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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. (6.417a)
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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 '08
 SHING

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