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Estimate the diffusion coefficient of carbon monoxide through air in which the mole fractions of each constituents are
O 2 = 0.18
N 2 = 0.72
CO = 0.10
The gas mixture is at 300 K and 2 atmosphere total pressure. The diffusivity values are
D CO = 18.5 * 10 -6 m2/s at 273 K, 1 atm
D CN = 19.2 * 10 -6 m2/s at 288 K, 1 atm
The subscripts C is carbon monoxide, O is oxygen and N is nitrogen
O 2 = 0.18
N 2 = 0.72
CO = 0.10
The gas mixture is at 300 K and 2 atmosphere total pressure. The diffusivity values are
D CO = 18.5 * 10 -6 m2/s at 273 K, 1 atm
D CN = 19.2 * 10 -6 m2/s at 288 K, 1 atm
The subscripts C is carbon monoxide, O is oxygen and N is nitrogen
- a)10.29 * 10 -6 m2/s
- b)18.5 * 10 -6 m2/s
- c)17.5 * 10 -6 m2/s
- d)13.5 * 10 -6 m2/s
Correct answer is option 'A'. Can you explain this answer?
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Estimate the diffusion coefficient of carbon monoxide through air in w...
Explanation: (T 1/T 2) 3/2 (p 2/p 1) = D 1/D 2.
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Estimate the diffusion coefficient of carbon monoxide through air in w...
Estimating the Diffusion Coefficient of Carbon Monoxide through Air
Given:
Mole fraction of O2 = 0.18
Mole fraction of N2 = 0.72
Mole fraction of CO = 0.10
Temperature (T) = 300 K
Total pressure (P) = 2 atm
Diffusivity values:
DCO= 18.5 x 10^-6 m^2/s at 273 K, 1 atm
DCN= 19.2 x 10^-6 m^2/s at 288 K, 1 atm
Calculating the Diffusion Coefficient of Carbon Monoxide
To estimate the diffusion coefficient of carbon monoxide through air, we can use the Chapman-Enskog equation, which relates the diffusion coefficient to the molecular properties of the gas mixture. The equation is given by:
Dij = (1/3) * λm,i * λm,j * (2kT/πM)^(1/2)
where:
Dij = diffusion coefficient of species i through species j
λm,i = mean free path of species i
λm,j = mean free path of species j
k = Boltzmann constant
T = temperature
M = molar mass of the gas mixture
For a binary gas mixture, the diffusion coefficient of species i through species j is given by:
Dij = (1/3) * λm,i * λm,j * (2kT/πM)^(1/2)
= (1/3) * (1/σi, j)(1/2) * (1/NA * (VkBT/2πMi)^(1/2)) * (1/NA * (VkBT/2πMj)^(1/2)) * (2kT/π(Mi + Mj))^(1/2)
where:
σi, j = collision diameter of species i and j
NA = Avogadro's number
V = molar volume of the gas mixture
Mi, Mj = molar masses of species i and j
Substituting the given values, we get:
Dco,air = (1/3) * (1/σco,air)(1/2) * (1/NA * (VkBT/2πMco)^(1/2)) * (1/NA * (VkBT/2πMair)^(1/2)) * (2kT/π(Mco + Mair))^(1/2)
= (1/3) * (1/3.65 x 10^-10 m)(1/2) * (1/6.022 x 10^23 mol^-1 * (1.38 x 10^-23 J/K * 300 K/2π*28.01 g/mol)^(1/2)) * (1/6.022 x 10^23 mol^-1 * (1.38 x 10^-23 J/K * 300 K/2π*28.01 g/mol + 1.38 x 10^-23 J/K * 300 K/2π*32.00 g/mol)^(1/2)) * (2 * 1.38 x 10^-23 J/K * 300 K/π(28.01 g/mol +
Given:
Mole fraction of O2 = 0.18
Mole fraction of N2 = 0.72
Mole fraction of CO = 0.10
Temperature (T) = 300 K
Total pressure (P) = 2 atm
Diffusivity values:
DCO= 18.5 x 10^-6 m^2/s at 273 K, 1 atm
DCN= 19.2 x 10^-6 m^2/s at 288 K, 1 atm
Calculating the Diffusion Coefficient of Carbon Monoxide
To estimate the diffusion coefficient of carbon monoxide through air, we can use the Chapman-Enskog equation, which relates the diffusion coefficient to the molecular properties of the gas mixture. The equation is given by:
