# 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, $$T_2=[\frac{C_v + (\frac{P_{ext}}{P_1})R}{C_v + (\frac{P_{ext}}{P_2})R}]T_1$$
The ideal gas law describes the behavior of an "ideal gas" only under thermodynamic equilibrium conditions. For a rapid irreversible expansion or compression, the gas is not close to thermodynamic equilibrium, and its behavior is much more complicated. It depends not only on the volume of the gas but also on the rate of change of volume. To a very crude approximation, the force F that the gas exerts on its surroundings (which, by Newton's 3rd law is also equal to the force the surroundings exert on the gas $P_{ext}A$) can be described by: $$\frac{F}{A}=\frac{nRT}{V}-\frac{k}{V}\frac{dV}{dt}=P_{ext}$$where k is proportional to the viscosity of the gas. This viscosity term in the equation is responsible for the irreversibility of the behavior. Note that at very slow rates of volumetric change, the equation reduces to the usual form of the ideal gas law.
If the external force of the surroundings on the gas is constant, the complexities associated with the above equation can be circumvented, and the work done by the gas on its surroundings can be directly calculated from $W=P_{ext}(V_2-V_1)$. If the expansion is stopped before the gas has had a chance to equilibrate, the gas can ultimately equilibrate to a final pressure $p_2$ different from $P_{ext}$. But that will not change the amount of work that has been done, since all displacement hss ceased once the final volume $V_2$ has been attained.
Between the initial and final thermodynamic equilibrium states of the system, we have from the 1st law of thermodynamics, $$\Delta U=nC_v(T_2-T_1)=-P_{ext}(V_2-V_1)$$