Two moles a monoatomic gas is mixed with six moles of a diatomic gas. The molar specific heat of the mixture at constant volume is:

### Related Formula Molar specific heat at constant volume for a gas mixture: C_Vtext, mix = fracn_1 C_V1 + n_2 C_V2n_1 + n_2$$C_{V\text{, mix}} = \frac{n_1 C_{V1} + n_2 C_{V2}}{n_1 + n_2}$$ For a monoatomic gas: C_V1 = frac32R$$C_{V1} = \frac{3}{2}R$$ For a diatomic gas: C_V2 = frac52R$$C_{V2} = \frac{5}{2}R$$ ### Core Logic Given values: n_1 = 2$n_1 = 2$ (monoatomic), n_2 = 6$n_2 = 6$ (diatomic). Substitute these inputs directly into the mixture equation: C_Vtext, mix = frac2 times left(frac32Rright) + 6 times left(frac52Rright)2 + 6$$C_{V\text{, mix}} = \frac{2 \times \left(\frac{3}{2}R\right) + 6 \times \left(\frac{5}{2}R\right)}{2 + 6}$$ ### Step 1: Simplify Expression C_Vtext, mix = frac3R + 15R8 = frac18R8 = frac94R$$C_{V\text{, mix}} = \frac{3R + 15R}{8} = \frac{18R}{8} = \frac{9}{4}R$$ ### Pattern Recognition Weighted average rule based on internal degrees of freedom: total internal energy changes scale additively with mole numbers. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Kinetic Theory of Gases Class 11 Physics: Thermodynamics