Irreversible quasi-static processes?

[Rahulx084]

TL;DR

Thermodynamic process

I read an example where if I go from initial to final state extremely fast (gas inside a piston cylinder assembly) , the gas inside it will be very unhappy, its not going to stay in equilibrium, parts of the system are going to be at different pressure and parts of it at something other different , so its an irreversible process and to get back at initial state will require some inputs from outside.

I have read that every reversible process is a quasi static process but every quasi static process is not reversible, then what are those quasi static processes which are not reversible? Are they irreversible process?
Aren't irreversible process are fast?

 

[Lord Jestocost]

See @Chestermiller's comment #2 in https://www.physicsforums.com/threads/is-a-quasi-static-but-irreversible-process-possible.909904/

 

[Chestermiller]

See @Chestermiller's comment #2 in https://www.physicsforums.com/threads/is-a-quasi-static-but-irreversible-process-possible.909904/

Another example of a process that can be quasi-static but irreversible is one involving kinetic friction.

 

[Rahulx084]

Displacement work dw=pdv , is this relationship applicable to all kinds of processes? For example displacement work during isothermal process is W=nRTln(Vf/Vi) , can we use this relationship in an irreversible isothermal process? As we have derived this from dw=pdv

 

[Chestermiller]

Displacement work dw=pdv , is this relationship applicable to all kinds of processes? For example displacement work during isothermal process is W=nRTln(Vf/Vi) , can we use this relationship in an irreversible isothermal process? As we have derived this from dw=pdv

Do you think that, in an irreversible process, and ideal gas obeys the ideal gas law?

 

[Rahulx084]

If we carry out the process at high temperature and low pressure,cant it be assumed to be ideal gas?
Okay let's say even if we can't use ideal gas law , still this relationship exists dw=pdv , is this applicable for irreversible?
And if not , then what are the conditions where it is applicable? If quasi static, then we just discussed that there are slow irreversible processes.
Does that mean this is only applicable to reversible processes only?

 

[Chestermiller]

If we carry out the process at high temperature and low pressure,cant it be assumed to be ideal gas?
Okay let's say even if we can't use ideal gas law , still this relationship exists dw=pdv , is this applicable for irreversible?

In the equation dw=pdv, the p you are supposed to use is $p_{ext}$, where $p_{ext}$ is the pressure exerted by whatever the gas is pushing against. In the case of a reversible process, this is also equal to the pressure calculated from the ideal gas law. You can externally force $p_{ext}$ to be whatever you want it to be (manually). But in the case of an irreversible process, it depends, not only on the volume change but also the rate at which the volume is changing.

And if not , then what are the conditions where it is applicable? If quasi static, then we just discussed that there are slow irreversible processes.
Does that mean this is only applicable to reversible processes only?

If your system is the gas only, then even with piston friction, you can get the work that the gas does (if the expansion is quasi static) using the ideal gas law. In that case, the gas alone would be experiencing a reversible process (although receiving frictional heat from the piston). So, in effect, that does mean that the ideal gas law only applies to reversible processes.

 

[Rahulx084]

But having frictional dissipation isn't one of the causes of irreversiblity?
Also can thermodynamics define irreversible processes ? We know that thermodynamic processes are the sequence of process via equilibrium, and irreversible processes are not known to be in equilibrium.

 

[Chestermiller]

But having frictional dissipation isn't one of the causes of irreversiblity?

It depends on what you define as your system. In the example I gave, if your system is the gas alone, then for that system, the process is reversible. But, if your system is the combination of gas plus piston, then for that system, the process is irreversible. That is because the irreversibility is the result of friction in the piston part of the system.

Also can thermodynamics define irreversible processes ? We know that thermodynamic processes are the sequence of process via equilibrium, and irreversible processes are not known to be in equilibrium.

Yes, thermodynamics can identify irreversible processes. For a closed system, a process it suffers is irreversible if the change in entropy of the system is greater than the integral of $dq/T_B$ over the process path, where $T_B$ is the temperature at the system boundary (with its surroundings) where the heat transfer dq occurs.

 

[Rahulx084]

Okay yes , thermodynamics can identify irreversible process via the theory entropy ,like classius inequality. Thanks for making me remember this, It slipped through my mind .
But can a irreversible process be a thermodynamic process provided we know that thermodynamic processes are sequence of equilibrium process?

 

[Chestermiller]

Okay yes , thermodynamics can identify irreversible process via the theory entropy ,like classius inequality. Thanks for making me remember this, It slipped through my mind .
But can a irreversible process be a thermodynamic process provided we know that thermodynamic processes are sequence of equilibrium process?

