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 :

<p>To calculate the molar specific heat at constant volume of a gas mixture, we can use a weighted average based on the molar specific heats of the individual gases and their respective amounts (moles).</p> <p>Let's call $C_{V,m}$ the molar specific heat at constant volume of the mixture, $n_1$ the number of moles of the monoatomic gas, $n_2$ the number of moles of the diatomic gas, $C_{V,m1}$ the molar specific heat at constant volume of the monoatomic gas, and $C_{V,m2}$ the molar specific heat at constant volume of the diatomic gas.</p> <p>For a monoatomic ideal gas, the molar specific heat at constant volume is:</p> <p>$C_{V,m1} = \frac{3}{2}R$</p> <p>For a diatomic ideal gas, if we assume the gas is rigid and does not exhibit vibrational modes, the molar specific heat at constant volume is:</p> <p>$C_{V,m2} = \frac{5}{2}R$</p> <p>The weighted average of the molar specific heat for the mixture is:</p> <p>$C_{V,m} = \frac{(n_1 \cdot C_{V,m1} + n_2 \cdot C_{V,m2})}{n_1 + n_2}$</p> <p>Putting the values into the equation, we get:</p> <p>$C_{V,m} = \frac{(2 \cdot \frac{3}{2}R + 6 \cdot \frac{5}{2}R)}{2 + 6}$</p> <p>$C_{V,m} = \frac{(3R + 15R)}{8}$</p> <p>$C_{V,m} = \frac{18R}{8}$</p> <p>$C_{V,m} = \frac{9}{4}R$</p> <p>Therefore, the molar specific heat of the mixture at constant volume is $\frac{9}{4}R$. The correct answer is:</p> <p>Option D</p> <p>$\frac{9}{4}R$</p>