6.67 - th Solutions to Chemical and Engineering...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 ...
View Full Document

Ask a homework question - tutors are online