irreversible adiabatic process against constant external pressure? My book has these formulae for irreversible adiabatic process against a constant external pressure $p_ext$ : $$ T_f = [\frac{C_v+(p_{ext}/p_1)R}{C_v+(p_{ext}/p_2)R}]T_i$$. It is also mentioned that when external pressure = p2 $$ T_f = [\frac{C_v+(p_{ext}/p_1)R}{C_p}]T_i$$. How do I get these formulae? 
I am getting confused because they are saying external pressure. In general when irreversible work don't we take into account only the external pressure and not the internal pressure? And why exactly is this so? I have assumed so far that this is because the internal pressure differs only slightly from the external pressure. Is this so? Would someone please explain this?
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, $$T_2=[\frac{C_v + (\frac{P_{ext}}{P_1})R}{C_v + (\frac{P_{ext}}{P_2})R}]T_1$$