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03 - Solutions

# 03 - Solutions - Solutions for Homework 2 Chapter 19...

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Solutions for Homework 2 Chapter 19 Problems 17. Find the exhaust temperature from the original Carnot efficiency, and then recalculate the intake temperature for the new Carnot efficiency, using the same exhaust temperature. e 1 = 1 ! T L T H1 " T L = T H1 1 ! e ( ) = 580K + 273K ( ) 1 ! 0.32 ( ) = 580.0K e 2 = 1 ! T L T H2 " T H2 = T L 1 ! e 2 = 580.0 K 1 ! 0.38 = 936 K = 663 o C # 660 o C 20. ( a ) We use the ideal gas law and the adiabatic process relationship to find the values of the pressure and volume at each of the four points. P a = 8.8atm ; T a = 623K ; V a = nRT a P a = 1.00mol ( ) 0.0821L & atm mol & K ( ) 623K ( ) 8.8atm = 5.81L ! 5.8L T b = 623K ; V b = 2 V a = 2 5.81L ( ) = 11.62L ! 11.6L P b = P a V a V b = 1 2 P a = 4.4atm T c = 483K ; P b V b ! = P c V c " nRT b V b V b = nRT c V c V c " V c = V b T b T c # \$ % & ( 1 ) 1 = ( ) 623K 483K # \$ % & ( 3/2 = 17.02L * 17.0L P c = nRT c V c = 1.00mol ( ) 0.0821L & atm mol & K ( ) 483K ( ) = 2.33atm * 2.3atm T d = 483K ; V d = V a T a T d ! " # \$ % & 1 ( 1 = 5.81L ( ) 623K 483K ! " # \$ % & = 8.51L ) 8.5L P d = nRT d V d = 1.00mol ( ) 0.0821L & atm mol & K ( ) 483K ( ) 8.51L = 4.66atm ) 4.7atm To summarize: P a = 8.8atm ; V a = ; P b = 4.4atm ; V b = P c = 2.3atm ; V c = ; P d = 4.7atm ; V d = ( b ) Isotherm ab: ! E int ab = 0 ;

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Q ab = W ab = nRT a ln V b V a = 1.00mol ( ) 8.314J mol & K ( ) 623K ( ) ln2 = 3590J ! 3600J Adiabat bc: Q bc = 0 ; ! E int bc = nC V T c " T b ( ) = 3 2 nR T c " T b ( ) = 3 2 1.00mol ( ) 8.314J mol & K ( ) " 140K ( ) = " 1746J # " 1700J ; W bc = Q bc " ! E int bc = # Isotherm cd: ! E int cd = 0 ; Q cd = W cd = nRT c ln V d V c = 1.00mol ( ) 8.314J mol & K ( ) 483K ( ) ln 1 2 = ! 2783J " ! 2800J Adiabat da: Q da = 0 ; ! E int bc = nC V T c " T b ( ) = 3 2 nR T c " T b ( ) = 3 2 1.00mol ( ) 8.314J mol & K ( ) ( ) = # ; W bc = Q bc " ! E int bc = # " To summarize: ab: ! E int = 0 ; Q = 3600J ; W = bc: ! E int = " 1700J ; Q = 0 ; W = cd: ! E int = 0 ; Q = " 2800J ; W = " da: ! E int = 1700J ; Q = 0 ; W = " ( c ) Using Eq. 20-1: e = W Q input = + ! 2783J !
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03 - Solutions - Solutions for Homework 2 Chapter 19...

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