# What features do reversible work have over the irreversible kind?

At 6:47 of this video lecture, the professor defines enthalpy for a constant pressure process as

$$q_{p}= \Delta U + p \Delta V$$

but, I can not understand why the work he implicitly starts referring work as an exact differential. Is this due to it being the reversible kind of work?

and around 36:55 of this lecture, an even stranger thing happens,

$$dU \neq dW$$

unless it is reversible process, but why?

What exactly is the distinguishing difference between reversible and non-reversible work, and what are the consequences of these differences?

In this stack, a similar question is asked and while the answer does make sense, the professor says that the process is adiabatic around 36:15, then, writes the first law. Now, by the definition of first law, isn't

$$dU = dW$$

Always? or is the first law a statement which changes on what situation which you place it in?

• Have you looked up the first law? dU=dW+dQ (inexact differentials for the path functions). – electronpusher Sep 4 at 22:42
• Often, when I have trouble understanding new concepts, I dream up and solve a specific focusing problem(s) to see how it all plays out in practice. If you want, I will formulate for you comparable focusing problems involving reversible and irreversible adiabatic expansion so you can solve them and see for yourself how the results compare. Any interest? – Chet Miller Sep 4 at 23:57
• I brought up a point about the definition of it at the last part @electronpusher – Buraian Sep 5 at 2:39
• @ChetMiller sure, would be interestetd – Buraian Sep 5 at 2:39

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 $$P_1$$.

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?
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?
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), $$P=\left[\frac{V_1}{V_2\gamma+V_1(\gamma-1)}\right]P_1$$