Final answer:
The molar heat capacity at constant volume for a mixture of 2 moles of monoatomic ideal gas and 3 moles of diatomic ideal gas is 21½R or 4 R when calculated using the equipartition of energy principle.
Step-by-step explanation:
To calculate the molar heat capacity at constant volume for a mixture containing 2 moles of monoatomic and 3 moles of diatomic ideal gas, we use the equipartition theorem which states that each degree of freedom contributes ½R to the molar heat capacity at constant volume. A monoatomic ideal gas has 3 degrees of translational freedom, contributing 3½R, and a diatomic ideal gas has 3 translational and 2 rotational degrees of freedom, contributing 5½R. Since vibrational motions are generally not considered at room temperature due to low energy, they are not accounted for here. Thus, two moles of monoatomic gas contribute 2 × 3½R, and three moles of diatomic gas contribute 3 × 5½R. Adding these contributions together, the total molar heat capacity at constant volume is:
Cv = (2 × 3½R) + (3 × 5½R)
Cv = 6½R + 15½R
Cv = 21½R
In terms with the universal gas constant R,
Cv = 21½ × 8.314 J/(mol·K)
Cv = 4 R, where R is the universal gas constant.