O PROBLEMA: Durante uma parada numa planta cloro química, um vazamento de cloro líquido é detectado. Gotículas, que podem ser consideradas esféricas, de cloro são lançadas na atmosfera. Considerando o diâmetro das gotículas igual a
0,2 \mathrm{~cm} , sendo a Pressão atmosférica de 1 atm e temperatura ambiente de
25^{\circ} \mathrm{C} , nestas condições a pressão de vapor do cloro é de
27,48 \mathrm{mmHg} e a densidade
1,05 \mathrm{~g} / \mathrm{cm}^{3} , Massa Molar do
\mathrm{Cl}_{2}=71 \mathrm{~g} / \mathrm{mol} . Determine:
a) Qual a difusividade do cloro no ar utilizando a equação de Champman?
b) Qual a taxa de difusão em g/dia?
c) Em quanto tempo a gotícula se dissipará completamente?
Tabular parse error: Open and close tags mismatch!,
\begin{tabular}{|l|l|}
\hline \begin{tabular}{l}
A equação de Chapman-Enkosg:
\[D_{A B}=\frac{1,858 \times 10^{-3} T^{3 / 2}\left[\frac{1}{M_{A}}+\frac{1}{M_{B}}\right]^{1 / 2}}{P \sigma_{A B}^{2} \Omega_{D}}\] \\
onde: \(D_{A B}\left(\mathrm{~cm}^{2} / \mathrm{s}\right)\) \\
\(\mathbf{M}_{\mathbf{A}} \boldsymbol{e} \mathbf{M}_{\mathbf{B}} \boldsymbol{=}\) pesos moleculares \\
\(\mathbf{P}=\) pressão absoluta (atm) \\
\(\sigma_{A B}=\) diâmetro de colisão, parâmetro de Leonard-Jones (A) \\
\(\Omega_{D}=\) integral de colisão
\end{tabular} & \[\sigma_{A B}=\frac{\sigma_{A}+\sigma_{B}}{2}\]
\varepsilon_{\mathrm{AB}}=\sqrt{\varepsilon_{\mathrm{A}} \varepsilon_{\mathrm{B}}} \
\hline
\end{tabular}
Table E. 1
Potential Parameters and Critica
Table E. 2 Collision Integrals for Use with the Lennard-Jones (6-12) Pot Prediction of Transport Properties of Gases at Low Densities
{ }^{2 / 3 / 4}
Lennard-Jones parameters
\Omega_{\mu}=\Omega_{\text {s }}
\mathbf{\Omega}_{\boldsymbol{n}} \boldsymbol{=} \mathbf{\Omega}_{\boldsymbol{k}}
Molecular
\sigma
or
conductivity)
diffusivity)
\kappa T / \sigma_{A B}
conductivity)
Substance
Weight
0.30
2.840
2.649
2.7
1.0691
0.35
2.676
2.468
2.8
1.0583
Light elements:
0.40
2.531
2.314
2.9
1.0482
\mathrm{H}_{2}
2.016
2.915
38.0
a
0.50
2.284
2.066
3.1
1.0300
-
0.55
2.178
1.965
3.2
1.0217
0.60
2.084
1.877
3.3
1.0139
Noble gases:
0.65
1.999
1.799
3.4
1.0066
0.70
1.922
1.729
3.5
0.9996
Ne
20.183
2.789
35.7
a
0.75
1.853
1.667
3.6
0.9931
Ne
39.948
3.432
122.4
b
0.80
1.790
1.562
3.7
0.9868
Kr
83.80
3.675
170.0
b
0.90
1.682
1.517
3.9
0.9753
Xe
4.009
234.7
{ }^{b}
0.95
1.636
1.477
4.0
0.9699
1.00
1.593
1.440
4.1
0.9647
Simple polyatomic gases:
1.05
1.554
1.406
4.2
0.9598
1.10
1.518
1.375
4.3
0.9551
3.617
97.0
{ }^{a}
1.15
1.485
1.347
4.4
0.9506
\mathrm{N}_{2}
28.01
3.667
99.8
{ }^{b}
1.20
1.455
1.320
4.5
0.9462
\mathrm{O}_{2}
32.00
3.433
113.
