Trying to understand quasi-static processes

In summary, the conversation discusses the distinction between quasi-static and reversible transitions in a thermodynamics course. Quasi-static means the system transitions through equilibrium states, while reversible means both system and surroundings can return to their initial state. The concept of equilibrium states is further explained and the question of how to avoid friction in a quasi-static process is raised. The conversation also explores the effects of friction and non-uniform kinetic energy on the temperature of a gas.
  • #1
johng23
294
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 each other 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.
 
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  • #2
johng23 said:
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 each other 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
 
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1) What are quasi-static processes?

Quasi-static processes refer to a type of thermodynamic process in which the system changes gradually and slowly enough that it can be considered approximately in thermodynamic equilibrium at all times.

2) How are quasi-static processes different from other thermodynamic processes?

Quasi-static processes are distinguished by their slow pace, which allows for the system to reach equilibrium at each step. Other processes, such as adiabatic or isothermal processes, occur at a faster rate and do not allow for equilibrium to be reached at all times.

3) What are some examples of quasi-static processes?

Examples of quasi-static processes include the expansion or compression of a gas in a piston, heat transfer between two systems, and changes in pressure or volume in a closed container.

4) Why is it useful to study quasi-static processes?

Studying quasi-static processes allows scientists to understand the behavior of thermodynamic systems, which can be applied to real-world situations such as engines, refrigerators, and power plants. It also helps in the development of more efficient and sustainable technologies.

5) What are the limitations of quasi-static processes?

Quasi-static processes assume that the system is always in thermodynamic equilibrium, which may not be the case in real-world situations. Additionally, the slow pace of these processes may not accurately reflect the behavior of systems undergoing rapid changes.

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