Introduction:
In this question, we are given a mixture of one mole of a monoatomic ideal gas and one mole of a diatomic ideal gas. We need to determine the molar specific heat of the mixture at constant volume.

Explanation:
To solve this problem, we need to consider the individual molar specific heats of the monoatomic and diatomic gases separately.

Monoatomic Ideal Gas:
A monoatomic ideal gas consists of atoms that do not have any internal degrees of freedom. The molar specific heat of a monoatomic ideal gas at constant volume (Cv) is given by the equation:

Cv = (3/2)R

Here, R is the ideal gas constant. The monoatomic gas has three degrees of freedom associated with the three translational directions in space.

Diatomic Ideal Gas:
A diatomic ideal gas consists of molecules that have two atoms bonded together. In addition to the translational degrees of freedom, diatomic molecules also possess rotational degrees of freedom. The molar specific heat of a diatomic ideal gas at constant volume (Cv) is given by the equation:

Cv = (5/2)R

Here, R is the ideal gas constant. The diatomic gas has five degrees of freedom, which include three translational and two rotational degrees of freedom.

Mixture of Monoatomic and Diatomic Gases:
When the monoatomic and diatomic gases are mixed together, the total molar specific heat at constant volume (Cv) is the weighted sum of the individual molar specific heats of the gases, based on their mole fractions.

In this case, since we have equal moles of monoatomic and diatomic gases, the mole fractions of each gas in the mixture are 0.5. Therefore, the molar specific heat of the mixture at constant volume is given by:

Cv = (0.5 * (3/2)R) + (0.5 * (5/2)R)
= (3/4)R + (5/4)R
= 2R

Since R is the ideal gas constant, the molar specific heat of the mixture at constant volume is 2. Therefore, the correct answer is '2'.

Summary:
When one mole of a monoatomic ideal gas is mixed with one mole of a diatomic ideal gas, the molar specific heat of the mixture at constant volume is 2. This is because the molar specific heat of the monoatomic gas is (3/2)R and the molar specific heat of the diatomic gas is (5/2)R. The total molar specific heat of the mixture is the weighted sum of these values based on the mole fractions of each gas in the mixture. Since the mole fractions are equal in this case, the molar specific heat of the mixture is 2R.