Dij = (1/3) * λm,i * λm,j * (2kT/πM)^(1/2)
where:
Dij = diffusion coefficient of species i through species j
λm,i = mean free path of species i
λm,j = mean free path of species j
k = Boltzmann constant
T = temperature
M = molar mass of the gas mixture
For a binary gas mixture, the diffusion coefficient of species i through species j is given by:
Dij = (1/3) * λm,i * λm,j * (2kT/πM)^(1/2)
= (1/3) * (1/σi, j)(1/2) * (1/NA * (VkBT/2πMi)^(1/2)) * (1/NA * (VkBT/2πMj)^(1/2)) * (2kT/π(Mi + Mj))^(1/2)
where:
σi, j = collision diameter of species i and j
NA = Avogadro's number
V = molar volume of the gas mixture
Mi, Mj = molar masses of species i and j
Substituting the given values, we get:
Dco,air = (1/3) * (1/σco,air)(1/2) * (1/NA * (VkBT/2πMco)^(1/2)) * (1/NA * (VkBT/2πMair)^(1/2)) * (2kT/π(Mco + Mair))^(1/2)
= (1/3) * (1/3.65 x 10^-10 m)(1/2) * (1/6.022 x 10^23 mol^-1 * (1.38 x 10^-23 J/K * 300 K/2π*28.01 g/mol)^(1/2)) * (1/6.022 x 10^23 mol^-1 * (1.38 x 10^-23 J/K * 300 K/2π*28.01 g/mol + 1.38 x 10^-23 J/K * 300 K/2π*32.00 g/mol)^(1/2)) * (2 * 1.38 x 10^-23 J/K * 300 K/π(28.01 g/mol +
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Estimate the diffusion coefficient of carbon monoxide through air in which the mole fractions of each constituents areO2= 0.18N2= 0.72CO = 0.10The gas mixture is at 300 K and 2 atmosphere total pressure. The diffusivity values areDCO= 18.5 * 10-6m2/s at 273 K, 1 atmDCN= 19.2 * 10-6m2/s at 288 K, 1 atmThe subscripts C is carbon monoxide, O is oxygen and N is nitrogena)10.29 * 10-6m2/sb)18.5 * 10-6m2/sc)17.5 * 10-6m2/sd)13.5 * 10-6m2/sCorrect answer is option 'A'. Can you explain this answer?
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Estimate the diffusion coefficient of carbon monoxide through air in which the mole fractions of each constituents areO2= 0.18N2= 0.72CO = 0.10The gas mixture is at 300 K and 2 atmosphere total pressure. The diffusivity values areDCO= 18.5 * 10-6m2/s at 273 K, 1 atmDCN= 19.2 * 10-6m2/s at 288 K, 1 atmThe subscripts C is carbon monoxide, O is oxygen and N is nitrogena)10.29 * 10-6m2/sb)18.5 * 10-6m2/sc)17.5 * 10-6m2/sd)13.5 * 10-6m2/sCorrect 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 Estimate the diffusion coefficient of carbon monoxide through air in which the mole fractions of each constituents areO2= 0.18N2= 0.72CO = 0.10The gas mixture is at 300 K and 2 atmosphere total pressure. The diffusivity values areDCO= 18.5 * 10-6m2/s at 273 K, 1 atmDCN= 19.2 * 10-6m2/s at 288 K, 1 atmThe subscripts C is carbon monoxide, O is oxygen and N is nitrogena)10.29 * 10-6m2/sb)18.5 * 10-6m2/sc)17.5 * 10-6m2/sd)13.5 * 10-6m2/sCorrect 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 Estimate the diffusion coefficient of carbon monoxide through air in which the mole fractions of each constituents areO2= 0.18N2= 0.72CO = 0.10The gas mixture is at 300 K and 2 atmosphere total pressure. The diffusivity values areDCO= 18.5 * 10-6m2/s at 273 K, 1 atmDCN= 19.2 * 10-6m2/s at 288 K, 1 atmThe subscripts C is carbon monoxide, O is oxygen and N is nitrogena)10.29 * 10-6m2/sb)18.5 * 10-6m2/sc)17.5 * 10-6m2/sd)13.5 * 10-6m2/sCorrect answer is option 'A'. Can you explain this answer?.
