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 - Page 512: 36


(a) This is an isobaric process. (b) The final temperature is $118^{\circ}C$ (c) There are $6.23\times 10^{-3}$ moles of gas.

Work Step by Step

(a) The pressure remains constant so this is an isobaric process. (b) We can convert the initial temperature to Kelvin. $K = C + 273$ $K = 900 + 273$ $K = 1173$ The initial temperature is 1173 K. We can find an expression for the initial temperature. $PV_1 = nRT_1$ $T_1 = \frac{PV_1}{nR}$ We can find the final temperature. $PV_2 = nRT_2$ $T_2 = \frac{PV_2}{nR}$ $T_2 = \frac{P(V_1/3)}{nR}$ $T_2 = \frac{1}{3}~\frac{PV_1}{nR}$ $T_2 = \frac{T_1}{3}$ $T_2 = \frac{1173~K}{3}$ $T_2 = 391~K$ We can convert the final temperature to Celsius. $C = K - 273$ $C = 391-273$ $C = 118$ The final temperature is $118^{\circ}C$ (c) We can use the initial conditions to find the number of moles of gas. $PV = nRT$ $n = \frac{PV}{RT}$ $n = \frac{(2)(1.013\times 10^5~Pa)(300\times 10^{-6}~m^3)}{(8.314~m^3~Pa/mol~K)(1173~K)}$ $n = 6.23\times 10^{-3}~mol$ There are $6.23\times 10^{-3}$ moles of gas.
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