Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 18 - A Macroscopic Description of Matter - Exercises and Problems: 35

Answer

(a) This is an isochoric process. (b) The initial temperature is $1555^{\circ}C$ The final temperature is $336^{\circ}C$

Work Step by Step

(a) Since the volume does not change, this is an isochoric process. (b) We can find the initial temperature as: $P_1V = nRT_1$ $T_1 = \frac{P_1V}{nR}$ $T_1 = \frac{(3)(1.013\times 10^5~Pa)(200\times 10^{-6}~m^3)}{(0.0040~mol)(8.314~m^3~Pa/mol~K)}$ $T_1 = 1828~K$ We then convert this temperature to Celsius: $C = K - 273$ $C = 1828-273$ $C = 1555$ The initial temperature is $1555^{\circ}C$. Next, we find the final temperature: $P_2V = nRT_2$ $T_2 = \frac{P_2V}{nR}$ $T_2 = \frac{(1)(1.013\times 10^5~Pa)(200\times 10^{-6}~m^3)}{(0.0040~mol)(8.314~m^3~Pa/mol~K)}$ $T_2 = 609~K$ We can convert this temperature to Celsius. $C = K - 273$ $C = 609-273$ $C = 336$ The final temperature is $336^{\circ}C$.
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