The correct relation between $$\gamma = \dfrac{C_P}{C_V}$$ and temperature $$T$$ is:

The ratio $$\gamma = \frac{C_P}{C_V}$$ for an ideal gas.

For an ideal gas, $$C_P = C_V + R$$, where $$R$$ is the universal gas constant.

$$C_V = \frac{f}{2}R$$ where $$f$$ is the degrees of freedom.

$$\gamma = \frac{C_P}{C_V} = 1 + \frac{2}{f}$$

For an ideal gas, $$f$$ is a constant (depends only on the type of gas - monatomic, diatomic, etc.), and does not depend on temperature.

Therefore $$\gamma$$ is independent of temperature: $$\gamma \propto T^0$$.

(Note: In reality, at very high temperatures, vibrational modes get activated, changing $$f$$, but in the standard kinetic theory treatment, $$\gamma$$ is constant.)

The correct answer is Option (2): $$\boxed{\gamma \propto T^0}$$.