Thermodynamic processes are not necessarily a sequence of equilibrium states. Reversible processes are. The only requirement of a thermodynamic process is that the initial and final states are thermodynamic equilibrium states.

 

[Rahulx084]

Okay so here is my take from all this discussion
1)Thermodynamic process doesn't necessarily mean to be in equilibrium throughout , only condition is to be the initial and final state at equilibrium.
2) Reversible processes are the sequence of process via equilibrium states and its is carried quasi statically
3) Irreversible processes can also be carried quasi statically
4)dw=pdv can be used for irreversible processes but then p is the external pressure and its not equal to the internal gas pressure
5) we can use ideal gas law in reversible processes only cause then only the external and internal pressure are same as it is carried out very slowly.
6) Thermodynamics identify both the reversible and irreversible processes.

 

[Chestermiller]

Okay so here is my take from all this discussion

4)dw=pdv can be used for irreversible processes but then p is the external pressure and its not equal to the internal gas pressure
5) we can use ideal gas law in reversible processes only cause then only the external and internal pressure are same as it is carried out very slowly.

This is not quite correct. The external pressure and the internal force per unit area (at the piston face) are, by Newton's 3rd law, equal. However, the internal force per unit area (which also includes viscous stresses, even in the ideal gas limit) is not described the the ideal gas law (or other real gas equation of state) for an irreversible process. In a reversible process, viscous stresses are negligible, and the equation of state for the gas can be used.

 

[Rahulx084]

This is not quite correct. The external pressure and the internal force per unit area (at the piston face) are, by Newton's 3rd law, equal. However, the internal force per unit area (which also includes viscous stresses, even in the ideal gas limit) is not described the the ideal gas law (or other real gas equation of state) for an irreversible process. In a reversible process, viscous stresses are negligible, and the equation of state for the gas can be used.

Thank you very much sir . I have one last doubt , you said p is the Pext , but when the gas is expanding, expansion process in a piston cylinder assembly. Isnt the inside gas is doing work? So here p is going to internal or external is dw=pdv.

 

[Chestermiller]

Thank you very much sir . I have one last doubt , you said p is the Pext , but when the gas is expanding, expansion process in a piston cylinder assembly. Isnt the inside gas is doing work? So here p is going to internal or external is dw=pdv.

If the expansion is reversible (quasi-static), you can use the ideal gas law to get the pressure of the gas at the piston face where the work is being done (and it is equal to the external pressure of the piston face on the gas). If the expansion is irreversible, an ideal gas does not satisfy the ideal gas equation during the expansion, so you can't use it to determine the gas pressure at the piston face. But, if you somehow know the external force (say by imposing it manually or removing a known weight from the piston), you can still get the work. In both the reversible and irreversible cases, the work is $dW=P_{ext}dV$, but, in the irreversible case, the force per unit area at the piston face:

1. Is not equal to the value you would calculate from the ideal gas law
2. Is not isotropic (as described by Pascal's law)
3. Is determined, not just by the gas volume but also the rate of change of gas volume
4. Deviates from the ideal gas law by the presence of viscous stress

 

[Rahulx084]

Sir , I get it now . I was just studying thermodynamics and I read something. Let's say we have an adiabatic irreversible process. So from first law , dU=0 , but its not correct, so it says "you can have an irreversible process and can measure work, it can be 0 or any other value, but if its not reversible its not dU".
1) Does that mean we use first law for reversible process only?
2) Let's say for isochoric process , we get du=dq , does that mean here also dq must be reversible ? , I mean to ask the above statement is true for adiabatic process or other processes as well?

 

[Chestermiller]

Sir , I get it now . I was just studying thermodynamics and I read something. Let's say we have an adiabatic irreversible process. So from first law , dU=0 , but its not correct, so it says "you can have an irreversible process and can measure work, it can be 0 or any other value, but if its not reversible its not dU".
1) Does that mean we use first law for reversible process only?
2) Let's say for isochoric process , we get du=dq , does that mean here also dq must be reversible ? , I mean to ask the above statement is true for adiabatic process or other processes as well?

None of this makes any sense to me, and most of it is wrong.

Also, please do not use differentials for irreversible processes.

 

[Rahulx084]

I think I very poorly framed the question. Here is the better version.
For an adiabatic process and irreversible process, the reason of irreversiblity is that, the Pext=0 , so the piston going to slap up once the stopper is removed (very fast) , in a piston cylinder assembly. In this case we see heat transfer and work done both are 0 , so from first law the change in internal energy must be 0 .But there is this statement I read which says "You can have an adiabatic irreversible process where work is either 0 or something else, but its not equal to change in internal energy unless the process is reversible (like what we see in entropy).
So my query are:
1)Is this above statement correct?

