Answer
monoatomic
Work Step by Step
We know that;
$\gamma=\frac{\ln(\frac{P_i}{P_f})}{\ln(\frac{V_f}{V_i})}$
We plug in the known values to obtain:
$\gamma=\frac{\ln(\frac{1.0}{1.0\times 10^5})}{\ln(\frac{1.0\times 10^3}{1.0\times 10^6})}=\frac{5}{3}$
The value of $\gamma$ suggests that the gas is monoatomic.
