OK. Here are the focus problems I recommend considering:
I have an ideal gas of pressure, volume, and temperature $P_1$, $V_1$, and $T_1$, respectively, in an insulated cylinder with a massless, frictionless piston. Initially, the external pressure is also $P_1$.
REVERSIBLE ADIABATIC EXPANSION
I gradually lower the external pressure (reversibly) until the volume has increased to $V_2$. Determine the final pressure $P_2$ and final temperature $T_2$. Determine the amount of work done on the surroundings W and the change in internal energy $\Delta U$. How does the amount of work compare with the change in internal energy?
IRREVERSIBLE ADIABATIC EXPANSION:
I suddenly lower the external pressure to a new value P and hold it constant at this value until the system re-equilibrates. In terms of P, what is the final volume and final temperature? What value of P would be required for the final volume to be the same as it was in the reversible case, $V_2$, and what would be the final temperature under these circumstances? What would be the work done on the surroundings W and what would be the change in internal energy $\Delta U$. How does the irreversible work compare with the irreversible change in internal energy? How does the work done on the surroundings in this irreversible case compare with the work done in the reversible case?
SOLUTION TO THE IRREVERSIBLE CASE:
The first law tells us that, for an adiabatic process, Q = 0 and $$\Delta U=-W$$So, for the irreversible expansion described here: $$nC_v(T-T_1)=-P(V-V_1)$$where n is the number of moles of gas. Substituting the ideal gas law in this equation for the initial and final thermodynamic equilibrium states gives: $$nC_v(T-T_1)=-P\left(\frac{nRT}{P}-\frac{nRT_1}{P_1}\right)$$This allows us to find the final temperature T in terms of the final pressure P:
$$T=\left[\frac{1+(\gamma-1)\frac{P}{P_1}}{\gamma}\right]T_1$$where $\gamma=\frac{C_p}{C_v}$. From the ideal gas law, $$\frac{PV}{T}=\frac{P_1V_1}{T_1}$$So, if $V=V_2$ (the final volume that we got in the reversible case), $$P=\left[\frac{V_1}{V_2\gamma+V_1(\gamma-1)}\right]P_1$$