1. Use ideal gas relations

For any ideal gas, $C_P - C_V = R$ where $R$ is the universal gas constant.

So this difference is the same for all ideal gases, independent of whether the gas is monoatomic or diatomic.

Hence, Option A is false.

2. Write $C_V$ and $C_P$ for monoatomic and diatomic ideal gases

For a monoatomic ideal gas: $C_V = \frac{3}{2}R, \qquad C_P = C_V + R = \frac{5}{2}R$

For a diatomic ideal gas (neglecting vibrational modes, as usual): $C_V = \frac{5}{2}R, \qquad C_P = C_V + R = \frac{7}{2}R$

3. Check Option B: $C_P + C_V$

For monoatomic gas: $C_P + C_V = \frac{5}{2}R + \frac{3}{2}R = 4R$

For diatomic gas: $C_P + C_V = \frac{7}{2}R + \frac{5}{2}R = 6R$

Since $6R > 4R$ Option B is true.

4. Check Option C: $\dfrac{C_P}{C_V}$

For monoatomic gas: $\gamma = \frac{C_P}{C_V} = \frac{\frac{5}{2}R}{\frac{3}{2}R} = \frac{5}{3}$

For diatomic gas: $\gamma = \frac{C_P}{C_V} = \frac{\frac{7}{2}R}{\frac{5}{2}R} = \frac{7}{5}$

Now, $\frac{5}{3} \approx 1.67, \qquad \frac{7}{5} = 1.4$

So, $\frac{C_P}{C_V}$ is smaller for a diatomic gas than for a monoatomic gas.

Hence, Option C is false.

5. Check Option D: $C_P \cdot C_V$

For monoatomic gas: $C_P C_V = \frac{5}{2}R \cdot \frac{3}{2}R = \frac{15}{4}R^2$

For diatomic gas: $C_P C_V = \frac{7}{2}R \cdot \frac{5}{2}R = \frac{35}{4}R^2$

Since $\frac{35}{4}R^2 > \frac{15}{4}R^2$ Option D is true.

6. Final answer

The correct options are: $\boxed{B, D}$

7. Comparison with stored answer

Stored correct answer: B, D

This matches our derived answer exactly.