Irreversible Quasistatic PV Work

[Red_CCF]

Yes, so far. The acceleration can be expressed in terms of x as d²x/dt². So:
$$m\frac{d^2x}{dt^2}=P_IA$$
If we multiply both sides of this equation by dx/dt, and integrate with respect to time, we get:
$$W(t)=\frac{1}{2}mv^2(t)$$
where W(t) is the work done by the gas on the piston up to time t, v(t) = dx/dt, and the right hand side is recognized as the kinetic energy of the piston.

I got a little mixed up, how did you go from the first to the second equation? Also, for this integral to work one has to assume that a and P_I are non-zero for at least some time during the time interval; how is this known?

Here are some questions to consider:
What happens when the piston collides with the far end of the cylinder (if the collision is elastic)? Does the piston keep moving forever (even if the piston is frictionless)? If the frictionless piston eventually stops, what causes it to slow down and stop?

Chet

If the piston collides at at the end of the cylinder, I would assume that some energy is transferred to the cylinder even if the collision is elastic (i.e. if the cylinder was on a frictionless surface it would gain kinetic energy). If there are no irreversibilities within the gas itself, I think the piston would initially move back until the gas pressure slows it down and pushes it towards the far wall again, and this behaviour should oscillate with decreasing range of motion for the piston. I think this behaviour would continue as t → ∞, so maybe the piston never stops?

Thanks very much

 

[Chestermiller]

I got a little mixed up, how did you go from the first to the second equation?

$$m\frac{d^2x}{dt^2}=P_IA$$
Multiply both sides by dx/dt:
$$m\frac{dx}{dt}\frac{d^2x}{dt^2}=P_IA\frac{dx}{dt}$$
Substitute v for dx/dt on the left side of the equation and dV/dt = Adx/dt on the right side of the equation:
$$mv\frac{dv}{dt}=P_I\frac{dV}{dt}$$
Now, integrate with respect to t:
$$\frac{m}{2}v^2(t)=\int{P_IdV}$$

Also, for this integral to work one has to assume that a and P_I are non-zero for at least some time during the time interval; how is this known?

For the case of a piston with mass, P_I is never going to be zero. The gas is always going to be accelerating the piston toward the far end of the cylinder.

If the piston collides at at the end of the cylinder, I would assume that some energy is transferred to the cylinder even if the collision is elastic (i.e. if the cylinder was on a frictionless surface it would gain kinetic energy).

No. The cylinder is assumed to be anchored to the lab, so, in an elastic collision, the piston retains all its kinetic energy, and its velocity simply reverses sign.

If there are no irreversibilities within the gas itself, I think the piston would initially move back until the gas pressure slows it down and pushes it towards the far wall again, and this behaviour should oscillate with decreasing range of motion for the piston. I think this behaviour would continue as t → ∞, so maybe the piston never stops?
Thanks very much

By "irreversibilities within the gas," I assume you mean viscous dissipation. Without viscous dissipation, the cylinder and the gas would continue moving back and forth forever, interchanging compression with kinetic energy. In our case, the air within the cylinder does have viscosity, and eventually, the piston velocity will slow down to zero, and the piston will come to rest at the far end of the cylinder. This will be the condition of our system at final steady state. We already said earlier this when we analyzed the problem taking the combination of the piston and the gas as our system.

So, in our adiabatic irreversible problem, we now know that the work done by the gas on the piston is zero, and the heat transferred is zero, so the change in internal energy is zero, and the change in temperature is therefore also zero. So, even in the case of a finite mass for the piston, we still get the same answer.

Chet

 

[Red_CCF]

Maybe doping it out at the molecular level will help. I'm not too good at doing this kind of thing, so no guarantees. Before the membrane is removed, the gas has a Boltzmann distribution, and the molecules are flying around in all directions. Molecules are bouncing off the membrane and reversing direction, so that this change in momentum translates into pressure on the membrane. Further back, molecules are bouncing off each other, and this also translates into pressure. Now the membrane is removed, and no molecules are bounding back the other way in the region that was previously adjacent to the interface. Whatever molecules would have hit the membrane now continue on. This causes the molecules that were next to the membrane to get further apart and to no longer interact with each other. But, slightly further back, the molecules do not yet know that the interface has been removed. So they continue exhibiting the same molecular interactions until the dilatation region reaches them.
Chet

I imagine that the molecules shooting out once the membrane is gone aren't all parallel to each other so some should collide with each other and create pressure; is the number density in the first layer so low that this can be ignored?

For the case of a piston with mass, P_I is never going to be zero. The gas is always going to be accelerating the piston toward the far end of the cylinder.

Out of curiosity, is there a theoretical velocity (order of magnitude wise) in which the piston can move such that P_I = 0; I am guessing it is comparable to gas expanding against a massless piston? Is this affected by the initial pressure?

So, in our adiabatic irreversible problem, we now know that the work done by the gas on the piston is zero, and the heat transferred is zero, so the change in internal energy is zero, and the change in temperature is therefore also zero. So, even in the case of a finite mass for the piston, we still get the same answer.

Why is W = 0 when the piston has mass, as I thought the equation derived for this case was $$W(t)=\frac{1}{2}mv^2(t)$$?

A couple of aside questions:

How was Newon's Second law used to justify P_ext = P_I?

In general, is it valid to treat friction heat addition as Q in first law or for ΔS = ∫δQ/T + σ, or does Q have to be heat transfer via finite temp gradient between the system and surrounding? In my OP you mentioned that first law was simply W = ∫P_extdV= E_2 - E_1 if piston was included in the system; I took this to mean that the work that turned into frictional heat that went into the system still counted as work and thus ∫δQ/T = 0, is this correct?

Thanks very much

 

[Chestermiller]

I imagine that the molecules shooting out once the membrane is gone aren't all parallel to each other so some should collide with each other and create pressure; is the number density in the first layer so low that this can be ignored?

Yes. That's the idea. It's just a limiting case.

