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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 P1=0.9atm and P2=0.1atm respectively. Calculate the value of Molar flux (in mol/m2.s) with respect to an observer moving with molar average velocity. Calculate this value at the end of the path where P2=0.1atm. Use R=8.205*10-5m3atmK-1mol-1.
- a)2.713
- b)1.713
- c)0.1713
- d)0.01713
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
Oxygen diffuses through a stagnant layer of air, 1mm thick, ambient te...
Explanation: This is a case of diffusion of A through non-diffusing B.
Molar flux of oxygen, NA= (DAB*P ln[(P-P2)/(P-P1)])/RTl
Putting all the required values in the above equation.
NA=1.904 mol/m2.s
Molar average velocity, U= (NA+NB)/C
U=NA/C (NB=0)
U=NART/P
U=4.702*10-2m/s
Now, UA=UC/CA
UA=U/yA=0.4702 m/s (yA is mole fraction of A at the end)
JA=CA(uA-U)
=PA*( uA-U)/RT
=1.713mol/m2.s.
Molar flux of oxygen, NA= (DAB*P ln[(P-P2)/(P-P1)])/RTl
Putting all the required values in the above equation.
NA=1.904 mol/m2.s
Molar average velocity, U= (NA+NB)/C
U=NA/C (NB=0)
U=NART/P
U=4.702*10-2m/s
Now, UA=UC/CA
UA=U/yA=0.4702 m/s (yA is mole fraction of A at the end)
JA=CA(uA-U)
=PA*( uA-U)/RT
=1.713mol/m2.s.
Most Upvoted Answer
Oxygen diffuses through a stagnant layer of air, 1mm thick, ambient te...
°C, and a pressure of 101.3 kPa. The partial pressure of oxygen in the layer is 20.9% of the total pressure, which is 21.2 kPa. The diffusion coefficient of oxygen in air at 28°C is 0.22 cm²/s.
Using Fick's Law of Diffusion, we can calculate the rate of diffusion of oxygen through the stagnant layer:
J = -D*(dC/dx)
Where J is the flux (molecules/cm²s), D is the diffusion coefficient (cm²/s), C is the concentration of oxygen (molecules/cm³), and x is the distance across the layer (cm).
Since the layer is stagnant, there is no convective flow, so the flux is constant across the layer. Therefore, we can integrate the above equation to obtain the total flux through the layer:
J = -D*(C₂ - C₁)/d
Where C₂ is the concentration of oxygen at the surface (21.2 kPa), C₁ is the concentration at the interface with the bulk air (0 kPa), and d is the thickness of the layer (0.1 cm).
Plugging in the values, we get:
J = -0.22*(21.2 - 0)/0.1 = -46.64 molecules/cm²s
The negative sign indicates that oxygen is diffusing from the surface into the layer, as expected.
To convert the flux to a rate of diffusion, we need to multiply by the surface area of the layer. Let's assume a square layer with sides of 1 cm, so the surface area is 1 cm². Then:
Rate of diffusion = J*A = -46.64 molecules/s
This means that 46.64 oxygen molecules are diffusing through the layer every second. Note that this is a very small rate of diffusion, since the layer is only 1 mm thick. If the layer were thicker, the rate of diffusion would increase proportionally.
Using Fick's Law of Diffusion, we can calculate the rate of diffusion of oxygen through the stagnant layer:
J = -D*(dC/dx)
Where J is the flux (molecules/cm²s), D is the diffusion coefficient (cm²/s), C is the concentration of oxygen (molecules/cm³), and x is the distance across the layer (cm).
Since the layer is stagnant, there is no convective flow, so the flux is constant across the layer. Therefore, we can integrate the above equation to obtain the total flux through the layer:
J = -D*(C₂ - C₁)/d
Where C₂ is the concentration of oxygen at the surface (21.2 kPa), C₁ is the concentration at the interface with the bulk air (0 kPa), and d is the thickness of the layer (0.1 cm).
Plugging in the values, we get:
J = -0.22*(21.2 - 0)/0.1 = -46.64 molecules/cm²s
The negative sign indicates that oxygen is diffusing from the surface into the layer, as expected.
To convert the flux to a rate of diffusion, we need to multiply by the surface area of the layer. Let's assume a square layer with sides of 1 cm, so the surface area is 1 cm². Then:
Rate of diffusion = J*A = -46.64 molecules/s
This means that 46.64 oxygen molecules are diffusing through the layer every second. Note that this is a very small rate of diffusion, since the layer is only 1 mm thick. If the layer were thicker, the rate of diffusion would increase proportionally.
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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 P1=0.9atm and P2=0.1atm respectively. Calculate the value of Molar flux (in mol/m2.s) with respect to an observer moving with molar average velocity. Calculate this value at the end of the path where P2=0.1atm. Use R=8.205*10-5m3atmK-1mol-1.a)2.713b)1.713c)0.1713d)0.01713Correct answer is option 'B'. Can you explain this answer?
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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 P1=0.9atm and P2=0.1atm respectively. Calculate the value of Molar flux (in mol/m2.s) with respect to an observer moving with molar average velocity. Calculate this value at the end of the path where P2=0.1atm. Use R=8.205*10-5m3atmK-1mol-1.a)2.713b)1.713c)0.1713d)0.01713Correct answer is option 'B'. Can you explain this answer? for Chemical Engineering 2024 is part of Chemical Engineering preparation. The Question and answers have been prepared according to the Chemical Engineering exam syllabus. Information about 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 P1=0.9atm and P2=0.1atm respectively. Calculate the value of Molar flux (in mol/m2.s) with respect to an observer moving with molar average velocity. Calculate this value at the end of the path where P2=0.1atm. Use R=8.205*10-5m3atmK-1mol-1.a)2.713b)1.713c)0.1713d)0.01713Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for Chemical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 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 P1=0.9atm and P2=0.1atm respectively. Calculate the value of Molar flux (in mol/m2.s) with respect to an observer moving with molar average velocity. Calculate this value at the end of the path where P2=0.1atm. Use R=8.205*10-5m3atmK-1mol-1.a)2.713b)1.713c)0.1713d)0.01713Correct answer is option 'B'. Can you explain this answer?.
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 P1=0.9atm and P2=0.1atm respectively. Calculate the value of Molar flux (in mol/m2.s) with respect to an observer moving with molar average velocity. Calculate this value at the end of the path where P2=0.1atm. Use R=8.205*10-5m3atmK-1mol-1.a)2.713b)1.713c)0.1713d)0.01713Correct answer is option 'B'. Can you explain this answer? for Chemical Engineering 2024 is part of Chemical Engineering preparation. The Question and answers have been prepared according to the Chemical Engineering exam syllabus. Information about 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 P1=0.9atm and P2=0.1atm respectively. Calculate the value of Molar flux (in mol/m2.s) with respect to an observer moving with molar average velocity. Calculate this value at the end of the path where P2=0.1atm. Use R=8.205*10-5m3atmK-1mol-1.a)2.713b)1.713c)0.1713d)0.01713Correct answer is option 'B'. Can you explain this answer? covers all topics & solutions for Chemical Engineering 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 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 P1=0.9atm and P2=0.1atm respectively. Calculate the value of Molar flux (in mol/m2.s) with respect to an observer moving with molar average velocity. Calculate this value at the end of the path where P2=0.1atm. Use R=8.205*10-5m3atmK-1mol-1.a)2.713b)1.713c)0.1713d)0.01713Correct answer is option 'B'. Can you explain this answer?.
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