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).

Thermodynamics: JEE Main 2022 Physics question

JEE Main 2022 · 29 Jul, Morning shift · Q63

JEE Main 202229 Jul, Morning shiftPhysicsMediumThermodynamics163× since 2002

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} \mathrm{R} \,\mathrm{J} / \mathrm{mol} \,\mathrm{K}

; then the value of

\alpha

will be _________. (Assume that the given diatomic gas has no vibrational mode).

Your answerType your answer. Integers and decimals are accepted.

JEE Main 2022 Thermodynamics Question with Answer (29 Jul Morning shift, Q63) | JEE Arc

JEE Main 2022 · 29 Jul, Morning shift · Q63

JEE Main 202229 Jul, Morning shiftPhysicsMediumThermodynamics163× since 2002

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} \mathrm{R} \,\mathrm{J} / \mathrm{mol} \,\mathrm{K}

; then the value of

\alpha

will be _________. (Assume that the given diatomic gas has no vibrational mode).

Your answerType your answer. Integers and decimals are accepted.