Out of curiosity, is there a theoretical velocity (order of magnitude wise) in which the piston can move such that P_I = 0; I am guessing it is comparable to gas expanding against a massless piston?

Sure. That's what it is. We would have to solve the gas dynamics problem in detail to pin all this down accurately.

Is this affected by the initial pressure?

Probably.

Why is W = 0 when the piston has mass, as I thought the equation derived for this case was $$W(t)=\frac{1}{2}mv^2(t)$$?

This is a correct equation for time t. However, in our analysis, we are looking for W at infinite time W(∞). By that time, as we said earlier, because of viscous dissipation in the gas, the piston velocity v is going to be equal to zero, and the piston is going to stay at the dead end of the cylinder.

A couple of aside questions:

How was Newton's Second law used to justify P_ext = P_I?

For a massless frictionless piston, the forces on both sides of the piston are equal.

In general, is it valid to treat friction heat addition as Q in first law or for ΔS = ∫δQ/T + σ, or does Q have to be heat transfer via finite temp gradient between the system and surrounding?

I don't quite understand this question. What are you including in your "system"? Gas or gas plus piston?

In my OP you mentioned that first law was simply W = ∫P_extdV= E_2 - E_1 if piston was included in the system; I took this to mean that the work that turned into frictional heat that went into the system still counted as work and thus ∫δQ/T = 0, is this correct?

I think there should a minus sign in front of the W and the integral. What you said is correct regarding the frictional heat.

Chet

 

[Red_CCF]

This is a correct equation for time t. However, in our analysis, we are looking for W at infinite time W(∞). By that time, as we said earlier, because of viscous dissipation in the gas, the piston velocity v is going to be equal to zero, and the piston is going to stay at the dead end of the cylinder.

As the piston oscillates and the gas does work on the piston and vice versa back and forth, the end result is that at the end no net work is done by either one? I noticed that W in that equation is always positive, but isn't there a period where the piston does work on the gas (after colliding at the wall) such that there should be a sign change?

I understand that viscous dissipation is a loss, but like the example in my OP, if some of the applied work is turned into friction heat and still goes into the system, the system's temperature and pressure will go up; how does one show that this gain in pressure and hence ∫PdV over a cycle is less than ∫PdV if the process was reversible without using second law?

For a massless frictionless piston, the forces on both sides of the piston are equal.

Is this because a massless object cannot have a finite net force acted up on it?

If the piston does have mass but frictionless, how does P_ext relate to P_I,gas in quasistatic and non-quasistatic compression/expansion?

I don't quite understand this question. What are you including in your "system"? Gas or gas plus piston?
Chet

I was thinking in a general case where friction heat is generated at the system boundary (perhaps the piston-cylinder boundary if the piston is included in the system), or internal irreversibilities that causes heat generation (ie viscous dissipation). In these cases does ∫δQ/T = 0 even if the system is insulated?

Thanks very much

 

[Chestermiller]

As the piston oscillates and the gas does work on the piston and vice versa back and forth, the end result is that at the end no net work is done by either one?

Yes.

I noticed that W in that equation is always positive, but isn't there a period where the piston does work on the gas (after colliding at the wall) such that there should be a sign change?

There would be a decrease in W(t) during this period. I'm not sure if W(t) ever actually changes sign, but, fortunately, we don't have to resolve this by solving the gas dynamics equations. All we need to know is that, in the end, W = 0.

I understand that viscous dissipation is a loss, but like the example in my OP, if some of the applied work is turned into friction heat and still goes into the system, the system's temperature and pressure will go up; how does one show that this gain in pressure and hence ∫PdV over a cycle is less than ∫PdV if the process was reversible without using second law?

You need to solve the gas dynamics equations to flesh out all the details. But, for our purposes we don't need to do this. During the transient phase of the process, the gas expands as it does work on the piston, and, as a result it cools. But, after the kinetic energy of the piston (and gas) is dissipated as heat, the gas gets reheated back to its original temperature.

Is this because a massless object cannot have a finite net force acted up on it?

Yes. Otherwise it would have to have an infinite acceleration, which it doesn't.

If the piston does have mass but frictionless, how does P_ext relate to P_I,gas in quasistatic and non-quasistatic compression/expansion?

In quasistatic, P_ext and P_I,gas differ by an insignificant amount (for the horizontal cylinder case). In non-quasistatic, (P_I,gas-P_ext)A = ma if the cylinder is horizontal, and (P_I,gas-P_ext)A = m(g+a) if the cylinder is vertical.

I was thinking in a general case where friction heat is generated at the system boundary (perhaps the piston-cylinder boundary if the piston is included in the system), or internal irreversibilities that causes heat generation (ie viscous dissipation). In these cases does ∫δQ/T = 0 even if the system is insulated?

Sure. If both the gas and the piston are included in the system, then Q = 0 if the system is insulated (by definition). But, there is extra work done to move the piston against the friction.

Chet

 

[Red_CCF]

There would be a decrease in W(t) during this period. I'm not sure if W(t) ever actually changes sign, but, fortunately, we don't have to resolve this by solving the gas dynamics equations. All we need to know is that, in the end, W = 0.

I was thinking that, when the piston first reaches the far end of the cylinder by the gas, the gas is doing work on it and when it hits the wall and moves back, it would be doing work onto the gas and hence there should be a sign reversal?

With the W(t) equation, v = 0 when the piston is slowed to a stop by the gas as it pushes back onto the gas, how does W(t) = 0 in those scenarios since the piston doesn't push the gas all the way back to its original state due to dissipation?

You need to solve the gas dynamics equations to flesh out all the details. But, for our purposes we don't need to do this. During the transient phase of the process, the gas expands as it does work on the piston, and, as a result it cools. But, after the kinetic energy of the piston (and gas) is dissipated as heat, the gas gets reheated back to its original temperature.

What I'm confused about is how does internal irreversibility like viscous dissipation during compression/expansion cause a work loss. Does it require a higher average applied pressure over the same volume change and if so why? Also, if some of the work applied turned into frictional heat and still goes into the system, how is that a loss?