Estimate the diffusion coefficient of carbon monoxide through air in which the mole fractions of each constituents areO2= 0.18N2= 0.72CO = 0.10The gas mixture is at 300 K and 2 atmosphere total pressure. The diffusivity values areDCO= 18.5 * 10-6m2/s at 273 K, 1 atmDCN= 19.2 * 10-6m2/s at 288 K, 1 atmThe subscripts C is carbon monoxide, O is oxygen and N is nitrogena)10.29 * 10-6m2/sb)18.5 * 10-6m2/sc)17.5 * 10-6m2/sd)13.5 * 10-6m2/sCorrect 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 Estimate the diffusion coefficient of carbon monoxide through air in which the mole fractions of each constituents areO2= 0.18N2= 0.72CO = 0.10The gas mixture is at 300 K and 2 atmosphere total pressure. The diffusivity values areDCO= 18.5 * 10-6m2/s at 273 K, 1 atmDCN= 19.2 * 10-6m2/s at 288 K, 1 atmThe subscripts C is carbon monoxide, O is oxygen and N is nitrogena)10.29 * 10-6m2/sb)18.5 * 10-6m2/sc)17.5 * 10-6m2/sd)13.5 * 10-6m2/sCorrect 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 Estimate the diffusion coefficient of carbon monoxide through air in which the mole fractions of each constituents areO2= 0.18N2= 0.72CO = 0.10The gas mixture is at 300 K and 2 atmosphere total pressure. The diffusivity values areDCO= 18.5 * 10-6m2/s at 273 K, 1 atmDCN= 19.2 * 10-6m2/s at 288 K, 1 atmThe subscripts C is carbon monoxide, O is oxygen and N is nitrogena)10.29 * 10-6m2/sb)18.5 * 10-6m2/sc)17.5 * 10-6m2/sd)13.5 * 10-6m2/sCorrect answer is option 'A'. Can you explain this answer?.
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Estimate the diffusion coefficient of carbon monoxide through air in which the mole fractions of each constituents areO2= 0.18N2= 0.72CO = 0.10The gas mixture is at 300 K and 2 atmosphere total pressure. The diffusivity values areDCO= 18.5 * 10-6m2/s at 273 K, 1 atmDCN= 19.2 * 10-6m2/s at 288 K, 1 atmThe subscripts C is carbon monoxide, O is oxygen and N is nitrogena)10.29 * 10-6m2/sb)18.5 * 10-6m2/sc)17.5 * 10-6m2/sd)13.5 * 10-6m2/sCorrect answer is option 'A'. Can you explain this answer?, a detailed solution for Estimate the diffusion coefficient of carbon monoxide through air in which the mole fractions of each constituents areO2= 0.18N2= 0.72CO = 0.10The gas mixture is at 300 K and 2 atmosphere total pressure. The diffusivity values areDCO= 18.5 * 10-6m2/s at 273 K, 1 atmDCN= 19.2 * 10-6m2/s at 288 K, 1 atmThe subscripts C is carbon monoxide, O is oxygen and N is nitrogena)10.29 * 10-6m2/sb)18.5 * 10-6m2/sc)17.5 * 10-6m2/sd)13.5 * 10-6m2/sCorrect answer is option 'A'. Can you explain this answer? has been provided alongside types of Estimate the diffusion coefficient of carbon monoxide through air in which the mole fractions of each constituents areO2= 0.18N2= 0.72CO = 0.10The gas mixture is at 300 K and 2 atmosphere total pressure. The diffusivity values areDCO= 18.5 * 10-6m2/s at 273 K, 1 atmDCN= 19.2 * 10-6m2/s at 288 K, 1 atmThe subscripts C is carbon monoxide, O is oxygen and N is nitrogena)10.29 * 10-6m2/sb)18.5 * 10-6m2/sc)17.5 * 10-6m2/sd)13.5 * 10-6m2/sCorrect answer is option 'A'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice Estimate the diffusion coefficient of carbon monoxide through air in which the mole fractions of each constituents areO2= 0.18N2= 0.72CO = 0.10The gas mixture is at 300 K and 2 atmosphere total pressure. The diffusivity values areDCO= 18.5 * 10-6m2/s at 273 K, 1 atmDCN= 19.2 * 10-6m2/s at 288 K, 1 atmThe subscripts C is carbon monoxide, O is oxygen and N is nitrogena)10.29 * 10-6m2/sb)18.5 * 10-6m2/sc)17.5 * 10-6m2/sd)13.5 * 10-6m2/sCorrect answer is option 'A'. Can you explain this answer? tests, examples and also practice Chemical Engineering tests.
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