Answer
$10^{-15}Pa$
Work Step by Step
We can find the required pressure as follows:
$P=\frac{nRT}{V}$
WE plug in the known values to obtain:
$P=\frac{(10^6)(8.314\times 10^8)(100)}{6.024\times 10^{24}}$
$P=1.38\times 10^{-16}atm=10^{-15}Pa$
Physics Technology Update (4th Edition)
by Walker, James S.
Published by
Pearson
ISBN 10:
0-32190-308-0
ISBN 13:
978-0-32190-308-2
Chapter 17 - Phases and Phase Changes - Problems and Conceptual Exercises - Page 603: 11
Update this answer!
You can help us out by revising, improving and updating this answer.
Update this answerAfter 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.