In quasistatic, P_ext and P_I,gas differ by an insignificant amount (for the horizontal cylinder case). In non-quasistatic, (P_I,gas-P_ext)A = ma if the cylinder is horizontal, and (P_I,gas-P_ext)A = m(g+a) if the cylinder is vertical.

And for the vertical case it would be P_I,gas = mg/A_piston+ P_ext?

How does friction between the gas and cylinder during compression/expansion change the relationship between P_I,gas and P_ext; is it F_friction + P_I,gasA = P_extA?

Thanks very much

 

[Chestermiller]

I was thinking that, when the piston first reaches the far end of the cylinder by the gas, the gas is doing work on it and when it hits the wall and moves back, it would be doing work onto the gas and hence there should be a sign reversal?

It would cause a sign reversal on the rate at which work is being done on the piston, not on the cumulative amount of work done on the piston.

With the W(t) equation, v = 0 when the piston is slowed to a stop by the gas as it pushes back onto the gas, how does W(t) = 0 in those scenarios since the piston doesn't push the gas all the way back to its original state due to dissipation?

This is not the proper venue for discussing the details of the gas dynamics. But, maybe an analogy will help. Think of a mass and spring connected in series on a flat frictionless table, with the other end of the spring attached to the wall (which is right next to the table). Initially, the spring is compressed, and at time t = 0, the mass is released, and the system is allowed to oscillate. It will oscillate forever in simple harmonic motion, and the kinetic energy of the mass will continually vary with time as the spring alternately stores and releases elastic energy. The spring is analogous to our gas, and the mass is analogous to our piston. Now introduce a dissipative element into the picture. Connect a viscous damper between the wall and the mass, in parallel with the spring. The characteristic of the viscous damper is that the force it exerts on the mass is proportional to the velocity of the mass (including sign), but in the opposite direction. The combination of spring and viscous damper is now analogous to our gas, rather than just the spring. Now, if the mass is released at time zero, instead of having simple harmonic motion, the mass will undergo damped oscillation, and the amplitude of the oscillation will decrease with time, so that, in the end, the mass will stop moving. This is not an outrageously bad analogy of what happens with the gas and piston. Even though, in the case of a gas, it is incapable of exerting tension on the piston, the elastic collisions of the piston with the far cylinder wall do the job of reversing the piston direction.

What I'm confused about is how does internal irreversibility like viscous dissipation during compression/expansion cause a work loss.

Does it require a higher average applied pressure over the same volume change and if so why?

Solve the spring-mass-damper oscillation problem. That should help. See how the decrease in kinetic energy of the mass comes about.

Also, if some of the work applied turned into frictional heat and still goes into the system, how is that a loss?

It's not a loss. It's just a conversion of work into internal energy. During the initial expansion, the gas is doing work on the piston, and parts of the gas cool. When, in the end, the cumulative work done on the piston drops to zero, the viscous heating has returned the gas to its original temperature.

And for the vertical case it would be P_I,gas = mg/A_piston+ P_ext?

Yes, for quasistatic.

How does friction between the gas and cylinder during compression/expansion change the relationship between P_I,gas and P_ext; is it F_friction + P_I,gasA = P_extA?

Yes, for quasistatic horizontal.

Chet

 

[Red_CCF]

It would cause a sign reversal on the rate at which work is being done on the piston, not on the cumulative amount of work done on the piston.

Solve the spring-mass-damper oscillation problem. That should help. See how the decrease in kinetic energy of the mass comes about.

I am currently imagining this as, when the piston moves back to compress the gas, once v = 0 all of the work energy the gas gave to it during expansion has been returned, but some are added as heat (viscous dissipation/damper), while some as PdV work (spring), which the gas applies back to the piston. Each cycle the amount of PdV work exchanged decreases until it is 0 when the piston stops. Is this correct?

It's not a loss. It's just a conversion of work into internal energy. During the initial expansion, the gas is doing work on the piston, and parts of the gas cool. When, in the end, the cumulative work done on the piston drops to zero, the viscous heating has returned the gas to its original temperature.

In the P-V graph in my OP, compared to the reversible case, the irreversible case over the same volume change requires a higher exerted pressure during compression, and extracts a lower pressure during expansion. If viscosity is the only irreversibility, I kind of get how a higher pressure is needed during the compression stage (viscous dissipation generates heat -> higher T -> higher P during the process). But this doesn't explain why pressure is lower during expansion?

Yes, for quasistatic horizontal.

In this case, unlike our previous examples, P_ext ≠ P_I due to friction between the gas and wall. Should I use P_I,gas (if the system gas only and insulated) to calculate ∫PdV or is P_ext more appropriate since the system is insulated hence the frictional heat is going into the gas?

From a Newton's third law perspective, what are the action-reaction pairs? I see F_fric between gas-cylinder, but the piston is exerting P_ext on the gas, yet P_I,gas < P_ext is applied to the piston by the gas, where does the difference come in?

Thanks very much

 

[Chestermiller]

Hi Red_CCF,

I want to cover a couple of things in this post. First of all, I would like to temporarily table our discussion of the frictional heating associated with the piston. I feel we need to focus on the other issue we have been discussing. We can come back to the piston friction soon.

Secondly, I wanted to make you aware of the fact that, when discussing the force per unit area at the interface (what we have been calling P_I), we have been employing the elementary thermodynamics approximation of the situation. This description assumes that P_I is the thermodynamic pressure at the interface, determined by the static PVT behavior of the gas:

P_I=ρ_IRT_I/M

where T_I is the temperature of the gas at the interface, M is the molecular weight of the gas, and ρ_I is the gas density at the interface. This description is OK for a quasistatic process where the gas is not deforming rapidly, but for an irreversible process, it is not.

