Kinetic Theory

JEE Main level, numerical answer, medium

81 of 168

Question 81 of 168

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 α24RJmolK−1\dfrac{{\alpha }^{2}}{4}RJmol{K}^{-1}; then the value of α\alpha will be _____ . (Assume that the given diatomic gas has no vibrational mode.)

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Answer: 3

Molecular specific heat at constant volume in terms of degrees of freedom is given by CV=f2R{C}_{V}=\dfrac{f}{2}R
Molecular specific heat of mixture at constant volume is CV/mix=n1CV1+n2CV2n1+n2{C}_{V}/mix=\dfrac{{n}_{1}{C}_{{V}_{1}}+{n}_{2}{C}_{{V}_{2}}}{{n}_{1}+{n}_{2}}
=1×3R2+3×5R21+3=\dfrac{1\times \dfrac{3R}{2}+3\times \dfrac{5R}{2}}{1+3}
=9R4=α24R=\dfrac{9R}{4}=\dfrac{{\alpha }^{2}}{4}R
⇒α=3\Rightarrow \alpha =3

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