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

Answer: (A)

$\frac{9}{4}R$

Answer 2

For a monatomic gas, the molar specific heat at constant volume is $C_V = \frac{3}{2}R$, and for a diatomic gas, $C_V = \frac{5}{2}R$. The total heat capacity $C_V$ of the mixture is given by the weighted average formula:

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

Where:

• $n_1 = 2$ is the number of moles of the monatomic gas.
• $n_2 = 6$ is the number of moles of the 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 values into the equation:

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

Thus, the molar specific heat of the mixture at constant volume is $\frac{9}{4}R$.