The correct way to describe the boundary condition at the piston interface is to use the compressive stress σ_I at the interface (i.e., to replace P_I with σ_I). For a non-quasistatic deformation, the compressive stress at the interface is the linear sum of two contributions: the thermodynamic pressure ρ_IRT_I/M plus a viscous stress contribution -2μ∂v/∂x, where ∂v/∂x is the rate of change of gas velocity with respect to position in the gas (in the immediate vicinity of the interface). So, mechanistically, the force per unit area at the interface is the result both of the compression/expansion deformation of the gas, plus the rate of deformation of the gas. The viscous part describes the effect of the rate of deformation of the gas. This part is insignificant for a quasistatic deformation.

In an irreversible expansion or compression, the viscous portion of the stress is present everywhere within the gas, and not just at the interface. It results in a dissipation of mechanical energy, and does not only give rise to an increase in the local temperature of the gas. For our purposes, its most important effect is the dissipation of mechanical energy, and the temperature rise is somewhat secondary.

Now, let's get back to the spring-mass-damper analogy that I was talking about. This analogy should give you an idea of what is happening mechanicsically in our gas/piston system. The combination of the spring and damper is supposed to be analogous to our gas. It exhibits a linear combination of elastic and dissipative behavior. If F is the force exerted by the combination of spring and damper on the mass, then:

F = k(x₀-x) - Cdx/dt

where x₀ is the location of the end of the spring when there is no tension in it, and C is the dissipative damper constant. F is analogous to σ_IA, k(x₀-x) is analogous to the thermodynamic PVT stress contribution, and - Cdx/dt is analogous to the viscous stress contribution. From the force balance on the mass:

md²x/dt²= F

or,

$$m\frac{d^2x}{dt^2}=k(x_0-x)-C\frac{dx}{dt}$$

The work done by the "gas" on the "piston" is given by dW = Fdx. If we multiply both sides of the above equation by dx/dt, we obtain:

$$\frac{dW}{dt}+\frac{k}{2}\frac{d(x_0-x)^2}{dt}=-Cv^2$$

Note from this equation that the combination of kinetic energy of the mass plus compressional energy of the spring is decreasing with time, irrespective of the direction that the mass is moving. Only when the velocity of the mass reaches zero does the compressional energy of the spring reach its final value (assuming, as in our problem, that there is an elastic wall causing the mass to rebound periodically, and located such that the spring never goes into tension...the initial location of the mass in our situation requires the spring to be in compression).

At infinite time, W = 0, so, according to the above equation:

$$\frac{k}{2}[(x_0-x_∞)^2-(x_0-x_{init})^2]=-\int_0^∞{Cv^2dt}$$
That is, the energy stored in the spring at infinite time is less than the energy stored in the spring initially.

I think I'll stop here and let you digest what I've written.

Chet

 

[Red_CCF]

Thank you for such a detailed explanation.

For a non-quasistatic deformation, the compressive stress at the interface is the linear sum of two contributions: the thermodynamic pressure ρ_IRT_I/M plus a viscous stress contribution -2μ∂v/∂x, where ∂v/∂x is the rate of change of gas velocity with respect to position in the gas (in the immediate vicinity of the interface). So, mechanistically, the force per unit area at the interface is the result both of the compression/expansion deformation of the gas, plus the rate of deformation of the gas. The viscous part describes the effect of the rate of deformation of the gas. This part is insignificant for a quasistatic deformation.

What is the origin of -2μ∂v/∂x and what is the physical mechanism by which the rate of compression induces stress such that they can be superpositioned to P_I instead of changing P_I? Is this effect negligible in irreversible but quasistatic cases?

In an irreversible expansion or compression, the viscous portion of the stress is present everywhere within the gas, and not just at the interface. It results in a dissipation of mechanical energy, and does not only give rise to an increase in the local temperature of the gas. For our purposes, its most important effect is the dissipation of mechanical energy, and the temperature rise is somewhat secondary.

If all other irreversibilities are nonexistent and only viscous dissipation exists, the viscous stress term would be the reason why a higher pressure (in compression) is needed for work compared to reversible cases? Is this why textbooks say that P_ext should always (reversible or irreversible) be used for calculating work?

Now, let's get back to the spring-mass-damper analogy that I was talking about. This analogy should give you an idea of what is happening mechanicsically in our gas/piston system. The combination of the spring and damper is supposed to be analogous to our gas. It exhibits a linear combination of elastic and dissipative behavior. If F is the force exerted by the combination of spring and damper on the mass, then:

F = k(x₀-x) - Cdx/dt

where x₀ is the location of the end of the spring when there is no tension in it, and C is the dissipative damper constant. F is analogous to σ_IA, k(x₀-x) is analogous to the thermodynamic PVT stress contribution, and - Cdx/dt is analogous to the viscous stress contribution. From the force balance on the mass:

md²x/dt²= F

or,

$$m\frac{d^2x}{dt^2}=k(x_0-x)-C\frac{dx}{dt}$$

The work done by the "gas" on the "piston" is given by dW = Fdx. If we multiply both sides of the above equation by dx/dt, we obtain:

$$\frac{dW}{dt}+\frac{k}{2}\frac{d(x_0-x)^2}{dt}=-Cv^2$$

Note from this equation that the combination of kinetic energy of the mass plus compressional energy of the spring is decreasing with time, irrespective of the direction that the mass is moving. Only when the velocity of the mass reaches zero does the compressional energy of the spring reach its final value (assuming, as in our problem, that there is an elastic wall causing the mass to rebound periodically, and located such that the spring never goes into tension...the initial location of the mass in our situation requires the spring to be in compression).

At infinite time, W = 0, so, according to the above equation:

$$\frac{k}{2}[(x_0-x_∞)^2-(x_0-x_{init})^2]=-\int_0^∞{Cv^2dt}$$
That is, the energy stored in the spring at infinite time is less than the energy stored in the spring initially.

I think I'll stop here and let you digest what I've written.

Chet

A couple of questions on this:

Is dW/dt supposed to represent the (rate of) kinetic energy (change) of the piston and how?

For this spring system, is there a simple way of showing ∫dW = 0 for t = 0 to ∞ or is the v = 0? If -Cv^2 = 0 (no damper), would the last equation be not integrable?

