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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 m2/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.
- a)4.73h
- b)14.6h
- c)5.87h
- d)16.2h
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
In an accident a container containing ethanol(s.g 0.789) falls down in...
Explanation: This is the case of diffusion of through non diffusing B.
NA=DP/RTZ*ln[(P-P1)/(P-P2)] NA=(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/m2.s=9.259*10-5kg/m2.s
Amount of ethanol = 2*10-3*1=0.002m3=1.578kg
Time for evaporation= 1.578/9.259*10-5=17042.87s=4.73h.
NA=DP/RTZ*ln[(P-P1)/(P-P2)] NA=(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/m2.s=9.259*10-5kg/m2.s
Amount of ethanol = 2*10-3*1=0.002m3=1.578kg
Time for evaporation= 1.578/9.259*10-5=17042.87s=4.73h.
Most Upvoted Answer
In an accident a container containing ethanol(s.g 0.789) falls down in...
To find the time required for the ethanol layer to disappear completely, we need to calculate the rate of diffusion of ethanol through the air film.
First, let's calculate the concentration gradient of ethanol. The pressure difference between the ethanol-air interface and the bulk air is given as 0.02718 - 0.02244 = 0.00474 bar.
Using the ideal gas law, we can convert this pressure difference to a concentration difference:
ΔC = (0.00474 bar) / (RT)
where R is the ideal gas constant and T is the temperature in Kelvin.
Let's convert the temperature from Celsius to Kelvin:
T = 25.2 + 273.15 = 298.35 K
Now we can calculate the concentration difference:
ΔC = (0.00474 bar) / (0.0831 L·bar/mol·K * 298.35 K) ≈ 0.000190 mol/L
Next, let's calculate the average concentration of ethanol in the ethanol layer. The specific gravity of ethanol is given as 0.789, which means it has a density of 0.789 g/mL or 789 kg/m^3.
The thickness of the ethanol layer is 2 mm, which is equal to 0.002 m.
The volume of the ethanol layer is then:
V_ethanol = (1 m) * (1 m) * (0.002 m) = 0.002 m^3
The mass of the ethanol in the layer is:
m_ethanol = V_ethanol * density = 0.002 m^3 * 789 kg/m^3 = 1.578 kg
The number of moles of ethanol in the layer is:
n_ethanol = m_ethanol / molar mass = 1.578 kg / 46.07 g/mol = 34.24 mol
The average concentration of ethanol in the layer is then:
C_ethanol = n_ethanol / V_ethanol = 34.24 mol / 0.002 m^3 = 17,120 mol/m^3
Now we can calculate the rate of diffusion using Fick's law of diffusion:
J = -D * ΔC / δ
where J is the rate of diffusion, D is the diffusivity of ethanol in air, ΔC is the concentration difference, and δ is the thickness of the air film.
Plugging in the values:
J = -(2.567 * 10^-5 m^2/s) * (0.000190 mol/L) / (0.0025 m)
J ≈ -0.0194 mol/(m^2·s)
The negative sign indicates that ethanol is diffusing from the ethanol-air interface into the air film.
Finally, to find the time required for the ethanol layer to disappear completely, we need to calculate the volume of the ethanol layer and divide it by the rate of diffusion.
The volume of the ethanol layer is:
V_ethanol = (1 m) * (1 m) * (0.002 m) = 0.002 m^3
The time required is then:
t = V_ethanol / (-J) = 0.002 m^3 / (0.0194 mol/(m^2·s)) ≈ 0.103 s
Therefore, it will take approximately 0.103 seconds for the ethanol
First, let's calculate the concentration gradient of ethanol. The pressure difference between the ethanol-air interface and the bulk air is given as 0.02718 - 0.02244 = 0.00474 bar.
Using the ideal gas law, we can convert this pressure difference to a concentration difference:
ΔC = (0.00474 bar) / (RT)
where R is the ideal gas constant and T is the temperature in Kelvin.
Let's convert the temperature from Celsius to Kelvin:
T = 25.2 + 273.15 = 298.35 K
Now we can calculate the concentration difference:
ΔC = (0.00474 bar) / (0.0831 L·bar/mol·K * 298.35 K) ≈ 0.000190 mol/L
Next, let's calculate the average concentration of ethanol in the ethanol layer. The specific gravity of ethanol is given as 0.789, which means it has a density of 0.789 g/mL or 789 kg/m^3.
The thickness of the ethanol layer is 2 mm, which is equal to 0.002 m.
The volume of the ethanol layer is then:
V_ethanol = (1 m) * (1 m) * (0.002 m) = 0.002 m^3
The mass of the ethanol in the layer is:
m_ethanol = V_ethanol * density = 0.002 m^3 * 789 kg/m^3 = 1.578 kg
The number of moles of ethanol in the layer is:
n_ethanol = m_ethanol / molar mass = 1.578 kg / 46.07 g/mol = 34.24 mol
The average concentration of ethanol in the layer is then:
C_ethanol = n_ethanol / V_ethanol = 34.24 mol / 0.002 m^3 = 17,120 mol/m^3
Now we can calculate the rate of diffusion using Fick's law of diffusion:
J = -D * ΔC / δ
where J is the rate of diffusion, D is the diffusivity of ethanol in air, ΔC is the concentration difference, and δ is the thickness of the air film.
Plugging in the values:
J = -(2.567 * 10^-5 m^2/s) * (0.000190 mol/L) / (0.0025 m)
J ≈ -0.0194 mol/(m^2·s)
The negative sign indicates that ethanol is diffusing from the ethanol-air interface into the air film.
Finally, to find the time required for the ethanol layer to disappear completely, we need to calculate the volume of the ethanol layer and divide it by the rate of diffusion.
The volume of the ethanol layer is:
V_ethanol = (1 m) * (1 m) * (0.002 m) = 0.002 m^3
The time required is then:
t = V_ethanol / (-J) = 0.002 m^3 / (0.0194 mol/(m^2·s)) ≈ 0.103 s
Therefore, it will take approximately 0.103 seconds for the ethanol
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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*10-5m2/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.a)4.73hb)14.6hc)5.87hd)16.2hCorrect answer is option 'A'. Can you explain this answer?
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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*10-5m2/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.a)4.73hb)14.6hc)5.87hd)16.2hCorrect answer is option 'A'. 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 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*10-5m2/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.a)4.73hb)14.6hc)5.87hd)16.2hCorrect answer is option 'A'. 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 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*10-5m2/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.a)4.73hb)14.6hc)5.87hd)16.2hCorrect answer is option 'A'. Can you explain this answer?.
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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*10-5m2/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.a)4.73hb)14.6hc)5.87hd)16.2hCorrect answer is option 'A'. Can you explain this answer?, a detailed solution for 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*10-5m2/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.a)4.73hb)14.6hc)5.87hd)16.2hCorrect answer is option 'A'. Can you explain this answer? has been provided alongside types of 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*10-5m2/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.a)4.73hb)14.6hc)5.87hd)16.2hCorrect answer is option 'A'. Can you explain this answer? theory, EduRev gives you an
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