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A gas mixture containing 21% O2 and 79% N2 flows through a pipe of circular cross-section of diameter 2cm at atmospheric pressure. The velocity of O2 is 0.04m/s and N2 is 0.035m/s.
Q. What is the value of molar average velocity (m/s) of the mixture?
- a)0.036
- b)0.035
- c)0.034
- d)0.033
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
A gas mixture containing 21% O2and 79% N2flows through a pipe of circu...
Explanation: Molar average velocity= (C1u1+C2u2+……….)/C
=y1u1+y2u2…..
= 0.21*0.04+0.79*0.035
=0.036
=y1u1+y2u2…..
= 0.21*0.04+0.79*0.035
=0.036
Most Upvoted Answer
A gas mixture containing 21% O2and 79% N2flows through a pipe of circu...
To find the molar average velocity of the gas mixture, we need to consider the velocities of the individual gases (O2 and N2) and their respective mole fractions.
Given data:
- Gas mixture composition: 21% O2 and 79% N2
- Diameter of the pipe: 2 cm
- Velocities: O2 = 0.04 m/s, N2 = 0.035 m/s
1. Calculating the mole fractions:
The mole fraction of each gas can be calculated using the given composition. Let's denote the mole fraction of O2 as xO2 and N2 as xN2.
Given that the gas mixture contains 21% O2 and 79% N2, we can write:
xO2 = 0.21 and xN2 = 0.79
2. Calculating the molar average velocity:
The molar average velocity of the mixture can be calculated using the formula:
V̅ = ∑(xi * Vi)
Where V̅ is the molar average velocity, xi is the mole fraction of the gas, and Vi is the velocity of the gas.
In this case, we need to calculate the sum of the product of each mole fraction and velocity for O2 and N2.
V̅ = (xO2 * VO2) + (xN2 * VN2)
Substituting the values:
V̅ = (0.21 * 0.04) + (0.79 * 0.035)
V̅ = 0.0084 + 0.02765
V̅ = 0.03605 m/s
Therefore, the molar average velocity of the gas mixture is approximately 0.036 m/s.
Hence, the correct answer is option A) 0.036.
Given data:
- Gas mixture composition: 21% O2 and 79% N2
- Diameter of the pipe: 2 cm
- Velocities: O2 = 0.04 m/s, N2 = 0.035 m/s
1. Calculating the mole fractions:
The mole fraction of each gas can be calculated using the given composition. Let's denote the mole fraction of O2 as xO2 and N2 as xN2.
Given that the gas mixture contains 21% O2 and 79% N2, we can write:
xO2 = 0.21 and xN2 = 0.79
2. Calculating the molar average velocity:
The molar average velocity of the mixture can be calculated using the formula:
V̅ = ∑(xi * Vi)
Where V̅ is the molar average velocity, xi is the mole fraction of the gas, and Vi is the velocity of the gas.
In this case, we need to calculate the sum of the product of each mole fraction and velocity for O2 and N2.
V̅ = (xO2 * VO2) + (xN2 * VN2)
Substituting the values:
V̅ = (0.21 * 0.04) + (0.79 * 0.035)
V̅ = 0.0084 + 0.02765
V̅ = 0.03605 m/s
Therefore, the molar average velocity of the gas mixture is approximately 0.036 m/s.
Hence, the correct answer is option A) 0.036.
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A gas mixture containing 21% O2and 79% N2flows through a pipe of circular cross-section of diameter 2cm at atmospheric pressure. The velocity of O2is 0.04m/s and N2is 0.035m/s.Q. What is the value of molar average velocity (m/s) of the mixture?a)0.036b)0.035c)0.034d)0.033Correct answer is option 'A'. Can you explain this answer?
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A gas mixture containing 21% O2and 79% N2flows through a pipe of circular cross-section of diameter 2cm at atmospheric pressure. The velocity of O2is 0.04m/s and N2is 0.035m/s.Q. What is the value of molar average velocity (m/s) of the mixture?a)0.036b)0.035c)0.034d)0.033Correct 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 A gas mixture containing 21% O2and 79% N2flows through a pipe of circular cross-section of diameter 2cm at atmospheric pressure. The velocity of O2is 0.04m/s and N2is 0.035m/s.Q. What is the value of molar average velocity (m/s) of the mixture?a)0.036b)0.035c)0.034d)0.033Correct 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 A gas mixture containing 21% O2and 79% N2flows through a pipe of circular cross-section of diameter 2cm at atmospheric pressure. The velocity of O2is 0.04m/s and N2is 0.035m/s.Q. What is the value of molar average velocity (m/s) of the mixture?a)0.036b)0.035c)0.034d)0.033Correct answer is option 'A'. Can you explain this answer?.
A gas mixture containing 21% O2and 79% N2flows through a pipe of circular cross-section of diameter 2cm at atmospheric pressure. The velocity of O2is 0.04m/s and N2is 0.035m/s.Q. What is the value of molar average velocity (m/s) of the mixture?a)0.036b)0.035c)0.034d)0.033Correct 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 A gas mixture containing 21% O2and 79% N2flows through a pipe of circular cross-section of diameter 2cm at atmospheric pressure. The velocity of O2is 0.04m/s and N2is 0.035m/s.Q. What is the value of molar average velocity (m/s) of the mixture?a)0.036b)0.035c)0.034d)0.033Correct 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 A gas mixture containing 21% O2and 79% N2flows through a pipe of circular cross-section of diameter 2cm at atmospheric pressure. The velocity of O2is 0.04m/s and N2is 0.035m/s.Q. What is the value of molar average velocity (m/s) of the mixture?a)0.036b)0.035c)0.034d)0.033Correct answer is option 'A'. Can you explain this answer?.
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