When you say that the spring energy is lower at t = ∞ than t = 0, is this equivalent of saying that the final gas pressure is lower than the initial and thus has less potential to push the piston if the cylinder wall was removed?

Thanks very much

 

[Chestermiller]

What is the origin of -2μ∂v/∂x and what is the physical mechanism by which the rate of compression induces stress such that they can be superpositioned to P_I instead of changing P_I?

It originates with Newton's Law of viscosity. It is part of the 3D tensorial version of the law. See Bird, Stewart, and Lightfoot, Transport Phenomena (a wonderful book that should be in everyone's library) for more details.

Is this effect negligible in irreversible but quasistatic cases?

Yes. But, in such cases, there can also be irreversibility due to conductive heat transport and chemical reaction.

If all other irreversibilities are nonexistent and only viscous dissipation exists, the viscous stress term would be the reason why a higher pressure (in compression) is needed for work compared to reversible cases?

No. During the irreversible deformation, the inertia of the gas also contributes to the force on the piston (the gas itself is experiencing spatially inhomogeneous acceleration, and is exchanging kinetic energy, in the same kind of way that the kinetic energy of the piston is contributing). However, at final steady state, of course, the gas is again stationary, and all its kinetic energy has been dissipated.

Is this why textbooks say that P_ext should always (reversible or irreversible) be used for calculating work?

Not exactly. In the irreversible case, you basically have very little control over what is happening in the gas (as you say), but you do have control over what you impose on its boundary. That's why you use P_ext, since you have control over that. In the reversible case, P_ext is the same as P of the gas (if, by P_ext, you mean the total force per unit area imposed on the boundary from outside, and the gas is your system).

Is dW/dt supposed to represent the (rate of) kinetic energy (change) of the piston and how?

Yes. We derived this a few posts ago, when we multiplied md²x/dt² by dx/dt to obtain mdv²/dt.

For this spring system, is there a simple way of showing ∫dW = 0 for t = 0 to ∞ or is the v = 0?

By solving the differential equation, you find that v = 0 at infinite time. This automatically means that ∫dW is equal to zero, since the change in kinetic energy of the piston is then zero.

If -Cv^2 = 0 (no damper), would the last equation be not integrable?

If there were no damping, then the system would oscillate forever. The equation would be integrable, and it would tell you that the sum of the kinetic energy plus the compressional energy is constant.

When you say that the spring energy is lower at t = ∞ than t = 0, is this equivalent of saying that the final gas pressure is lower than the initial and thus has less potential to push the piston if the cylinder wall was removed?

Yes.

I'll wait to see if you have any additional questions or clarifications. I also want to add one more detail to the spring damper analog model, so you can see how the mass of the gas comes into play. Tell me when you are ready to hear about this.

Chet

 

[Red_CCF]

It originates with Newton's Law of viscosity. It is part of the 3D tensorial version of the law. See Bird, Stewart, and Lightfoot, Transport Phenomena (a wonderful book that should be in everyone's library) for more details.

I always think of gas applying a force on something via pressure; how does this mechanism work such that an additional force is applied without increasing gas pressure?

No. During the irreversible deformation, the inertia of the gas also contributes to the force on the piston (the gas itself is experiencing spatially inhomogeneous acceleration, and is exchanging kinetic energy, in the same kind of way that the kinetic energy of the piston is contributing). However, at final steady state, of course, the gas is again stationary, and all its kinetic energy has been dissipated.

Assuming the process is irreversible but quasistatic, can gas inertia be neglected? I'm interested in looking at viscous dissipation by itself, and why it would required pressure is higher during compression yet lower during expansion compared to a reversible case.

Not exactly. In the irreversible case, you basically have very little control over what is happening in the gas (as you say), but you do have control over what you impose on its boundary. That's why you use P_ext, since you have control over that. In the reversible case, P_ext is the same as P of the gas (if, by P_ext, you mean the total force per unit area imposed on the boundary from outside, and the gas is your system).

In a general case (quasistatic or not), if the gas is the system, should ∫P_extdV always be used, but P_ext may or may not equal to P_I due to friction or viscous stresses etc.? In your blog, what was the assumption behind the system type such that P_I may be used for work calculations?

By solving the differential equation, you find that v = 0 at infinite time. This automatically means that ∫dW is equal to zero, since the change in kinetic energy of the piston is then zero.

Does the path dependence of work affect the way it's integrated? Can we just subtract the piston's KE at the beginning and end (0 both ways) and say that the net work done on it is 0?

I'll wait to see if you have any additional questions or clarifications. I also want to add one more detail to the spring damper analog model, so you can see how the mass of the gas comes into play. Tell me when you are ready to hear about this.

Please do.

Thanks very much

 

[Chestermiller]

I always think of gas applying a force on something via pressure; how does this mechanism work such that an additional force is applied without increasing gas pressure?

Newton's law of viscosity shows that the stress within a fluid (a gas is an incompressible fluid) is a function not only of the amount that the fluid has deformed, but also the rate at which it is deforming. For a so-called Newtonian fluid, the stress is a linear function of both the volumetric strain and the rate of strain. Almost all gases and most liquids are described extremely accurately by the Newtonian fluid model (properly expressed mathematically in 3D tensorial form). The thermodynamic pressure corresponds to the portion of the stress resulting from volumetric strain, and the viscous stresses describe the portion of the stress resulting from rate of strain. So, in summary, the pressure does not tell the whole story.

Assuming the process is irreversible but quasistatic, can gas inertia be neglected?

Yes.

I'm interested in looking at viscous dissipation by itself, and why it would required pressure is higher during compression yet lower during expansion compared to a reversible case.

One answer to this question is to say "see BSL for the molecular mechanism associated with viscosity." Another answer would be to consider a very viscous fluid bar that is being pulled on both ends. The rate of deformation of the bar results in actual tension in the bar, which is the same as negative compressive stress. Once you stop the deformation (i.e., zero rate of deformation), the tension goes away. From a fluid mechanics standpoint, liquids and gases are both described very accurately by the Newtonian fluid model.

