Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 15 - The Laws of Thermodynamics - Problems - Page 439: 12

Answer

a) See line A in image b) $W=2550J$ $\Delta U=3825J$ c)See curve and line B in image d) $\Delta U=3825J$

Work Step by Step

a) See line A in image b) $W=P\Delta V=(425\frac{N}{m^2})(6.00m^3)=2550J$ $\Delta U=\frac{3}{2}nR\Delta T=\frac{3}{2}(PV_1-PV_2)=3825J$ c)See curve and line B in image d) The change in internal energy for isothermal reaction is 0 $\Delta U$is only dependent on $\Delta T$, so it is the same as b) 2. $\Delta U=3825J$
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