Responde Aí

Enunciado

Vapor d'água muda de um estado inicial a 450 ∘ C e 3.000 k P a para um estado final a 140 ∘ C e 235 k P a . Determine Δ H e Δ S :

(a) Com dados da tabela de vapor.(b) Com equações para um gás ideal.(c) Com correlações generalizadas apropriadas.

Passo 1

(a) Tab e la  F.2,  3000 k P a  e   450 d e g ⁡ C   :

H 1 : = 3344.6 ⋅ J g m S 1 : = 7.0854 ⋅ J g m ⋅ K

Tabela F.2, interpole 235 k P a e 140   d e g C :

H 2 : = 2744.5 ⋅ J g m S 2 : = 7.2003 ⋅ J g m ⋅ K

Δ H : = H 2 - H 1 Δ H = - 600.1 J g m

Δ S : = S 2 - S 1 Δ S = 0.115 J g m ⋅ K

Passo 2

(b) T 1 : = ( 450 + 273.15 ) ⋅ K T 2 : = ( 140 + 273.15 ) ⋅ K

T 1 = 723.15 K T 2 = 413.15 K

P 1 : = 3000 ⋅ k P a P 2 : = 235 ⋅ k P a

Eqs . (6.95) & (6.96) para um gás ideal :

molwt   : = 18 g m m o l

Δ H i g : = R ⋅ I C P H T 1 , T 2 , 3.470,1.450 ⋅ 10 - 3 , 0.0,0.121 ⋅ 10 5 ) m o l w t

Δ S i g : = R ⋅ ICPS ⁡ T 1 , T 2 , 3.470,1.450 ⋅ 10 - 3 , 0.0,0.121 ⋅ 10 5 ) - l n ⁡ P 2 P 1  molwt 

Δ H i g = - 620.6 J g m Δ S i g = 0.0605 J g m ⋅ K

Passo 3

(c) T c : = 647.1 ⋅ K P c : = 220.55 ⋅ b a r ω : = 0.345

T r 1 : = T 1 T c P r 1 : = P 1 P c T r 2 : = T 2 T c P r 2 : = P 2 P c

T r 1 = 1.11752 P r 1 = 0.13602 T r 2 = 0.63846 P r 2 = 0.01066

A correlação generalizada do coeficiente virial é adequada aqui

Δ H : = Δ H i g + R ⋅ T c ⋅ H R B ) ( T r 2 , B r 2 , ) - ( H R B ) ( T r 1 , d r 1 m o l w t

Δ H = - 593.95 J g m

Δ S : = Δ S i g + R ⋅ δ R B ) ( T r 2 , B r 2 ) , - ( S R B ) T r 1 , B r 1 m o l w t

Δ S = 0.078 J g m ⋅ K

Resposta

( a ) Δ H = - 600.1 J g m

Δ S = 0.115 J g m ⋅ K

(b)   Δ H i g = - 620.6 J g m

Δ S i g = 0.0605 J g m ⋅ K

(c)   Δ H = - 593.95 J g m

Δ S = 0.078 J g m ⋅ K