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### 6.67

Course: CHE 330, Fall 2011
School: USC
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Word Count: 159

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to th Solutions Chemical and Engineering Thermodynamics, 4 ed 1023.5 = 27.93 (T 420 ) + So that the equation to be solved is ( 0.093 2 T 4202 2 ) The solution to this equation is T = 404.56 K so that the temperature change is -15.44 K 6.67 P( T) := R T V + R T B( T) Cv( T) := 27.93 + 0.093 T 8.314 2 V d B( T) R B( T) dT d P( T) + R + R T V 2 2 dT V V 5 R := 8.314 10 d B( T) B( T) dT d P(...

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to th Solutions Chemical and Engineering Thermodynamics, 4 ed 1023.5 = 27.93 (T 420 ) + So that the equation to be solved is ( 0.093 2 T 4202 2 ) The solution to this equation is T = 404.56 K so that the temperature change is -15.44 K 6.67 P( T) := R T V + R T B( T) Cv( T) := 27.93 + 0.093 T 8.314 2 V d B( T) R B( T) dT d P( T) + R + R T V 2 2 dT V V 5 R := 8.314 10 d B( T) B( T) dT d P( T) dV R ln( V) R R T dT V V B( T) := 297.6 256100 T DB( T) := B( T) + T 6 10 d B( T) dT d B( T) dT 2561 10000 T 2 4 DB( = 420) 2.976 10 This combination is a constant 4 B( 420) = 3.122 10 Solving for volumes RTP1 := R 420 ( 2 RTP1 + RTP1 4 RTP1 B( 420) V1 := 10 )0.5 3 V1 = 3.78 10 2 RTP2 := R 404.56 ( 2 RTP2 + RTP2 4 RTP2 B( 404.56) V2 := 0.1 )0.5 2 Using Eq. (6.4-17a) V2 V1 Term1 := 8.314 ln Term2 := Term3 := Term1 = 37.325 DB( 420) 8.314 Term2 = 0.655 V1 DB( 404.56 8.314 ) 3 Term3 = 7.349 10 V2 Term4 := 404.56 Cv( T) T Term4 = 2.171 dT 420 DelS := Term1 + Term2 Term3 + Term4 J DelS = 35.801 mol K V2 = 0.337
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x(EA)00.10.20.30.40.50.60.70.80.91x(B)10.90.80.70.60.50.40.30.20.10Pvap(EA)0.23210.23210.23210.23210.23210.23210.23210.23210.23210.23210.2321Pvap(B)0.29390.29390.29390.29390.29390.29390.29390.29390.29390.2939
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x1x2100.10.20.30.40.50.60.70.80.9110.90.80.70.60.50.40.30.20.10Py1y21.92236108301.904395812 0.0915097781.886430542 0.1847625281.868465271 0.2798085261.85050.37671.832534729 0.4754912271.814569458 0.5762386341.796604188
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