2) If yes then, can we only use first law for rev. Processes only.I mean like if a process is irreversible , do we have to make a reversible path between the initial and final point of the irreversible process and calculate the work done and heat transfer to get internal energy, like what we do while calculating entropy.

3)For isochoric process carried out irreversibly change in internal energy will be equal to heat transfer in the process, so do we have to here get a reversible path again between the initial and final points of irreversible process and get the heat transfer?

But this sounds so absurd to me cause we can have a lot of reversible path and every path will lead to different heat transfer or work transfer leading to different change in internal energy every time .

Here is from where I'm taking the reference :
MIT opencourseware 5.60 Thermodynamics and kinetics lecture 3

(at time 36:15)

 

[Chestermiller]

OK. I looked over this video and saw what the guy said. I was very unhappy with the way he presented the material, and it's no wonder you are confused. It seems that he has only a nodding acquaintance with the idea that the math must be precise in order to properly describe thermodynamics.

Please be advised that, in my judgment, differentials should be applied to state functions like internal energy U (e.g., dU) only for reversible processes, For irreversible processes, one should only focus on the changes in the state functions between the initial and final thermodynamic equilibrium states of the system, and in those cases, only use finite differences over the entire path (e.g., $\Delta U=U_2-U_1$). State functions are equilibrium properties of the material (and not directly related to any specific processes the material subjected to), and should only be considered for the initial and final thermodynamic equilibrium states of the system, not for intermediate non-equilibrium states (as occur in irreversible processes). For a reversible process, it is valid to consider differential changes in state functions because reversible processes are comprised of a continuous sequence of thermodynamic equilibrium states.

On the other hand, for path dependent quantities like work and heat, it is valid to use differentials both for reversible and for irreversible paths.

I think I very poorly framed the question. Here is the better version.
For an adiabatic process and irreversible process, the reason of irreversiblity is that, the Pext=0 , so the piston going to slap up once the stopper is removed (very fast) , in a piston cylinder assembly. In this case we see heat transfer and work done both are 0 , so from first law the change in internal energy must be 0 .But there is this statement I read which says "You can have an adiabatic irreversible process where work is either 0 or something else, but its not equal to change in internal energy unless the process is reversible (like what we see in entropy).
So my query are:
1)Is this above statement correct?

No. If the work is zero for an adiabatic irreversible path, the change in internal energy is zero. The first law tells us this. But, if the gas is not an ideal gas, there will be a change in temperature, because U is a function both of T and V, and V has changed. For an ideal gas, where U is a function only of temperature, the change in temperature is zero.

2) If yes then, can we only use first law for rev. Processes only.I mean like if a process is irreversible , do we have to make a reversible path between the initial and final point of the irreversible process and calculate the work done and heat transfer to get internal energy, like what we do while calculating entropy.

Since the answer to question 1 is NO, we don't need to address this question.

3)For isochoric process carried out irreversibly change in internal energy will be equal to heat transfer in the process, so do we have to here get a reversible path again between the initial and final points of irreversible process and get the heat transfer?

No. If you know the temperature change and the heat capacity at constant volume (a property of the material), then you know the heat transfer.

It's too bad this guy did such a bad job of explaining this. All he did was confuse you. It's not your fault, it's his.

 

[Rahulx084]

Thank you so much sir . Things are much clearer to me now .
I have one problem statement here

And here is the solution provided of it

Why while equating the irreversible work we are using formulae Cv(Tf-Ti) , which is derived for a reversible process? Is it because the question says work(irreversible)=1.5Work(reversible) or its because we have ideal gas in working ?

 

[Chestermiller]

Thank you so much sir . Things are much clearer to me now .
I have one problem statement here
View attachment 263525
And here is the solution provided of it
View attachment 263526
Why while equating the irreversible work we are using formulae Cv(Tf-Ti) , which is derived for a reversible process? Is it because the question says work(irreversible)=1.5Work(reversible) or its because we have ideal gas in working ?

For an ideal gas, because internal energy is a material property that is function of temperature only, the change in internal energy, regardless of process path is $$\Delta U=nC_v\Delta T$$. So $$nC_v\Delta T=W_{irrev}$$ for the irreversible path. Notice that I used $\Delta$ to represent the change between the initial and final states.

 

[Rahulx084]

Ohh yes , my bad sorry. Thank again sir . If I will face any problems further in this course , I will add it in this thread . :)