Essential University Physics: Volume 1 (4th Edition) Clone

Published by Pearson
ISBN 10: 0-134-98855-8
ISBN 13: 978-0-13498-855-9

Chapter 18 - Exercises and Problems - Page 343: 61

Answer

334 K

Work Step by Step

We find: $\frac{T_1^{\gamma}}{p_1^{\gamma-1}}=\frac{T_2^{\gamma}}{p_2^{\gamma-1}}$ Simplifying this equation, it follows: $T_2=\sqrt[1.3] {\frac{T_1^{\gamma}(p_2^{\gamma-1})}{p_1^{\gamma-1}}}$ $T_2=\sqrt[1.3] {\frac{273^{1.3}(240^{.3})}{100^{.3}}}=334K$
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.