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 _________. (Assume that the given diatomic gas has no vibrational mode). Numerical answer type. Work it out, then open the answer.

Answer: 3. CVmix = (1 × 3/2R + 3 × 5/2R)/(4) = 9R/4 (α²)/(4)R = 9/4R ⇒ α² = 9 So α = 3.

One mole of a monoatomic gas is mixed with three moles of a diatomic… | JEE Main 2022 (29 Jul, Shift 1)

Q33JEE Main 2022 · 29 JUL · SHIFT 1Medium

Specific Heats of Gases - Cp, Cv, γ & Mayer's Relation

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 $\frac{\alpha^2}{4}R\text{ J/mol K}$ ; then the value of $\alpha$ will be _________. (Assume that the given diatomic gas has no vibrational mode).

Numerical answer type. Work it out, then open the answer.

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Answer: 3

$C_V^{mix} = \dfrac{1 \times \frac{3}{2}R + 3 \times \frac{5}{2}R}{4} = \dfrac{9R}{4}$
$\dfrac{\alpha^2}{4}R = \dfrac{9}{4}R \Rightarrow \alpha^2 = 9$
So $\alpha = 3$ .

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