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Problem 2 (30 Points A cylindrical container filled with liquid 225 kPa stored in a large room. The diameter of the the lido 25 kPa is initially aniline-free and the room temperature container is 45 cm. The ambient air is id of the container is removed and the aniline evaporates slowly. a ndfiusion coefficient of aniline in air. Use the empirical equation aniline diffusion has not reached steady state at 15 cm above the aniline is ambient air 101.325 kPa is initially a) (12 Pts) Calculate the rom Fuller, Schettler, and coe b) (13 Pts) AsuGiddings,has not reached steady surface of the container after 4 min, calculate the concentration of aniline in the air at that position at that time. State your r assumption to solve this problem. much mass of aniline will have evaporated after 4 min? Chemical formula of aniline: CeHsNHa Composition of air: 21 v01% O2, 79 vol% N2 Molar masses: O, 32 g/mol, N2-28 g/mol Saturation vapor pressure of aniline at 20 C: 0.5 kPa R 8.314 J/mol K

Answer 1

a.

The required gas diffusion coefficient can be calculated by following the empirical formula (Fuller, Schettler, and Giddings, 1966/ Cussler’s book, 3rd ed., Eq.5.1-9):

D = 10^-3 * (T^1.75) (1/M₁ + 1/M₂)^1/2 / p [($\sum$_i( V_i1)^1/3 + ($\sum$_i (V_i2)^1/3 )]² - equation 1

where V_ij are the diffusion volumes of parts i of the molecule j, p is in atm and T is in K.

For air (molecule 1): M₁ = 0.21 *32 + 0.79 *28 = 28.84 g/mol

$\sum$_i( V_i1) = 20.1

(Cussler’s book, 3rd ed., Table 5.1-4)

For aniline (molecule 2): M₂= 93.13 g/mol

$\sum$_i (V_i2) = 6 *16.5 + 7 *1.98 + 1 *5.69- 20.2 = 98.35

Also: p = 1 atm T = 20+273=293 K So, from Eq.1 we get:

D = 10^-3 * (293^1.75) (1/28.84+ 1/93.13)^1/2 / 1 [($\sum$_i( 20.1)^1/3 + ($\sum$_i (98.35)^1/3 )]² =  8.22 *10 ^-2 cm²/s

b.

Since the diffusion takes place at non-steady state, the semi-infinite slab approach should be used. So, the concentration profile of aniline (species 1) in the air is described by (Cussler’s book, 3rd ed., Chapter 2.3-Unsteady Diffusion in a Semi-infinite Slab, Eq.2.3-15):

where: 4Dt Thus: Ci =C101 1-erf 4Dt where elsat

c₁₀ = 0.5 * 10^3 Pa / 8.314 J/mol.K * 293 K = 0.205 mol/m³

15cm = 1.69 ç=T4DI-1 - 8.22 10m4min 60min 4.8.22-10 _ . 4 min erf.-erf(1.69)-0.98307 So: Ci = Cio (1-0.98307) = 0.0 16%, = 3.48-10-3m01 / m 3

c.

Since the diffusion takes place at non-steady state, the flux across the interface (at z - 0) as a function of time is given by (Cussler's book, 3'd ed., Eq.2.3-18) 10 C10 and the flow rate across the interface at z-0 as a function of time is given by I lz=0 So, the amount of aniline that evaporates after time t is calculated by integrating J,frorm t=0 to t=t,:

where: 4 So: -on, =2-T(0.45m)- min 60-1.64.10mol 0.205mOl 4 Im And mı = 1.64-10-3m 01.93. 13-gl mol m,0.152g