Trying to understand quasi-static processes

  • Thread starter johng23
  • Start date
  • #1
292
1
In my thermo course, we made a distinction between quasi-static and reversible transitions (which apparently some don't make). Reversible means both system and surroundings need to be able to return to their initial state. Quasi-static only means that the system transitions to a new state by going through only equilibrium states. There is no friction or dissipative effects.

Intuitively, this makes sense but I'm not sure precisely how to differentiate an equilibrium state from a non-equilibrium state. The case I want to use to illustrate is an adiabatic container with rigid walls, with a paddle wheel going into it.

If I turn the paddle wheel, I can bring the system to any isometric state of higher energy by increasing the entropy. Is this a quasi-static process? Let's assume there is no friction between the paddle wheel and the wall. Then the general picture I have is that if I turn the paddle wheel slowly enough, the process is quasi-static since the distribution of gas molecules never appears macroscopically different than it otherwise would, even though you are gradually transferring energy to the system. But can I really reach an arbitrarily high energy this way? If I move the paddle wheel fast enough that there are pressure variations or heterogeneous velocity distributions in the container, at that point I am no longer going through equilibrium states. But why does it matter if energy is dissipated in these ways between molecules? I'm still transferring the same amount of energy to the system.

As another (maybe clearer) question, how do we avoid friction in a quasi-static process? If I slide two plates across each other, is it even conceptually possible to avoid friction? As I pull two atoms away from eachother which are attracted by van der Waals forces or otherwise, don't they increase distance up to a point and then reach some threshold where they pull apart and start to oscillate? That's how it seems like it would happen in my mind.
 

Answers and Replies

  • #2
Andrew Mason
Science Advisor
Homework Helper
7,621
363
In my thermo course, we made a distinction between quasi-static and reversible transitions (which apparently some don't make). Reversible means both system and surroundings need to be able to return to their initial state. Quasi-static only means that the system transitions to a new state by going through only equilibrium states. There is no friction or dissipative effects.

Intuitively, this makes sense but I'm not sure precisely how to differentiate an equilibrium state from a non-equilibrium state.
Here is my definition of a quasi-static process involving a system and its surroundings: it is a process in which the changes occur slowly enough so that there is always:
1. thermal equilibrium within in all system components,
2. thermal equilibrium between all system components and surroundings across whose boundaries heat flows and
3. mechanical equilibrium between all system components and surroundings between which mechanical work/energy is transferred.​


So, in the case of a quasi-static heat flow from a reservoir to a gas, the heat flow occurs due to infinitessimal temperature difference between the reservoir and gas. In the case of a gas doing work on its surroundings, the gas pressure is infinitessimally higher than the surroundings at all times. All components are always in equilibrium (ie. out of equilbrium by an infinitessimal amount).

If there is a finite difference in temperature between two components in thermal contact, these components are not in equilibrium. If there is a finite pressure difference between two components in mechanical contact, these components are not in equilibrium. These processes are, therefore, not quasi-static.

The case I want to use to illustrate is an adiabatic container with rigid walls, with a paddle wheel going into it.

If I turn the paddle wheel, I can bring the system to any isometric state of higher energy by increasing the entropy. Is this a quasi-static process? Let's assume there is no friction between the paddle wheel and the wall. Then the general picture I have is that if I turn the paddle wheel slowly enough, the process is quasi-static since the distribution of gas molecules never appears macroscopically different than it otherwise would, even though you are gradually transferring energy to the system. But can I really reach an arbitrarily high energy this way? If I move the paddle wheel fast enough that there are pressure variations or heterogeneous velocity distributions in the container, at that point I am no longer going through equilibrium states. But why does it matter if energy is dissipated in these ways between molecules? I'm still transferring the same amount of energy to the system.
If you accelerate the paddle slowly (quasi-statically), you will do some work on the gasbut once you reach a constant speed no more work will be done (ignoring friction). Because the paddle is rigid, you are not changing the volume of the gas between the paddles. So a slow constant speed turning of the paddle does no work. Ignoring friction, the paddle should keep turning without any addition of energy. All the work done on the paddle results in uniform kinetic energy of the gas and does not cause the temperature of the gas to increase. That kinetic energy can be fully recovered as work.

If you accelerate the paddle quickly, you will be compressing the gas immediately in front of the turning paddle and expanding the parts behind the paddle. Ultimately what you are doing is adding non-uniform kinetic energy to the gas. After you stop accelerating the paddle and the gas reaches kinetic equilibrium, its temperature will be higher than when you started.
As another (maybe clearer) question, how do we avoid friction in a quasi-static process? If I slide two plates across each other, is it even conceptually possible to avoid friction? As I pull two atoms away from eachother which are attracted by van der Waals forces or otherwise, don't they increase distance up to a point and then reach some threshold where they pull apart and start to oscillate? That's how it seems like it would happen in my mind.
You can minimize friction but you can't really avoid it. Friction losses are ultimately turned into heat.

For your paddle example, if you turn the paddle quickly, you create turbulence which is kinetic energy that does not follow a Maxwell-Boltzmann distribution (temperature distribution). As turbulence settles down, the molecular motion will follow the Maxwell-Boltzmann distribution and, therefore, appears as heat. You can think of the turbulence as "friction".

However, if you accelerate the paddle slowly so that there is not turbulence created (ie quasi-statically), the energy transferred from the paddle to the fluid is retained in the fluid as mechanical energy rather than heat. That mechanical energy can be recovered fully from the gas (ie. by putting a load on the paddle the turning gas does work on the paddle until the turning ends).

AM
 
Last edited:

Related Threads on Trying to understand quasi-static processes

  • Last Post
Replies
2
Views
3K
  • Last Post
Replies
3
Views
388
  • Last Post
Replies
3
Views
3K
Replies
5
Views
978
  • Last Post
Replies
21
Views
413
Replies
1
Views
1K
Replies
19
Views
2K
Replies
9
Views
5K
Replies
4
Views
8K
Top