Since there is no exchange of heat $(\Delta Q = 0)$ and the process is reversible, the process is an adiabatic process. This concept is extremely important for JEE Main, JEE Advanced and IIT JEE.
For an adiabatic process of an ideal gas:
$$ TV^{\gamma - 1} = \text{constant} $$
Given:
$$ \frac{V_2}{V_1} = 8 $$
$$ \frac{T_2}{T_1} = \frac{1}{4} $$
Using adiabatic relation:
$$ \frac{T_2}{T_1} = \left(\frac{V_1}{V_2}\right)^{\gamma - 1} $$
$$ \frac{1}{4} = \left(\frac{1}{8}\right)^{\gamma - 1} $$
Writing in powers of 2:
$$ \frac{1}{4} = \left(\frac{1}{2^3}\right)^{\gamma - 1} = 2^{-3(\gamma - 1)} $$
$$ 2^{-2} = 2^{-3(\gamma - 1)} $$
Comparing powers:
$$ 3(\gamma - 1) = 2 $$
$$ \gamma = \frac{5}{3} $$
The value $\gamma = \frac{5}{3}$ corresponds to a monoatomic gas.
Among the given options, Helium (He) is a monoatomic gas.
Therefore, the correct answer is:
He