University Physics with Modern Physics (14th Edition)

Published by Pearson
ISBN 10: 0321973615
ISBN 13: 978-0-32197-361-0

Chapter 17 - Temperature and Heat - Problems - Exercises - Page 574: 17.8

Answer

$444\ \mathrm{m}\mathrm{m}\mathrm{H}\mathrm{g}$

Work Step by Step

Use equation 17-5 $T=T_{\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{p}\mathrm{l}\mathrm{e}}\displaystyle \frac{p}{p_{\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{p}\mathrm{l}\mathrm{e}}}=(273.15)\frac{p}{p_{\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{p}\mathrm{l}\mathrm{e}}}$ Given: $P_{\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{p}\mathrm{l}\mathrm{e}}=325$ mm of mercury, Boiling point of water: $\mathrm{T}=373.15\mathrm{K}$ Solve 17-5 for p $p= (325 \mathrm{m}\mathrm{m}\mathrm{H}\mathrm{g}) (\displaystyle \frac{373.15\mathrm{K}}{273.15\mathrm{K}})=444 \mathrm{m}\mathrm{m}\mathrm{H}\mathrm{g}$
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.