Answer
The gas is monatomic. See explanation
Work Step by Step
Process $D \rightarrow A$ is adiabatic which means the pressure and volume remains the same before and after the process.
$p_DV_D^{\gamma} = p_AV_A^{\gamma}$
From the graph we can conclude that
$V_D = 8.0V_0$ and $p_D = \frac{p_0}{32}$
and $V_A = V_0$ and $p_A = p_0$
Now we need to know what is the value of $\gamma$ to categorize it in which type of gas it is. Substitute all the values and solve for $\gamma$
$\frac{p_0}{32} 8.0V_0^{\gamma} = p_0V_0^{\gamma}$
$ \frac{8.0V_0^{\gamma}}{V_0^{\gamma} } = \frac{32 p_ 0}{p_0}$
$ 8.0^{\gamma}= 32$
$\gamma = \frac{5}{3}$
From the result, it is known that the gas is monatomic since it has the $\gamma$ value of $\frac{5}{3}$
