ps02_sol

# 8 mole k dt 3226j 1373 q nh 21 800

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Unformatted text preview: p# dT = q1373 K &quot;2300 K = 1373 J \$ (2 moles)(44.8 mole % K )dT = 3,226J 1373 q&quot; #\$ = n%H&quot; #\$ = (2)(1, 800) = 3, 600 J We hav e 11,701 J of heat yet to use up. We determine the final temperat ure by integrating to us e up ! the ‘left over’ heat in the delt a phase: ! Tf q1409 K &quot;T f = \$ nC 1409 Tf # p dT = J \$ (2 moles)(47.3 mole % K )dT = 11, 701J 1409 94.6Tf &amp; 133, 291J = 11, 701J ' Tf = 1535K b. Because this proc ess occurs at constant pressure, we immediately know: ! &quot; dq % &quot; (H % C p = \$ rev ' = \$ ' # dT &amp; p # (T &amp; p dqrev = dH qrev = )H * )H = 100, 698 J c. We det ermine the total entropy change in a manner similar to the integrating procedure us ed in part (a): ! First, the appropriate relation to get us started is: 3.012 PS 2 2 of 6 10/4/05 &quot; dq % &quot; (S % C p = \$ rev ' = T\$ ' # dT &amp; p # (T &amp; p Cp dT T Tf C )S = * p dT Ti T dS = Note that the last equality and integral hold only for regions where no phase transition occurs. Using this ex pression, we integrat e over the temperatur...
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## This document was uploaded on 12/21/2013.

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