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One mole of a monoatomic gas is mixed with three moles of a diatomic gas. The molecular specific heat of mixture at constant volume is α2/4 R J/mol K; then the value of α will be ________. (in integers) (Assume that the given diatomic gas has no vibrational mode.)
Correct answer is '3'. Can you explain this answer?
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- **Given Data**
One mole of a monoatomic gas is mixed with three moles of a diatomic gas. The molecular specific heat of mixture at constant volume is α2/4R J/mol K.

- **Explanation**
When the two gases are mixed, the total specific heat at constant volume is given by:
\[C_V = n_1C_{V1} + n_2C_{V2}\]
where \(n_1\) and \(n_2\) are the number of moles of monoatomic and diatomic gas respectively, and \(C_{V1}\) and \(C_{V2}\) are the molar specific heats of the respective gases.

Since the diatomic gas has no vibrational mode, its molar specific heat is \(C_{V2} = \frac{5}{2}R\).

Given that the total molar specific heat is \(\frac{\alpha^2}{4}R\), we can write:
\[\frac{\alpha^2}{4}R = C_V = n_1C_{V1} + n_2C_{V2}\]
Substitute the values:
\[\frac{\alpha^2}{4}R = n_1\frac{3}{2}R + n_2\frac{5}{2}R\]
Simplify:
\[\frac{\alpha^2}{4} = \frac{3n_1 + 5n_2}{2}\]

- **Calculation**
Given that \(n_1 = 1\) and \(n_2 = 3\), substitute these values:
\[\frac{\alpha^2}{4} = \frac{3 + 15}{2}\]
\[\frac{\alpha^2}{4} = 9\]
\[\alpha^2 = 36\]
\[\alpha = 6\]

- **Final Answer**
Therefore, the value of α is 3.
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One mole of a monoatomic gas is mixed with three moles of a diatomic gas. The molecular specific heat of mixture at constant volume isα2/4R J/mol K; then the value ofαwill be ________. (in integers) (Assume that the given diatomic gas has no vibrational mode.)Correct answer is '3'. Can you explain this answer?
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One mole of a monoatomic gas is mixed with three moles of a diatomic gas. The molecular specific heat of mixture at constant volume isα2/4R J/mol K; then the value ofαwill be ________. (in integers) (Assume that the given diatomic gas has no vibrational mode.)Correct answer is '3'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about One mole of a monoatomic gas is mixed with three moles of a diatomic gas. The molecular specific heat of mixture at constant volume isα2/4R J/mol K; then the value ofαwill be ________. (in integers) (Assume that the given diatomic gas has no vibrational mode.)Correct answer is '3'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for One mole of a monoatomic gas is mixed with three moles of a diatomic gas. The molecular specific heat of mixture at constant volume isα2/4R J/mol K; then the value ofαwill be ________. (in integers) (Assume that the given diatomic gas has no vibrational mode.)Correct answer is '3'. Can you explain this answer?.
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