Answer

The proof is below.

#### Work Step by Step

We know the following: $dQ = nC_vdT$ We also can take the derivative of both sides of the equation $\Delta S = \frac{ Q}{T}$ to find : $d S = \frac{ dQ}{T}$ Combining these two equations, it follows: $dS = \frac{ nC_vdT}{T}$ $S = \int_{T_1}^{T_2}\frac{ nC_vdT}{T}$ $S = nC_v ln (\frac{T_2}{T_1})$

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