College Physics (4th Edition)

Published by McGraw-Hill Education
ISBN 10: 0073512141
ISBN 13: 978-0-07351-214-3

Chapter 14 - Problems - Page 534: 27


The new temperature of the gas will be $~43.9~^{\circ}C$

Work Step by Step

We can find the number of moles of nitrogen: $PV = nRT$ $n = \frac{PV}{RT}$ $n = \frac{(3.5)(1.013\times 10^5~Pa)(425\times 10^{-3}~m^3)}{(8.314~J/mol~K)(296.15~K)}$ $n = 61.2~mol$ We can find the change in temperature: $Q = c~n~\Delta T$ $\Delta T = \frac{Q}{c~n}$ $\Delta T = \frac{26,600~J}{(20.8~J/mol~C^{\circ})(61.2~mol)}$ $\Delta T = 20.9~C^{\circ}$ The new temperature of the gas will be $23^{\circ}C+20.9~C^{\circ}$ which is $43.9~^{\circ}C$.
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