1. deltaS = deltaQ / T for any infinitesimal process; in particular, for an isothermal process

2. Not all isothermal processes are reversible

3. Not all adiabatic processes are reversible

4. Non-PV work is any kind of non-mechanical work not based on volume changes: electrical, magnetic, gravitational, etc. that takes place and is exchanged between the system and its surroundings. Similar to P-V work, it is the product of a generalized force (e.g., electrical voltage) and a generalized displacement (e.g., electrical charge). Letting a system to evolve after releasing an internal constraint under controlled conditions allows extracting work from the system (maximal for a reversible process) and use it for driving another (appropriately) coupled process.

Isn’t a closed adiabatic process with no heat or mass exchange with the surroundings (an ideal case), always reversible? Can we use “delta S = Q/T” for such a process to get “delta S = 0” to make the process isentropic? Can you give an example of an adiabatic process which is not reversible and not isentropic?

An adiabatic process (Q = 0) is not necessarily reversible (deltaS = 0).

For reversible processes dS = dQ/T (adiabatic is equivalent to isentropic).

For irreversible processes dS > dQ/T (adiabatic is not equivalent to isentropic).

The free expansion of a gas may be adiabatic but not isentropic.

The diffusion of particles within a solution may be adiabatic but not isentropic.

I forgot to mention a very appropriate example since it involves heat transfer and entropy changes.

A system surrounded by adiabatic and rigid walls is constituted by two subsystems at different temperatures, initially separated by an adiabatic and rigid wall, but at some point that boundary is substituted by a diathermic and rigid wall. The initial state with two different temperatures evolves towards a final thermal equilibrium state in which the two subsystems have the same temperature, intermediate between the two initial temperatures. No work was done, but heat was transferred from one subsystem (that at higher temperature) to the other (that at lower temperature). The final equilibrium temperature is determined by applying the first law, and it will depend on the initial temperatures and the heat capacities of the subsystems.

The global heat is zero (heat transferred by one subsystem is taken by the other subsystem); there are no heat losses: Q = 0

The global entropy change is not zero. There is an overall positive entropy change accounting for the irreversibility of the process, deltaS > 0. The entropy of the global system increases. There is a change in entropy for each subsystem: that at higher initial temperature lowers its entropy, and that a lower initial temperature increases its entropy; but, overall, the entropy increases (the increase in entropy for the latter subsystem is larger than the decrease of entropy of the former subsystem).

The reverse process could be possible according to the first law (that is, the reverse process does not violate the first law), but not according to the second law (the reverse process violates the second law).

Thanks for your answer.

I think, third law of thermodynamics finds application in this example. As per third law of thermodynamics, entropy is directly proportional to temperature but as per second law of thermodynamics, change in entropy is inversely proportional to temperature.

This explains the second last paragraph of the answer.

Is it correct, Sir?

Can an isentropic process be isothermal? Can you give an example of such a process?

Well, the third law is required to calculate absolute values of entropy, but not for calculating entropy changes in a process.

The entropy change associated with the thermal equilibrium process mentioned above can be calculated knowing the initial temperatures, the heat capacities, and the final equilibrium temperature (determined by the first law).

An isentropic process is characterized by dS = 0.

An isothermal process is characterized by dT = 0.

For a monocomponent ideal gas there is no isentropic isothermal process, since dS = 0 and dT = 0 leaves no degrees of freedom for a process to occur. Besides, the isentropic and isothermal curves starting from a common temperature have different slopes and do not coincide. However, for an incompressible fluid the isentropic process (curve) coincides with the isothermal process (curve), since Cp and Cv are the same.

When considering dS = 0, you must always be careful for distinguishing between the entropy of the system or the global entropy involving the entropy of the system plus the entropy of the surrounding heat sink. The second law applies to the global entropy change.

Thanks for your answer.

can we have an isentropic isothermal process for a multicomponent ideal gas or a liquid or solid?

can you give examples of such processes?

For an ideal gas at T = 0 K, the isentropic process coincides with the isothermal process.

For a solid or an incompressible fluid any isotherm will coincide with an isentropic (because, Cp = Cv).

For a monocomponent system undergoing a first order phase transition, the temperature is constant but the entropy change is not zero (isothermal and not isentropic); however, in a second order phase transition the temperature and the entropy are constant (isothermal and isentropic).

Thanks for your answer.

All reversible adiabatic changes are isentropic. However all isentropic changes are not reversible adiabatic. They can be reversible isothermal also.

Is it correct, Sir?

If reversible (dS = dQ/T), an adiabatic process (dQ = 0) is isentropic (dS = 0).

If irreversible (dS > dQ/T), that is not true (dQ = 0, while dS > 0).

Thanks a lot for your help.

I think it is clear.

Can an irreversible isothermal process also be isentropic?

If irreversible, dS > 0, and, then, not isentropic.

Isentropic = reversible + adiabatic

delta S = Q/T only for a REVERSIBLE isothermal process. If the process is IRreversible, then delta S > Q/T. A simple example: free expansion of an ideal gas: Q = 0 but delta S > 0. Infinite slowness is usually necessary but not always sufficient for any process, adiabatic or otherwise, to be reversible. Some types of friction, such as sliding or rolling friction, do NOT vanish as speed vanishes, so if these types of friction are present, reversibility does NOT obtain EVEN WITH infinitely slow operation. A discharging battery and a falling weight are two simple examples of processes yielding non-PdV work.

Dear Dr Denur,

Thanks for your reply.

No irreversible process, isothermal or otherwise, can be isentropic overall. The total entropy increases in any irreversible process. The entropy of a system undergoing a reversible adiabatic process is constant; therefore this process is isentropic with respect to the system. In a reversible isothermal process, the entropy of the (system plus surroundings) is constant, therefore this process is isentropic with respect to the (system plus surroundings): the entropy changes of the system and surroundings are equal in magnitude but opposite in sign..

Dear Dr Denur,

Thanks for your reply.

As per Dr Campoy’s answer, in second order phase transitions, both temperature and entropy are constant. Is it a reversible isothermal and an isentropic change with respect to system or system + surroundings?

Hi! I was reading this question, and I just wanted to know why is it that ds> = dq/t

Where does this come from? Because they just gave us the formula in class and I would like to understand why is it like this.

Thanks!

Dear Gema

ds>dq/t ; for irreversible process in thermodynamic system

ds= dq/t ; for reversible process in thermodynamic system

Also, ds= change in entropy of system

dq/t = entropy transfer by heat

So ds> dq/t

Or

ds= dq/t + S gen.

Where S gen = entropy generated in irreversible process

Sgen= 0 ; for reversible process

Hope this clarifies your doubt