Answer: monoatomic

$C_V$ (heat absorbed per degree rise per mole) = $\frac{41.75J}{0.1 mol \times 20^\circ C} = 20.875 JK^{-1}mol^{-1}$

$C_P = C_V + R = 20.875 + 8.314 KJ^{-1}mol^{-1} = 29.189 JK^{-1}mol^{-1}$

$\frac{C_P}{C_V} = \frac{29.189}{20.875} = 1.40$

$\therefore$ The gas is monoatomic.