Three identical adiabatic containers A , B , and C hold helium,...

For an adiabatic process, the pressure and volume are related by $PV^{\gamma} = \text{constant}$, and temperature and volume are related by $TV^{\gamma-1} = \text{constant}$.

Let the initial pressure, volume, and temperature of the gases be $P_0$, $V_0$, and $T_0$ respectively. The final volume is $V = \dfrac{V_0}{2}$.

The final pressure is given by $P = P_0 \left(\dfrac{V_0}{V}\right)^{\gamma} = P_0 (2)^{\gamma}$.

The final temperature is given by $T = T_0 \left(\dfrac{V_0}{V}\right)^{\gamma-1} = T_0 (2)^{\gamma-1}$.

Helium and neon are monoatomic gases, so their ratio of specific heats is $\gamma = \dfrac{5}{3}$.

Oxygen is a diatomic gas, so its ratio of specific heats is $\gamma = \dfrac{7}{5}$.

Since $\gamma$ is the same for helium and neon, their final pressures and final temperatures will be identical.

Since $\gamma$ for oxygen is different, its final pressure and final temperature will differ from those of helium and neon.