I very strongly encourage you to read the first chapter or two in BSL, since you now seem highly motivated to learn this material. You won't be disappointed.

In a general case (quasistatic or not), if the gas is the system, should ∫P_extdV always be used, but P_ext may or may not equal to P_I due to friction or viscous stresses etc.?

As I said, if the gas is the system, P_I needs to be replaced by σ_I. This takes care of dealing with the viscous stresses that are present during irreversible. What we are dealing with here is the fact that, in most thermodynamics text developments, the authors play it fast and loose with the mathematics. They are trying to keep it simple (or are even unaware of the contribution of viscous stresses), but, as a result, some confusion ensues. If P_ext represents the force per unit area on the top of the piston, then ∫P_extdV is the work done on the system only if the massless piston is included as part of the system. Then, P_ext = F/A + σ_I, where F is the frictional force of the cylinder on the massless piston.

In your blog, what was the assumption behind the system type such that P_I may be used for work calculations?

In my Blog, I took a bit of literary license when I called P_I the pressure at the interface. I was trying to keep it simple for new students of thermodynamics, and was also trying to be somewhat consistent with what they would encounter in other elementary thermo developments. So, in a way, I was doing what I criticized above (playing a little fast and loose with the mathematics). However, I knew that it was going to be mostly novices that read my Blog, and I had some important points I wanted to make, particularly with regard to the Second Law. I didn't want to lose everyone who was reading the blog by starting to introduce concepts such as the stress tensor, deformational mechanics, and Newtonian fluids. These concepts are not typically included as part of the elementary thermo curriculum, and are not introduced until later when students start studying Fluid Mechanics, Solid Mechanics, and Continuum Mechanics.

Now that you personally have progressed beyond the novice level, you have begun to ask more perceptive questions that can be answered by starting to study these more advanced and precise disciplines.

Does the path dependence of work affect the way it's integrated?

No. The mathematics does not lie.

Can we just subtract the piston's KE at the beginning and end (0 both ways) and say that the net work done on it is 0?

Yes.

I'm going to stop here and try to field any further questions and clarifications you might have before proceeding to the more detailed mechanical analog that includes the inertia of the gas.

Chet

 

[Red_CCF]

Newton's law of viscosity shows that the stress within a fluid (a gas is an incompressible fluid) is a function not only of the amount that the fluid has deformed, but also the rate at which it is deforming. For a so-called Newtonian fluid, the stress is a linear function of both the volumetric strain and the rate of strain. Almost all gases and most liquids are described extremely accurately by the Newtonian fluid model (properly expressed mathematically in 3D tensorial form). The thermodynamic pressure corresponds to the portion of the stress resulting from volumetric strain, and the viscous stresses describe the portion of the stress resulting from rate of strain. So, in summary, the pressure does not tell the whole story.

Is gas considered incompressible in the sense that the Ma << 0.3? Is it possible to measure the force due to rate of strain alone like one would do with P_I?

As I said, if the gas is the system, P_I needs to be replaced by σ_I. This takes care of dealing with the viscous stresses that are present during irreversible. What we are dealing with here is the fact that, in most thermodynamics text developments, the authors play it fast and loose with the mathematics. They are trying to keep it simple (or are even unaware of the contribution of viscous stresses), but, as a result, some confusion ensues. If P_ext represents the force per unit area on the top of the piston, then ∫P_extdV is the work done on the system only if the massless piston is included as part of the system. Then, P_ext = F/A + σ_I, where F is the frictional force of the cylinder on the massless piston.

Is ∫σ_IdV easily solvable, as it requires the knowledge of ∂v/∂t as a function of volume? Is ∂v/∂t negative during compression?

With regards to my question about friction between the gas-cylinder (but not piston-cylinder), the pressure exerted onto the gas at the gas-piston interface is higher than P_I (or σ_I) if the piston wants to compress the gas. What are the Newton's Third Law action/reaction pairs as at the gas-piston interface, the forces applied by the two on each other in this case are not equal?

For a gas that undergoes a reversible or irreversible cycle, net work depends on the process and is not necessarily zero even though it begins and ends in the same state. How come for the piston this isn't the case?

Thanks very much

 

[Chestermiller]

Is gas considered incompressible in the sense that the Ma << 0.3?

This is a little off our topic. A gas is not usually considered incompressible. I would regard the deformation or flow of a gas as being incompressible if conditions are such that its density only changes by an amount insignificant in terms of what you are trying to calculate.

Is it possible to measure the force due to rate of strain alone like one would do with P_I?

For shear forces in some simple deformations it is, and for situations where the perpendicular contribution of the viscous stress at a boundary is zero (such that the only stress contribution is from the pressure P_I), it is. Ordinarily, it is not possible to elucidate the separate contributions of the pressure and the viscous stresses experimentally, and you are stuck with getting their combined stress. Of course, if the deformation is very slow, the viscous contribution is negligible.

Is ∫σ_IdV easily solvable, as it requires the knowledge of ∂v/∂t as a function of volume?

Actually, it should be dv/dx (rather than dv/dt), or, in more generality, the partial derivatives of all the velocity components with respect to all the spatial coordinates. You seem to be asking how one goes about solving fluid mechanics or gas dynamics problems, and, are they hard to solve. Some fluid mechanics and gas dynamics problems are fairly easy to solve, and can be handled analytically, while others are much more complicated, and require numerical solutions. Still others, involving turbulent flow, are still harder and require appropriate approximations. Usually, one will be dealing with the solution of a set of non-linear coupled partial differential equations. These equations are called the Navier-Stokes equations.

Is ∂v/∂t negative during compression?

Again, it should be dv/dx. If dv/dx is negative, the gas is being compressed. This would make the total compressive stress higher than just the thermodynamic pressure.

With regards to my question about friction between the gas-cylinder (but not piston-cylinder), the pressure exerted onto the gas at the gas-piston interface is higher than P_I (or σ_I) if the piston wants to compress the gas. What are the Newton's Third Law action/reaction pairs as at the gas-piston interface, the forces applied by the two on each other in this case are not equal?

