# A Ques. From Irreversible Process

Tags: irreversible, process, ques
 P: 21 Hello Lords of Physics, sorry for pestering you guys again but these days a question about irreversibility is pestering me , my question is Why we say that irreversibility in process is due to finite tempreture difference ? so if there is finite temp diff. why process is not reversible? please teach me with giving some examples , thanks a lot!!
 Sci Advisor HW Helper P: 8,953 Remember finite means non-zero as well as non-infinite. So a process which involves no change in temperature is reversable, one which changes temperature will need extra work putting in to make it go the other way. The topic is called entropy and gets quite complicated in the details.
 P: 21 Can anyone tell how isothermal reversible process takes place , means if we give heat to system , entropy will increase , and after doing work by system, does entropy will become zero again? but i heard that entropy once created can't be decreased! , so how the isothermal process is reversible? thanks!
 Sci Advisor HW Helper PF Gold P: 2,532 A Ques. From Irreversible Process In an isothermal reversible process, the system gains energy Q (in the form of heat from the environment) and loses energy W (in the form of work to the environment). The energies are equal, Q = W, so conservation of energy is satisfied. The system gains entropy Q/T by being heated and the environment loses entropy Q/T by heating the system. There is no entropy transfer associated with the work. Entropy is conserved because the process is reversible. Does this help?
 Emeritus Sci Advisor PF Gold P: 6,236 It is not very clear from the OP what is exactly the question, but I guess this might help: When heat "crosses" a finite (non-zero) temperature difference, then this is an irreversible process, in the sense that the total entropy of a thermally insulated system has augmented. The reason is that the entropy lowering dS1 = - Q / Thigh while the entropy increase dS2 = + Q / Tlow, and dS = dS1 + dS2 > 0 for the overall, insulated system. But this is not the only way to have irreversible processes.
 P: 86 Could you make some real-life examples of reversible and irreversible isothermal processes/phenomena?
Emeritus
PF Gold
P: 6,236
 Quote by Domenicaccio Could you make some real-life examples of reversible and irreversible isothermal processes/phenomena?
Reversible (or almost so: there are no truely reversible processes in nature): slow melting of ice in water when the ambiant temperature is around 0 degrees, and vice versa.

Irreversible: cooking water with a flame.
 P: 21 Actually lemme clear the question again , Actually when we add heat by hot reservoir in isothermal process , then there is net entropy decrease in entropy by doing work. but we say for reversible process that heat transfer should be done by very little temp. difference , in isothermal process temp is constant but still by giving finite heat by reservoir we breaking the law of reversibility.
HW Helper
PF Gold
P: 2,532
 Quote by daemonakadevil then there is net entropy decrease in entropy by doing work.
This is not true, there is no entropy decrease associated with doing work. There is an energy decrease, however, when a system does work on its surroundings.
P: 86
 Quote by vanesch Reversible (or almost so: there are no truely reversible processes in nature): slow melting of ice in water when the ambiant temperature is around 0 degrees, and vice versa. Irreversible: cooking water with a flame.
We are assuming here that the temperature doesn't change, right (isotermal)? So we are thinking that the entire ice-melting or water-boiling process considered is at constant T, and all the input energy goes into changing the phase/state from solid to liquid or from liquid to gas. Is this right?

So... what makes the first reversible and the second irreversible? Is it the "speed" in giving the heat?
Emeritus
PF Gold
P: 6,236
 Quote by Domenicaccio We are assuming here that the temperature doesn't change, right (isotermal)? So we are thinking that the entire ice-melting or water-boiling process considered is at constant T, and all the input energy goes into changing the phase/state from solid to liquid or from liquid to gas. Is this right? So... what makes the first reversible and the second irreversible? Is it the "speed" in giving the heat?
The fact that in the first case, the heat comes from a reservoir (the environment) at about the same temperature as the ice/water mixture (hence my "when the ambient temperature is about 0 degrees), while in the second case, the heat comes from a source (flame) at much higher temperature. The irreversible process is not the melting or the boiling, but the heat transfer.
 Sci Advisor HW Helper PF Gold P: 2,532 More generally, entropy is created (and a process becomes irreversible) whenever a flux occurs in response to a gradient in a generalized potential. Examples are: - Flow of heat (entropy & thermal energy) in response to a temperature gradient - Flow of charge in response to a voltage gradient - Flow of matter in response to a concentration gradient. There are also cross terms; e.g., flow of charged matter. Since a flux requires a gradient, no real process can be reversible.
Emeritus
PF Gold
P: 6,236
 Quote by Mapes More generally, entropy is created (and a process becomes irreversible) whenever a flux occurs in response to a gradient in a generalized potential. Examples are: - Flow of charge in response to a voltage gradient
I have a problem with this one. Voltage gradient = E-field. Now, do you consider the acceleration of a bunch of electrons in vacuum under a uniform E-field an irreversible process ?
 Sci Advisor HW Helper PF Gold P: 2,532 Yes; I think it would be impossible to restore the electrons to their original position, in real life, with 100% efficiency. But you could come arbitrarily close, by decelerating and accelerating the electrons v-e-r-y slowly, capturing, storing, and dispensing their kinetic energy.
Emeritus
PF Gold
P: 6,236
 Quote by Mapes Yes; I think it would be impossible to restore the electrons to their original position, in real life, with 100% efficiency. But you could come arbitrarily close, by decelerating and accelerating the electrons v-e-r-y slowly, capturing, storing, and dispensing their kinetic energy.
Imagine I have a cloud of electrons (not dense enough for the mutual repulsion to be of any effect), at point A, at rest. Now, I switch on brutally an E-field of 200 V/m in the x-direction for 10 seconds. Then I switch it off for 3 seconds. Next, I switch on an E-field of 200 V/m in the minus-x direction for 20 seconds. I switch it off for 3 seconds. I next switch on again an E-field in the +x direction for 10 seconds, and then switch it definitively off. Where are my electrons ?
 Sci Advisor HW Helper PF Gold P: 2,532 If the last E-field is 200 V/m, they are at point A, at rest. Some energy has been lost as radiation when the electrons were accelerating, correct? My undergrad training is in mechanical engineering, so it would take me a little while to make sure I'm doing the calculations correctly.
Emeritus