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
Air
28.97^{\prime}
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
28.01
28.01
3.590 110. { }^{a} 1.35 1.377 1.253 4.8 0.9341
\mathrm{F}_{2}
38.00
38.00
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}
159.82
253.81
253.81
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
\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}
(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}
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}

Passo 1

O enunciado trata da difusão de gotículas esféricas de cloro líquido (diâmetro D=0,2,cm) no ar à T=25^\circ\mathrm{C} e P=1,atm. Precisamos calcular:
  • (a) a difusividade D_{AB} do \mathrm{Cl}_2 em ar pela equação de Chapman–Enskog;
  • (b) a taxa de difusão (em g/dia) usando o modelo de filme estagnado e a lei de Fick;
  • (c) o tempo até que a gotícula se dissipe totalmente, integrando a equação de balanço de massa.
Dados principais:
\sigma_{\mathrm{Cl}_2}=4{,}115,\AA, \sigma_{\rm ar}=3{,}617,\AA, \varepsilon/k de \mathrm{Cl}_2=357,K, de ar =97,K, pressão de vapor P^V=27{,}48,mmHg =3,62\times10^{-2},atm, densidade \rho=1,05,g/cm ^3, massas molares M_{\mathrm{Cl}_2}=71,g/mol, M_{\rm ar}=28{,}97,g/mol.

Passo 2

Calcular \sigma_{AB}, \varepsilon_{AB}/k e a integral de colisão \Omega_D.
\sigma_{AB}=\frac{\sigma_{\mathrm{Cl}_2}+\sigma_{\rm ar}}{2} =\frac{4{,}115+3{,}617}{2}=3{,}866\\ \text{\AA}
\frac{\varepsilon_{AB}}{k} =\sqrt{357\cdot97}=186{,}09\\ \mathrm{K}
\frac{kT}{\varepsilon_{AB}} =\frac{298{,}15}{186{,}09}=1{,}6021 \quad\Longrightarrow\quad \Omega_D\approx1{,}168
Agora a equação de Chapman–Enskog para D_{AB} (cm ^2/s):
D_{AB} =\frac{1{,}858\times10^{-3}\,T^{3/2}\, \bigl[\tfrac1{M_{\mathrm{Cl}_2}}+\tfrac1{M_{\rm ar}}\bigr]^{1/2}} {P\,\sigma_{AB}^2\,\Omega_D}
Substituindo:
T^{3/2}=(298{,}15)^{3/2}\approx5{,}148\times10^3, \quad \Bigl[\tfrac1{71}+\tfrac1{28{,}97}\Bigr]^{1/2}\approx0{,}2205,
D_{AB} =\frac{1{,}858\times10^{-3}\times5{,}148\times10^3\times0{,}2205} {1\times(3{,}866)^2\times1{,}168} \approx0{,}1208\\ \mathrm{cm^2/s}.

Passo 3

Determinar a taxa molar de difusão N_{A} (mol/s) pelo modelo de filme estagnado e lei de Fick em regime estacionário:
N_{A} =-C\,D_{AB}\,\frac{d y_A}{d r}\quad\Longrightarrow\quad W_{A}=4\pi\,r_0\,C\,D_{AB}\,\ln\frac{1}{1-y_{A,0}}
onde
\;r_0=\tfrac{D}{2}=0{,}1,cm,
\;y_{A,0}=P^V/P=3{,}62\times10^{-2},
\;C=\dfrac{P}{RT} =\dfrac{1}{0,08206\times298{,}15}\approx4{,}085\times10^{-2},mol/L
\;\to4{,}085\times10^{-5},mol/cm ^3.
Logo,
W_{A} =4\pi(0{,}1)\,(4,085\times10^{-5})\,(0,1208)\,\ln\frac1{1-0{,}0362} \approx2,29\times10^{-7}\\ \mathrm{mol/s}.

Passo 4

Converter a taxa para g/dia:
W_m =W_{A}\times M_{\mathrm{Cl}_2} =2,29\times10^{-7}\times71 \approx1,62\times10^{-5}\\ \mathrm{g/s}
\Longrightarrow W_{\rm dia} =1,62\times10^{-5}\times86400 \approx1,40\\ \mathrm{g/dia}.

Passo 5

Integrar o balanço massa para achar o tempo de dissipação completa. Como
W_A =-\frac{1}{M}\frac{d m}{dt} =-\frac{1}{M}\,\rho\,4\pi\,r^2\,\frac{dr}{dt} \quad\text{e}\quad W_A=\text{constante},
temos
-\rho\,4\pi\,r^2\frac{dr}{dt}/M =4\pi\,r_0\,C\,D_{AB}\,\ln\frac1{1-y_{A,0}}
\frac{dr}{dt} =-\frac{M\,r_0\,C\,D_{AB}\,\ln\frac1{1-y_{A,0}}}{\rho\,r^2}.
Separando e integrando de r_0 até 0:
\int_{r_0}^{0}r^2dr =-\frac{M\,r_0\,C\,D_{AB}\,\ln\frac1{1-y_{A,0}}}{\rho} \int_{0}^{t}dt \;\Longrightarrow\; -\frac{r_0^3}{3} =-\frac{M\,r_0\,C\,D_{AB}\,\ln\frac1{1-y_{A,0}}}{\rho}\,t
t =\frac{\rho\,r_0^2}{3\,M\,C\,D_{AB}\,\ln\frac1{1-y_{A,0}}} =\frac{1{,}05\times(0{,}1)^2} {3\times71\times4{,}085\times10^{-5}\times0{,}1208\times0{,}0369} \approx1,92\times10^4\\ \mathrm{s} \approx5,3\\ \mathrm{h}.

Resposta

a) D_{AB}\approx0,1208\\ \mathrm{cm^2/s}
b) W\approx1,40\\ \mathrm{g/dia}
c) t\approx5{,}3\\ \mathrm{h}

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