Expansion under isobaric adiabatic condition
If, at time = 0, we suddenly drop the external force per unit area to $P_{\text{ext}}$ and hold it at this value until our adiabatic system equilibriates, the first law of thermodynamics for this situation reads:

$$nC_\mathrm v(T_2-T_1)=-P_{\text{ext}}(V_2-V_1)=-\frac{nRT_2}{V_2}(V_2-V_1)=-nRT_2\left[1-\frac{1}{8}\right]$$

Solving this equation for $T_2$ yields: $$T_2=\frac{T_1}{\left[1+\frac{7}{8}\frac{R}{C_\mathrm v}\right]}$$

But, for $C_v=3R$, this gives:$$T_2=0.774T_1=\pu{232 K}$$

So, $$\Delta H=3(4R)(232-300)=\pu{-6.78 kJ}$$

So the final temperature is higher than in the adiabatic reversible case, and the decrease in enthalpy is less.