Question

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 :

• A. \(\frac{9}{4}R\)
• B. \(\frac{7}{4}R\)
• C. \(\frac{3}{2}R\)
• D. \(\frac{5}{2}R\)

Answer 1

The Correct Option is A

The molar specific heat at constant volume for a monatomic gas is \(C_V = \frac{3}{2}R\), and for a diatomic gas, it is \(C_V = \frac{5}{2}R\). The total molar specific heat \(C_V\) for a mixture is calculated using the weighted average:

\[ C_V = \frac{n_1C_{V1} + n_2C_{V2}}{n_1 + n_2} \]

Where:

• \(n_1 = 2\) moles of monatomic gas.
• \(n_2 = 6\) moles of diatomic gas.
• \(C_{V1} = \frac{3}{2}R\) is the molar specific heat of the monatomic gas.
• \(C_{V2} = \frac{5}{2}R\) is the molar specific heat of the diatomic gas.

Substituting the given values:

\[ C_V = \frac{2 \times \frac{3}{2}R + 6 \times \frac{5}{2}R}{2 + 6} = \frac{3R + 15R}{8} = \frac{18R}{8} = \frac{9}{4}R \]

Therefore, the molar specific heat of the mixture at constant volume is \(\frac{9}{4}R\).