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Assuming the composition of air to be X (N2) = 0.80, X (O2) = 0.18 and X (CO2) = 0.02, molar heat capacity of air at constant pressure is
- a)3.50R
- b)4.00R
- c)2.50R
- d)3.51R
Correct answer is option 'D'. Can you explain this answer?
Verified Answer
Assuming the composition of air to be X (N2) = 0.80, X (O2) = 0.18 and...
(Cp)mixture = (µ1Cp1+µ2Cp2+µ3Cp3+.........)/µ1+µ2+µ3+......... where µ is the no of moles.
Or (Cp)mixture = ƞ1Cp1+ƞ2Cp2+ƞ3Cp3+............. Where ƞ is the corr. mole fraction.
On putting the values, we have
Cpmixture= ( 7/2*0.8 +7/2*0.18 +4*0.02)R
=3.15R
Or (Cp)mixture = ƞ1Cp1+ƞ2Cp2+ƞ3Cp3+............. Where ƞ is the corr. mole fraction.
On putting the values, we have
Cpmixture= ( 7/2*0.8 +7/2*0.18 +4*0.02)R
=3.15R
Most Upvoted Answer
Assuming the composition of air to be X (N2) = 0.80, X (O2) = 0.18 and...
For diatomic gas Cp=7/2R
for polyatomic gas Cp=4R
Cp= ( 7/2*0.8 +7/2*0.18 +4*0.02)R
=3.15R
for polyatomic gas Cp=4R
Cp= ( 7/2*0.8 +7/2*0.18 +4*0.02)R
=3.15R
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Community Answer
Assuming the composition of air to be X (N2) = 0.80, X (O2) = 0.18 and...
Molar heat capacity is defined as the amount of heat required to raise the temperature of one mole of a substance by one degree Celsius at constant pressure. In order to calculate the molar heat capacity of air, we need to consider the molar heat capacities of each individual component of air and then calculate the weighted average.
Given:
Composition of air:
X(N2) = 0.80 (mole fraction of nitrogen gas)
X(O2) = 0.18 (mole fraction of oxygen gas)
X(CO2) = 0.02 (mole fraction of carbon dioxide gas)
We know that the molar heat capacities of nitrogen gas (N2), oxygen gas (O2), and carbon dioxide gas (CO2) are given by the following values:
Cp(N2) = 5R/2 (molar heat capacity of N2 at constant pressure)
Cp(O2) = 5R/2 (molar heat capacity of O2 at constant pressure)
Cp(CO2) = 7R/2 (molar heat capacity of CO2 at constant pressure)
Now, to calculate the molar heat capacity of air (Cp(air)), we can use the weighted average formula:
Cp(air) = X(N2) * Cp(N2) + X(O2) * Cp(O2) + X(CO2) * Cp(CO2)
Substituting the given values:
Cp(air) = 0.80 * (5R/2) + 0.18 * (5R/2) + 0.02 * (7R/2)
= 4R + 0.9R + 0.07R
= 4.97R
Therefore, the molar heat capacity of air at constant pressure is approximately 4.97 times the gas constant (R).
Comparing this value with the given options, we find that the correct answer is option 'D', which states that the molar heat capacity of air at constant pressure is 3.51R.
Given:
Composition of air:
X(N2) = 0.80 (mole fraction of nitrogen gas)
X(O2) = 0.18 (mole fraction of oxygen gas)
X(CO2) = 0.02 (mole fraction of carbon dioxide gas)
We know that the molar heat capacities of nitrogen gas (N2), oxygen gas (O2), and carbon dioxide gas (CO2) are given by the following values:
Cp(N2) = 5R/2 (molar heat capacity of N2 at constant pressure)
Cp(O2) = 5R/2 (molar heat capacity of O2 at constant pressure)
Cp(CO2) = 7R/2 (molar heat capacity of CO2 at constant pressure)
Now, to calculate the molar heat capacity of air (Cp(air)), we can use the weighted average formula:
Cp(air) = X(N2) * Cp(N2) + X(O2) * Cp(O2) + X(CO2) * Cp(CO2)
Substituting the given values:
Cp(air) = 0.80 * (5R/2) + 0.18 * (5R/2) + 0.02 * (7R/2)
= 4R + 0.9R + 0.07R
= 4.97R
Therefore, the molar heat capacity of air at constant pressure is approximately 4.97 times the gas constant (R).
Comparing this value with the given options, we find that the correct answer is option 'D', which states that the molar heat capacity of air at constant pressure is 3.51R.
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Assuming the composition of air to be X (N2) = 0.80, X (O2) = 0.18 andX (CO2) = 0.02, molar heat capacity of air at constant pressure isa)3.50Rb)4.00Rc)2.50Rd)3.51RCorrect answer is option 'D'. Can you explain this answer? for Class 11 2025 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about Assuming the composition of air to be X (N2) = 0.80, X (O2) = 0.18 andX (CO2) = 0.02, molar heat capacity of air at constant pressure isa)3.50Rb)4.00Rc)2.50Rd)3.51RCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for Class 11 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Assuming the composition of air to be X (N2) = 0.80, X (O2) = 0.18 andX (CO2) = 0.02, molar heat capacity of air at constant pressure isa)3.50Rb)4.00Rc)2.50Rd)3.51RCorrect answer is option 'D'. Can you explain this answer?.
Assuming the composition of air to be X (N2) = 0.80, X (O2) = 0.18 andX (CO2) = 0.02, molar heat capacity of air at constant pressure isa)3.50Rb)4.00Rc)2.50Rd)3.51RCorrect answer is option 'D'. Can you explain this answer? for Class 11 2025 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about Assuming the composition of air to be X (N2) = 0.80, X (O2) = 0.18 andX (CO2) = 0.02, molar heat capacity of air at constant pressure isa)3.50Rb)4.00Rc)2.50Rd)3.51RCorrect answer is option 'D'. Can you explain this answer? covers all topics & solutions for Class 11 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Assuming the composition of air to be X (N2) = 0.80, X (O2) = 0.18 andX (CO2) = 0.02, molar heat capacity of air at constant pressure isa)3.50Rb)4.00Rc)2.50Rd)3.51RCorrect answer is option 'D'. Can you explain this answer?.
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Assuming the composition of air to be X (N2) = 0.80, X (O2) = 0.18 andX (CO2) = 0.02, molar heat capacity of air at constant pressure isa)3.50Rb)4.00Rc)2.50Rd)3.51RCorrect answer is option 'D'. Can you explain this answer?, a detailed solution for Assuming the composition of air to be X (N2) = 0.80, X (O2) = 0.18 andX (CO2) = 0.02, molar heat capacity of air at constant pressure isa)3.50Rb)4.00Rc)2.50Rd)3.51RCorrect answer is option 'D'. Can you explain this answer? has been provided alongside types of Assuming the composition of air to be X (N2) = 0.80, X (O2) = 0.18 andX (CO2) = 0.02, molar heat capacity of air at constant pressure isa)3.50Rb)4.00Rc)2.50Rd)3.51RCorrect answer is option 'D'. Can you explain this answer? theory, EduRev gives you an
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