{ }^{a}
1.30
1.401
1.274
4.7
0.9380
CO
3.590
110.
{ }^{a}
1.35
1.377
1.253
4.8
0.9341
\mathrm{F}_{2}
3.653
112.
1.40
1.355
1.234
4.9
0.9304
\mathrm{Cl}_{2}
4.115
357.
a
1.45
1.315
1.216
5.0
0.9268
\mathrm{Br}_{2}
4.268
520.
{ }^{a}
1.55
1.297
1.183
7.0
0.8727
1.60
1.280
1.168
8.0
0.8538
\mathrm{I}_{2}
550.
1.65
1.264
1.154
9.0
0.8380
n-\mathrm{C}_{9} \mathrm{H}_{20}
128.26
1.70
1.249
1.141
10.0
0.8244
7.463
351.
b
1.75
1.235
1.128
12.0
0.8018
Cyclohexane
84.16
6.143
313.
d
Benzene
78.11
5.443
387.
b
# Dados
\begin{aligned} & A \rightarrow C L_{2} \\\ & B \rightarrow A r \\\ & \text { Goticulas de Cl2 } \rightarrow D=0.2 \mathrm{~cm} \\\ & P_{a t m}=1 \mathrm{~atm} \\\ & T_{a m b}=25^{\circ} \mathrm{C}=298.15 \mathrm{~K} \\\ & P_{\mathrm{Cl}_{2}}^{V}=27.48 \mathrm{mmH}=3.62 \times 10^{-2} \mathrm{~atm} \\\ & P_{\mathrm{Cl}_{2}}=1.05 \mathrm{~g} / \mathrm{cm}^{3} \\\ & M M_{\mathrm{Cl}_{2}}=71 \mathrm{~g} / \mathrm{mol} \\\ & M M_{\mathrm{Ar}}=28.97 \mathrm{~g} / \mathrm{mol} \text { (Tabela E.1) } \end{aligned}
(a)
\sigma_{A B}=\frac{4,115 \AA+3,617 \AA}{2}=3.866 \AA Å Å Å
\begin{aligned} & \frac{\varepsilon_{A B}}{k}=\sqrt{357 k \cdot 97 k}=186.09 k \\\ & \log 0 ; \frac{k T}{\varepsilon_{A B}}=\frac{k}{\varepsilon_{A B}} T=\frac{1}{186.09 k} \quad 298.15 k=1.6021 \\\ & \frac{k T}{\varepsilon_{A B}}=1.6021 \rightarrow \Omega_{D}=1.168 \end{aligned}
substituindo os dados na equação de chapman-Enkosg, temos
\begin{aligned} & D_{A B}=\frac{1.858 \times 10^{-3} \cdot 298.15 \mathrm{k}}{1 \mathrm{~atm} \cdot(3.866 \AA)^{3 / 2}\left[\frac{1}{719 / \mathrm{mol}}+\frac{1}{28.979 / \mathrm{mol}}\right]^{1 / 2}} \\\ & D_{A B}=0.1208 \mathrm{~cm}^{2} / \mathrm{s} \end{aligned} Å
(a) Taxa de difusão em g/dia.
balanso de massa
Entrada-saida + Geraşõo
= Acýmulo
\left.N_{\mathrm{CL}_{2}, r} \cdot S\right|_{r}-\left.N_{\mathrm{CL}_{2}, r} \cdot S\right|_{r_{0}+\Delta r}=0 \quad \therefore Dividindo por
S \cdot \Delta r .
