6.67 - th Solutions to Chemical and Engineering...

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Unformatted text preview: th Solutions to 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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