We are given the ratio $\frac{R}{C_V} = 0.67 \approx \frac{2}{3}$. Rearranging this, we get $C_V = \frac{3}{2}R$. For an ideal gas, the molar heat capacity at constant volume ($C_V$) depends on the degrees of freedom ($f$) according to the relation $C_V = \frac{f}{2}R$. Comparing the values:

• For a monoatomic gas, $f=3$, so $C_V = \frac{3}{2}R$ .
• For a diatomic gas, $f=5$, so $C_V = \frac{5}{2}R$.
• For a polyatomic gas, $f \ge 6$, so $C_V \ge 3R$. Since the calculated $C_V$ matches that of a monoatomic gas, the gas is monoatomic. Alternatively, using the adiabatic index $\gamma$: $C_P - C_V = R \implies \gamma - 1 = \frac{R}{C_V} = 0.67$. Thus $\gamma = 1.67$, which corresponds to a monoatomic gas.