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The value of Cv for 1 mol of N2 gas predicted from the principle of equipartition of energy, ignoring vibrational contribution, is ______ J K–1 mol–1 (rounded up to two decimal places). [R = 8.3 JK–1mol–1]
Correct answer is between '20.00,21.00'. Can you explain this answer?
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The value of Cv for 1 mol of N2 gas predicted from the principle of eq...
The principle of equipartition of energy states that each degree of freedom of a molecule has an average energy of 1/2 RT. For a diatomic gas like N2, there are 5 degrees of freedom: 3 translational and 2 rotational. Therefore, the average energy per mole of N2 gas is:
(5/2) RT = (5/2) (8.31 J/mol K) (1 mol) = 20.8 J/K
The molar heat capacity at constant volume, Cv, is defined as the rate of change of internal energy with respect to temperature, holding volume constant. Since the internal energy of a gas depends only on its temperature, we can write:
Cv = dU/dT
Taking the derivative of the above expression, we get:
dU = Cv dT
Since U = (5/2) RT, we can write:
dU = (5/2) R dT
Comparing the two expressions for dU, we get:
Cv dT = (5/2) R dT
Cv = (5/2) R = (5/2) (8.31 J/mol K) = 20.8 J/K
Therefore, the value of Cv for 1 mol of N2 gas predicted from the principle of equipartition of energy, ignoring vibrational contribution, is 20.8 J/K.
(5/2) RT = (5/2) (8.31 J/mol K) (1 mol) = 20.8 J/K
The molar heat capacity at constant volume, Cv, is defined as the rate of change of internal energy with respect to temperature, holding volume constant. Since the internal energy of a gas depends only on its temperature, we can write:
Cv = dU/dT
Taking the derivative of the above expression, we get:
dU = Cv dT
Since U = (5/2) RT, we can write:
dU = (5/2) R dT
Comparing the two expressions for dU, we get:
Cv dT = (5/2) R dT
Cv = (5/2) R = (5/2) (8.31 J/mol K) = 20.8 J/K
Therefore, the value of Cv for 1 mol of N2 gas predicted from the principle of equipartition of energy, ignoring vibrational contribution, is 20.8 J/K.
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The value of Cv for 1 mol of N2 gas predicted from the principle of equipartition of energy, ignoring vibrational contribution, is ______ J K–1 mol–1 (rounded up to two decimal places). [R = 8.3 JK–1mol–1]Correct answer is between '20.00,21.00'. Can you explain this answer?
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The value of Cv for 1 mol of N2 gas predicted from the principle of equipartition of energy, ignoring vibrational contribution, is ______ J K–1 mol–1 (rounded up to two decimal places). [R = 8.3 JK–1mol–1]Correct answer is between '20.00,21.00'. Can you explain this answer? for IIT JAM 2024 is part of IIT JAM preparation. The Question and answers have been prepared according to the IIT JAM exam syllabus. Information about The value of Cv for 1 mol of N2 gas predicted from the principle of equipartition of energy, ignoring vibrational contribution, is ______ J K–1 mol–1 (rounded up to two decimal places). [R = 8.3 JK–1mol–1]Correct answer is between '20.00,21.00'. Can you explain this answer? covers all topics & solutions for IIT JAM 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The value of Cv for 1 mol of N2 gas predicted from the principle of equipartition of energy, ignoring vibrational contribution, is ______ J K–1 mol–1 (rounded up to two decimal places). [R = 8.3 JK–1mol–1]Correct answer is between '20.00,21.00'. Can you explain this answer?.
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