Essential University Physics: Volume 1 (3rd Edition)

Published by Pearson
ISBN 10: 0321993721
ISBN 13: 978-0-32199-372-4

Chapter 19 - Exercises and Problems: 46

Answer

The proof is below.

Work Step by Step

We know the following: $dQ = nC_pdT$ 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_pdT}{T}$ $ S = \int_{T_1}^{T_2}\frac{ nC_pdT}{T}$ $S = nC_p ln (\frac{T_2}{T_1})$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.