JEE Mains · Physics · STD 11 - 12 . kinetic theory of gases
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 \(\frac{\alpha^{2}}{4} R J / mol\,K\); then the value of \(\alpha\) will be \(.......\) (Assume that the given diatomic gas has no vibrational mode.)
- A \(2\)
- B \(5\)
- C \(8\)
- D \(3\)
Answer & Solution
Correct Answer
(D) \(3\)
Step-by-step Solution
Detailed explanation
\(C _{ V_{mix }} =\frac{ n _{1} Cv _{1}+ n _{2} Cv _{2}}{ n _{1}+ n _{2}}\) \(=\frac{1 \cdot \frac{3 R }{2}+3 \cdot \frac{5 R }{2}}{1+3}\) \(=\frac{9 R }{4}=\frac{\alpha^{2}}{4} R\) \(\alpha=3\)
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