\frac{\left.N_{c l_{2}, r}\right|_{r}}{\Delta r} \cdot \frac{\left.N_{c l_{2}, r}\right|_{r+\Delta r}}{\Delta r}=0
\frac{\left.N_{c L_{2}, r}\right|_{r}-\left.N_{c L_{2}, r}\right|_{r+\Delta r}}{\Delta r}=0
\frac{d N_{c L_{2}, r}}{d r}=0
Aplicando a lei de Fick para camada estagnada, temos:
N_{\mathrm{CL}_{2}, r}=-C \cdot D_{A B} \cdot \frac{d y_{C L_{2}}}{d r}+y_{C L_{2}}\left(N_{\mathrm{CL}_{2}, r}+N / A r, r\right)
N_{\mathrm{Cl}_{2}, r}=-C \cdot D_{A B} \cdot \frac{d y_{\mathrm{CL}_{2}}}{d r}+y_{\mathrm{CL}_{2}} \cdot N_{\mathrm{CL}_{2}, r}
N_{\mathrm{CL}_{2}, r}-Y_{\mathrm{CL}_{2}} \cdot N_{\mathrm{CL}_{2}, r}=-C \cdot D_{A B} \cdot \frac{d Y_{\mathrm{CL}_{2}}}{d r}
N_{\mathrm{CL}_{2}, r}\left(1-y_{\mathrm{CL}_{2}}\right)=-C \cdot D_{A B} \cdot \frac{d y_{\mathrm{CL}_{2}}}{d r}
N_{\mathrm{CL}_{2}, r} \cdot d r=-C \cdot D_{A B} \cdot \frac{d y_{\mathrm{CL}_{2}}}{1-y_{\mathrm{CL}_{2}}}
dado que
\quad N_{\mathrm{Cl}_{2}, r}=\frac{W_{\mathrm{Cl}_{2}, r}}{4 \pi r^{2}}
Logo;
\frac{W_{C L_{2}, r}}{4 \pi r^{2}} \cdot d r=-C \cdot D_{A B} \cdot \frac{d y_{C L_{2}}}{1-y_{C L_{2}}} .
\frac{W_{C L_{2}, r}}{r^{2}} \cdot d r=-4 \pi \cdot C \cdot D_{A B} \cdot \frac{d y_{C L_{2}}}{1-y_{C L_{2}}}
Condisões de Contorno
r=r_{A, O} \rightarrow y_{A}=Y_{A, O}
r=r_{A, \infty} \rightarrow y_{A}=Y_{A, \infty}=0
\begin{aligned} & W_{C L_{2}, r} \int_{r_{A, 0}}^{r_{A}, \infty} \frac{1}{r^{2}} d r=-4 \pi C \cdot D_{A B} \cdot \int_{y_{A, 0}}^{y_{A, \infty}} \frac{d y_{C L_{2}}}{1-y_{C L_{2}}} . \\\ & W_{C L_{2}, r}\left[\frac{-1}{r_{A, \infty}}+\frac{1}{r_{A, 0}}\right]=-4 \pi C \cdot D_{A B}\left[-\ln 1-\gamma_{A, \infty}+\ln 1-\gamma_{A, 0}\right] \\\ & W_{C L_{2}, r}\left[\frac{-1}{r_{A, \infty}}+\frac{1}{r_{A, 0}}\right]=-4 \pi C \cdot D_{A B}-\left[l_{\Pi} 1-\gamma_{A, \infty}-\ln 1-\gamma_{A, 0}\right] \\\ & W_{C L_{2}, r}\left[\frac{-\not f^{0}}{f_{A, \infty}}+\frac{1}{r_{A, 0}}\right]=4 \pi C \cdot D_{A B} \cdot l_{n} \frac{1-y_{A, \infty}}{1-y_{A, 0}} . \\\ & \frac{W_{C L_{2}, r}}{r_{A, 0}}=4 \pi C \cdot D_{A B} \cdot l_{n} \frac{1-y_{A, 00}}{1-y_{A, 0}} . \\\ & W_{\mathrm{CL}_{2}, r}=r_{A, 0} \cdot 4 \pi \cdot C \cdot D_{A B} \cdot l_{n} \frac{1}{1-y_{A, 0}} . \\\ & \text { dado que } y_{A, 0}=\frac{p^{V}}{p^{T}}=\frac{3.62 \times 10^{-2} \mathrm{~atm}}{1 \mathrm{~atm}}=3.62 \times 10^{-2} \\\ & C=\frac{n}{V}=\frac{n}{\frac{n R T}{P}}=\frac{P}{R T}=\frac{1 \text { atm }}{0.082 \frac{\text { Latm }}{\text { mol.K }} 298.15 \mathrm{~K}}=4.09 \times 10^{-2} \frac{\text { mol }}{\mathrm{L}} \\\ & C=4.09 \times 10^{-2} \frac{\mathrm{~mol}}{\mathrm{~L}} \cdot \frac{1 \mathrm{~L}}{1000 \mathrm{~cm}^{3}}=4.09 \times 10^{-5} \frac{\mathrm{~mol}}{\mathrm{~cm}^{3}} \\\ & \text { substituindo } \\\ & W_{\mathrm{Cl}_{2}, \mathrm{r}}=0.2 \mathrm{sm} \cdot 4 \pi \cdot 4.09 \times 10^{-5} \frac{\mathrm{~mol}}{\mathrm{~cm}^{3}} \cdot 0.1208 \frac{\mathrm{~cm}^{2}}{\mathrm{~s}} \cdot \ln \frac{1}{1-3.62 \times 10^{-2}} \\\ & W_{\mathrm{Cl}_{2}, \mathrm{r}}=4.58 \times 10^{-7} \frac{\mathrm{~mol}}{\mathrm{~s}} \cdot \frac{71 \mathrm{~g}}{\mathrm{~mol}} \cdot \frac{3600 \mathrm{~s}}{1 \mathrm{~h}} \cdot \frac{24 \mathrm{~h}}{1 \mathrm{dia}} \\\ & W_{\mathrm{Cl}_{2}, \mathrm{r}}=2.81 \mathrm{~g} / \mathrm{dia} \end{aligned}
(c) Partindo da equação
\rightarrow \frac{W_{C l_{2}, r}}{r_{A, 0}}=4 \pi C \cdot D_{A B} \cdot \ln \frac{1-y_{A, \infty}}{1-y_{A, 0}}
e sabendo que
W_{\mathrm{CL}_{2}, r}=\frac{d n}{d t}=\frac{d m}{M M_{\mathrm{CL}_{2}} \cdot d t}=\frac{\rho d v}{M M_{\mathrm{CL}_{2}} \cdot d t}
\begin{aligned} & W_{C L_{2}, r}=-\frac{\rho \cdot 4 \pi r^{2}}{M M_{C L_{2}}} \frac{d r}{d t} \\\ & \text { substituindo } \\\ & \frac{1}{上_{1,0}} \frac{\rho \cdot 4_{1} \pi^{r}}{M M_{C L_{2}}} \frac{d r}{d t}=4 \pi C \cdot D_{A B} \cdot l_{n} \frac{1}{1-y_{A, 0}} \quad \frac{d v}{d r}=4 \pi r^{2} \\\ & \frac{-\rho}{M M_{C L_{2}}} \cdot r d r=4 \pi r^{2} d r \end{aligned} 上
Condisões de Contorno
\begin{aligned} & r=r_{A, 0} \rightarrow t=0 \\\ & r=r_{A, \infty} \rightarrow t=t \\\ & \frac{-\rho}{M M_{C L_{2}}} \cdot \int_{r_{A, 0}}^{r_{A, \infty}} r d r=C \cdot D_{A B} \cdot l_{n} \frac{1}{1-y_{A, 0}} \cdot \int_{0}^{t} d t \\\ & \frac{-\rho}{\mathrm{MM}_{\mathrm{CL}_{2}}} \cdot \frac{-r_{A, 0}{ }^{2}}{2}=C \cdot D_{A B} \cdot \ln \frac{1}{1-y_{A, 0}} \cdot t \\\ & t=\frac{\frac{\rho}{M M_{\mathrm{CL}_{2}}} \cdot \frac{r_{A, D}{ }^{2}}{2}}{C \cdot D_{A B} \cdot l_{n} \frac{1}{1-y_{A, O}}} \\\ & t=\frac{1,05 \frac{9}{\mathrm{~cm}^{3}} \cdot \frac{1 \mathrm{~mol}}{71 \mathrm{~g}} \cdot \frac{(0,2 \mathrm{~cm})^{2}}{2}}{4.09 \times 10^{-5} \frac{\mathrm{~mol}}{\mathrm{~cm}^{3}} 0,1208 \frac{\mathrm{~cm}^{2}}{\mathrm{~s}} \cdot \ln \frac{1}{1-3,62 \times 10^{-2}}} \\\ & t=1623,60 \mathrm{~s} \approx 27 \mathrm{~min} \end{aligned}