I'm not quite sure what you are asking here. By "friction between the gas and cylinder," I assume you mean viscous shear stress at the cylinder wall. Is this what you mean? Newton's Third Law is always going to be satisfied. The force per unit area exerted by the gas on the wall is always going to be equal to the force per unit area exerted by the wall on the gas. The determination of the equation for this force per unit area in terms of the thermodynamic pressure and velocity gradients at the wall is where the 3D tensorial version of the Newtonian fluid model comes in.

For a gas that undergoes a reversible or irreversible cycle, net work depends on the process and is not necessarily zero even though it begins and ends in the same state. How come for the piston this isn't the case?

Because there is no dissipation within the piston. An ideal piston is rigid, and any forces applied to it can only result in changes in kinetic energy (i.e., conservation of mechanical energy). When it returns to zero velocity, the net work done on it must be zero.

Chet

 

[Red_CCF]

For shear forces in some simple deformations it is, and for situations where the perpendicular contribution of the viscous stress at a boundary is zero (such that the only stress contribution is from the pressure P_I), it is. Ordinarily, it is not possible to elucidate the separate contributions of the pressure and the viscous stresses experimentally, and you are stuck with getting their combined stress. Of course, if the deformation is very slow, the viscous contribution is negligible.

How would one measured the combined pressure and viscous stress?

I'm not quite sure what you are asking here. By "friction between the gas and cylinder," I assume you mean viscous shear stress at the cylinder wall. Is this what you mean? Newton's Third Law is always going to be satisfied. The force per unit area exerted by the gas on the wall is always going to be equal to the force per unit area exerted by the wall on the gas. The determination of the equation for this force per unit area in terms of the thermodynamic pressure and velocity gradients at the wall is where the 3D tensorial version of the Newtonian fluid model comes in.

I see an equal and opposite force between the wall and the gas, but for the gas and the piston, the piston exerts P_ext on the gas while the gas exerts only P_I (or σ_I) < P_ext as friction is pushing it, what makes up the difference?

As I said, if the gas is the system, P_I needs to be replaced by σ_I. This takes care of dealing with the viscous stresses that are present during irreversible. What we are dealing with here is the fact that, in most thermodynamics text developments, the authors play it fast and loose with the mathematics. They are trying to keep it simple (or are even unaware of the contribution of viscous stresses), but, as a result, some confusion ensues. If P_ext represents the force per unit area on the top of the piston, then ∫P_extdV is the work done on the system only if the massless piston is included as part of the system. Then, P_ext = F/A + σ_I, where F is the frictional force of the cylinder on the massless piston.

So the only reason that using P_ext becomes a problem is that the piston must be included as the system, but so long as piston-cylinder friction is negligible and piston is massless the result would be identical to that of a gas only system (as P_ext = force of piston on the gas)? I am wondering because in the examples we talked about where the above assumptions hold, I did not see any where using P_ext would have given the wrong solution. Also, in the previous examples with shear stress friction between gas and cylinder, wouldn't using σ_I give an incorrect answer as it does not account for gas-wall friction (assuming the heat all goes to the gas)?

With regards to free expansion in a vacuum mentioned earlier, a couple of textbooks explicitly state that PdV ≠ 0 but W = 0 by intuitive argument. Earlier you mentioned that P_I = 0 only as a limiting case. Is it possible to show that even if P_I ≠ 0 that W = 0 (assuming massless piston)? The only examples I have found involves using P_ext = 0, is this valid?

Lastly, slightly off topic, in my OP had I chosen to only include the gas as the system, would the frictional heat input be considered a Q in first law, despite it not being the traditional heat transfer via temp. gradient?

Thank you very much

 

[Chestermiller]

How would one measured the combined pressure and viscous stress?

With a stress transducer (or maybe two stress transducers, one to measure the normal stress and one to measure the shear stress).

I see an equal and opposite force between the wall and the gas, but for the gas and the piston, the piston exerts P_ext on the gas while the gas exerts only P_I (or σ_I) < P_ext as friction is pushing it, what makes up the difference?

I don't fully understand this question. Remember, first one has to identify what the system is (gas or gas+piston). We do this by visualizing an imaginary boundary, the interface I, between the system and the surroundings. P_I, or, more precisely, σ_I is the normal force per unit area at the interface. If we take the gas as our system, and, if the piston exerts a force per unit area of σ_I on the gas, then the gas exerts a force per unit area of σ_I on the piston. If we take the gas plus piston as our system, with the interface taken as the outer surface of the piston, and, if the surroundings exerts a force per unit area of P_ext on the system (i.e., the outer surface of the piston), then the system exerts an equal and opposite force per unit area of P_ext on the surroundings. In the latter case, it would also be fair to call σ_I, which represents the force per unit area at the interface between the system and surroundings, as P_ext. (We're getting killed by notational confusion).

The important thing to recognize is that all we are talking about here is the stress boundary condition between the system and the surroundings at the interface. The force per unit area exerted by the system on the surroundings at the interface must match the force per unit area exerted by the surroundings on the system at the interface. The beauty of the First Law is that, even with an irreversible process, in many cases all we need to know is the stress exerted by the surroundings on the system at the interface. In most cases, we don't need to solve the dynamic and heat transfer equations within the system for the stress at the interface exerted by the system (in order to calculate the work) if we can force the surroundings to impose whatever stress we desire at the interface.

So the only reason that using P_ext becomes a problem is that the piston must be included as the system, but so long as piston-cylinder friction is negligible and piston is massless the result would be identical to that of a gas only system (as P_ext = force of piston on the gas)?

Yes. But I don't regard P_ext as becoming a problem.

Also, in the previous examples with shear stress friction between gas and cylinder, wouldn't using σ_I give an incorrect answer as it does not account for gas-wall friction (assuming the heat all goes to the gas)?

