Chemistry: The Central Science (13th Edition)

Published by Prentice Hall
ISBN 10: 0321910419
ISBN 13: 978-0-32191-041-7

Chapter 5 - Thermochemistry - Exercises: 5.38

Answer

$\Delta E = 276 J$ $\Delta H = 490J$

Work Step by Step

First off we need to find $\Delta E$ and $\Delta E = q + w$. We are told that 0.49kJ of heat is added to the gas. Since the heat is added, we are adding energy to the system so the sign of q = + 0.49kJ = 490 J. Next we are told that the gas expands and does 214J of work on the surroundings. This means enegy is leaving the system so we have w = -214J. We also know that $w = -P\Delta V$ so then $P\Delta V = 214J$, a fact that we will use later. Now we have $\Delta E$ = 490J + (-214J) = 276J Next we have $\Delta H = \Delta E + P\Delta V$ We found that $P\Delta V = 214J$ so then $\Delta H = 276J + 214J = 490 J$
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