JEE Main 2022 · 29 Jul · Shift 1 · Physics Q54

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PhysicsInteger+4 / -1Question 54 of 90

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

Your answer

Answer and worked solution

Official answer: 3

Mixture. $C_V=\dfrac{1\cdot\frac32R+3\cdot\frac52R}{4}=\dfrac94R$ , so $\alpha^2=9$ , $\alpha=3$ .

Answer: 3

Topics:Specific heats of gases

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JEE Main 2022 · 29 Jul · Shift 1, Physics question 54