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QUESTION: 1

Oxygen diffuses through a stagnant layer of air, 1mm thick, ambient temperature 28°C and 1atm total pressure. The partial pressure of oxygen on two sides of layer is P_{1}=0.9atm and P_{2}=0.1atm respectively. Calculate the value of Molar flux (in mol/m^{2}.s) with respect to an observer moving with molar average velocity. Calculate this value at the end of the path where P_{2}=0.1atm. Use R=8.205*10^{-5}m^{3}atmK^{-1}mol^{-1}.

Solution:

Explanation: This is a case of diffusion of A through non-diffusing B.

Molar flux of oxygen, N_{A}= (D_{AB}*P ln[(P-P_{2})/(P-P_{1})])/RTl

Putting all the required values in the above equation.

N_{A}=1.904 mol/m^{2}.s

Molar average velocity, U= (N_{A}+N_{B})/C

U=N_{A}/C (N_{B}=0)

U=N_{A}RT/P

U=4.702*10^{-2}m/s

Now, U_{A}=UC/C_{A}

UA=U/y_{A}=0.4702 m/s (y_{A} is mole fraction of A at the end)

J_{A}=C_{A}(u_{A}-U)

=P_{A}*( u_{A}-U)/RT

=1.713mol/m^{2}.s.

QUESTION: 2

In order to prepare Lithium nitride, air is passed gently over a reactor. Suppose that Nitrogen diffuse to a length 5cm before getting reacted with Lithium. Nitrogen reacts with lithium quickly so that partial pressure of Nitrogen at that point is 0. Reactor is a cylinder of length 10cm and diameter 2cm. Assume air to be a mixture of Nitrogen and Oxygen only. The temperature is 298K and total pressure 1atm. Diffusivity of N_{2} in O_{2} is 2*10^{-5}m^{2}/s. Calculate the Molar flux (mol/m^{2}.s) of N_{2} at a distance of 2.5cm from the top with respect to an observer moving with twice the mass average velocity a direction towards the liquid surface. (R=8.205*10^{-5}m^{3}atmK^{-1}mol^{-1}). Assume z=0 at the top of the reactor.

Solution:

Explanation:

Case: Diffusion of A through non-diffusing B

N_{2} in air= 79%

O_{2} in air= 21%

Molar flux of N_{2}, N_{A}=0.025mol/m^{2}.s

Pressure of N_{2} at the point z=2.5cm=

( D_{AB}*P ln[(P-P_{5})/(P-P0)])/RTl=(D_{AB}*P ln[(P-P_{5})/(P-P_{2.5})])/RTl

P_{2.5}=0.542atm

Velocity of N_{2}= N_{A}RT/P_{2.5}=112.78*10^{-5}m/s

Molar flux of O_{2}-=0

Velocity of O_{2}=0

Average molar mass=0.542*28+0.458*32=29.832

Mass average velocity= (0.542*28*112.78*10-5)/29.832=57.37*10^{-5}m/s

Required molar flux at the distance of z=5, is

C_{A}(u_{A}-2V)= N_{A}-2P_{A}V/RT=4.242*10^{-4}mol/m^{2}.s.

QUESTION: 3

A binary liquid mixture is separated by distillation process. The more volatile A vaporizes while less volatile B gets transported in opposite direction. Latent heat of vaporization of A and B is 0.5unit and 1unit respectively. What will be the relation between molar flux of A and B

Solution:

Explanation: Latent heat of vaporization of A is provided from condensation of B. N_{A}∆H_{A}=N_{B}∆H_{B}.

QUESTION: 4

In an accident a container containing ethanol(s.g 0.789) falls down in a room of dimension 1*1*2(all in m). Ethanol forms a layer of 2mm. Ethanol vaporizes and diffuses through a stagnant film of air of thickness 2.5mm. What will be the time required for the water layer to disappear completely. The diffusivity of ethanol in air is 2.567*10^{-5} m^{2}/s. the total pressure is 1.013 bar and the pressure of ethanol in bulk air is 0.02244 bar and at the ethanol-air interface is 0.02718bar. The temperature is 25.2°C.

Solution:

Explanation: This is the case of diffusion of through non diffusing B.

N_{A}=DP/RTZ*ln[(P-P_{1})/(P-P_{2})] N_{A}=(2.567*10-5*1.013)/(0.08317*298.2*2.5*10-3)*ln[(1.013-0.02244)/(1.013-0.02718)] = 2.01*10^{-6} kmol/m^{2}.s=9.259*10^{-5}kg/m^{2}.s

Amount of ethanol = 2*10^{-3}*1=0.002m^{3}=1.578kg

Time for evaporation= 1.578/9.259*10^{-5}=17042.87s=4.73h.

QUESTION: 5

In the previous question if the floor has micro pores and water penetrates the floor at a constant rate of 0.1kg/m^{2}h, what will be the time required for the water layer to disappear?

Solution:

Explanation: Combined rate of disappearance= 0.1+9.259*10^{-5}*3600=0.433kg/m^{2}h

Time for disappearance= 1.578/0.433=3.64h.

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