Musings on Entropy — Part III — Reversible and Irreversible Processes
What makes a process reversible?
The general rule of thumb is that a process is reversible if the system can be moved from its current state back to its original state without the loss of energy due to friction or other non-conserved forces.
Let’s look at the fully reversible case first.
Carnot Cycle
The Carnot cycle consists of two isothermal processes and two adiabatic processes.
Each leg of the cycle can be analyzed using the 2nd Law of Thermodynamics,
- dU = dQ — pdW
In the graph below, the isothermal process, from position 1 to position 2, absorbs heat into the system — and hence absorbs entropy into the system. From the 2nd Law
- dU=dQ-pdW
But since T(1) = T(2)
- dU=0
The entropy change is given directly by
- dQh=-dS * Th
Where Qh is the heat absorbed during the isothermal expansion from point 1 to point 2
The isotherm linking points 3 and 4, by contrast, expel heat from the system — and hence expels entropy from the system.
- dQc=+dS * Tc
Where dQc is the heat expelled during the isothermal contraction from point 3 to point 4 in the graph.
The adiabatic legs linking points 2–3 and points 4–1 do not absorb or expel heat — and hence, do not absorb or expel entropy.
Condition for Reversibility — lack of friction
The Carnot cycle is entirely reversible if and only if all the entropy absorbed during the isotherm, linking points 1 and 2, is exactly expelled during the isotherm linking points 3 and 4.
When will this hold true?
This condition will hold if the system doesn’t experience any frictional forces during any part of the cycle.
Isotherms
If there are no frictional forces, then the available kinetic energy of the particles, defined as the kinetic energy difference between particles in the hot and cold baths, is 100 % converted to work. Similarly, along the lower isotherm, the available kinetic energy of the particles is 100% converted back to heat.
Adiabats
Similarly, the work performed during the two adiabatic legs is reversible if, during these legs, no work is done to overcome frictional or non-conserved forces.
To use Clausius’s formulation, a process is reversible if and only if the net entropy across the cycle is zero.
- Net Entropy= dS1+dS2+dS3+dS4+…dSn = 0
The net entropy for a full cycle is zero when the conversion of heat to work or work to heat is 100% efficient everywhere along the cycle.
By contrast, the process is not reversible if some portion of the absorbed heat goes towards countering frictional forces and does not perform work.
Similarly, the process is not reversible if the work cannot be fully converted to heat.
For the Carnot cycle, the conversion of heat to work in the first isotherm is 100% efficient. Similarly, the conversion of work back to heat, in the second isotherm is also 100% efficient.
- dS(1,2)+dS(3,4) = 0
- dQh/Th — dQc/Tc = 0
From an energy perspective the particles absorbed from the heat bath have energy proportional to Th. The absorbed particles are coerced to do work; the very same particles are then expelled at Tc.
For a Carnot cycle, 100% of the available heat is converted to work. The available heat is precisely the difference between the energy at Th and the energy ‘floor’ at Tc.
Analogy with Electric Systems
The analogy with electrical systems operating between two potential differences is (nearly) exact.
The electron at energy V1*e falls across a potential difference of V1-V2. The electron can do maximum work of W=(V1-V2)*e if it encounters no friction. After doing this work, the electron exits the system at the lower potential, V2 eg at the cold bath.
Quasi-static evolution and Frictional forces
It is often mentioned that a process is reversible if it is done in quasi-static manner. What does this mean?
Presumably, frictional forces are minimized if a process is done quasi-statically. As frictional forces are minimized, the process becomes increasingly reversible.
Conversely, if a process is done very rapidly, relative to the relaxation time of the system, it stands to reason that at least some of the input energy or heat will not be converted to work; instead this input heat will will be wasted and will instead exist in the form of ‘waste heat’. The waste(d) heat will not be converted into work. It will be used to ‘warm’ various parts of the system but this heat will not be converted to work.
If, conversely, a process is quasi-static BUT still encounters frictional forces, then the process is not reversible, regardless of the fact that the process was done quasi-statically. Similarly, if a process is done ‘very quickly’ but no energy is expended to overcome frictional or non-conserved forces, then, again, the process will nonetheless be reversible.
If system has undergone an irreversible process, work will need to be done to the system to return to its original state; in short, additional heat will need to be added to the system, with the heat being converted to work, in order to compensate for the heat that was spent overcoming frictional or non-conserved forces eg to compensate for the ‘waste heat’ or ‘wasted heat’.