irreversible adiabatic process against constant external pressure?
In this irreversible adiabetic process, only the work done on the system will be =$-P_{ext}dV$. Because in a reversible process we actually consider $P_{ext} = P_{int} + dP$, and hence we neglect the $dPdV $ term as it is negligible.  Considering this, if we go by the definition of adiabetic process and first law of thermodynamics, we will have $$0 = nC_vdT  + P_{ext}dV$$ as the heat changed in an adiabetic process is $0$. Now simply taking one term to other side and integrating we have $$nC_v \int_{T_1}^{T_2} dT = -P_{ext} \int_{V_1}^{V_2} dV$$  where $P_1,V_1,T_1$ are initial conditions, and $P_2, V_2,T_2$ are final conditions of the system.Now we put $V_1 = \frac{nRT_1}{P_1}$ and $V_2 = \frac{nRT_2}{P_2}$, so we will have $$C_v(T_2 -T_1) = -P_{ext}R(\frac{T_2}{P_2}-\frac{T_1}{P_1})$$ now taking $T_2$ associated terms to one side and $T_1$ related terms to other side we have $$T_2(C_v + \frac{P_{ext}R}{P_2}) = T_1( C_v + \frac{P_{ext}R}{P_1})$$ and thus we have our desired result as, $$T_2=[\frac{C_v + (\frac{P_{ext}}{P_1})R}{C_v + (\frac{P_{ext}}{P_2})R}]T_1$$