If there is viscous drag by the cylinder wall on the gas, this effect is transmitted to the interface, and influences σ_I. If, in the end, the total amount of work done by the system on the surroundings is zero, the presence of high viscous drag only influences the amount of time it takes for the system to reach its final equilibrium state (i.e., stop oscillating) and not the total amount of work.

With regards to free expansion in a vacuum mentioned earlier, a couple of textbooks explicitly state that PdV ≠ 0 but W = 0 by intuitive argument. Earlier you mentioned that P_I = 0 only as a limiting case. Is it possible to show that even if P_I ≠ 0 that W = 0 (assuming massless piston)? The only examples I have found involves using P_ext = 0, is this valid?

I'm not familiar with exactly what they said in your references. If you want to provide the exact quotes in context, I would be glad to comment. All I can say is that I'm confident in what I am saying. If P_I ≠ 0 (assuming a massless piston), W≠0. We need to be careful precisely what we call P_ext and P_I for each system. I've been calling P_ext the pressure on the other side of the piston, but they might be calling P_ext what I call P_I.

Lastly, slightly off topic, in my OP had I chosen to only include the gas as the system, would the frictional heat input be considered a Q in first law, despite it not being the traditional heat transfer via temp. gradient?

If the heat makes its way into the gas (and not through the piston into the surroundings), then, in this case the frictional heat input is considered Q for the system. How would it make its way into the gas? Conduction would be involved, but I won't break it down into how this happens in detail here.

Chet

 

[Red_CCF]

Actually, it should be dv/dx (rather than dv/dt), or, in more generality, the partial derivatives of all the velocity components with respect to all the spatial coordinates. You seem to be asking how one goes about solving fluid mechanics or gas dynamics problems, and, are they hard to solve. Some fluid mechanics and gas dynamics problems are fairly easy to solve, and can be handled analytically, while others are much more complicated, and require numerical solutions. Still others, involving turbulent flow, are still harder and require appropriate approximations. Usually, one will be dealing with the solution of a set of non-linear coupled partial differential equations. These equations are called the Navier-Stokes equations.

Is it reasonable to find an spatial average of σ_I over the piston face by using the dv/dx of the piston and not necessarily the local dv/dx of the gas (which likely leads to the use of CFD)?

The important thing to recognize is that all we are talking about here is the stress boundary condition between the system and the surroundings at the interface. The force per unit area exerted by the system on the surroundings at the interface must match the force per unit area exerted by the surroundings on the system at the interface. The beauty of the First Law is that, even with an irreversible process, in many cases all we need to know is the stress exerted by the surroundings on the system at the interface. In most cases, we don't need to solve the dynamic and heat transfer equations within the system for the stress at the interface exerted by the system (in order to calculate the work) if we can force the surroundings to impose whatever stress we desire at the interface.

So in engineering practice, it is easier/more practical to use P_ext (the pressure the piston exerts on the gas if the system is gas only, or surrounding onto piston if the system includes the piston) as opposed to P_I or σ_I that most of our discussion revolved around?

I don't fully understand this question. Remember, first one has to identify what the system is (gas or gas+piston). We do this by visualizing an imaginary boundary, the interface I, between the system and the surroundings. P_I, or, more precisely, σ_I is the normal force per unit area at the interface. If we take the gas as our system, and, if the piston exerts a force per unit area of σ_I on the gas, then the gas exerts a force per unit area of σ_I on the piston. If we take the gas plus piston as our system, with the interface taken as the outer surface of the piston, and, if the surroundings exerts a force per unit area of P_ext on the system (i.e., the outer surface of the piston), then the system exerts an equal and opposite force per unit area of P_ext on the surroundings. In the latter case, it would also be fair to call σ_I, which represents the force per unit area at the interface between the system and surroundings, as P_ext. (We're getting killed by notational confusion).

In my example I took the gas as the system. If I am compressing a piston with gas-cylinder wall friction, work is needed to overcome wall shear stress. I am thinking the force balance to be P_{ext,piston on gas}A_piston = σ_IA_piston + $\tau$_gas-cylinderA_cylinder. I saw an action-reaction pair of $\tau$_gas-cylinder between the gas and cylinder. I also see σ_I of the gas onto the piston and P_ext as piston onto the gas. The problem is for the piston to move and overcome friction P_ext > σ_I, and the gas-piston forces onto each other are not equal; I believe that I have a conceptual error about how the forces interact.

If there is viscous drag by the cylinder wall on the gas, this effect is transmitted to the interface, and influences σ_I. If, in the end, the total amount of work done by the system on the surroundings is zero, the presence of high viscous drag only influences the amount of time it takes for the system to reach its final equilibrium state (i.e., stop oscillating) and not the total amount of work.

This probably has the same answer my question above. If my force balance above was correct, had I used W =∫σ_IdV, then I would not have accounted for the extra energy used to overcome $\tau$_gas-cylinder and thus be incorrect?

I'm not familiar with exactly what they said in your references. If you want to provide the exact quotes in context, I would be glad to comment. All I can say is that I'm confident in what I am saying. If P_I ≠ 0 (assuming a massless piston), W≠0. We need to be careful precisely what we call P_ext and P_I for each system. I've been calling P_ext the pressure on the other side of the piston, but they might be calling P_ext what I call P_I.

This is from Pippard's Classical Thermodynamics (figure attached) at the end of his development of the TdS equations

... illustrated by the experiment pictured in fig 3., the expansion of gas into a vacuum under isolated conditions. Here PdV has a definite, non-zero value, but w = 0; similarly q = 0 but TdS has a non-zero value, which could be determined by compressing the gas reversibly to its initial state and determining how much heat must be extracted during the compression. We should find that it is just equal to PdV, and that in consequence the entropy increase during the irreversible expansion was PdV/T.

However the wiki article on free expansion appear to agree with you:

A free expansion is typically achieved by opening a stopcock that allows the gas to expand into a vacuum. Although it would be difficult to achieve in reality, it is instructive to imagine a free expanion caused by moving a piston faster than virtually any atom. No work is done because there is no pressure on the piston. No heat energy leaves or enters the piston. Nevertheless, there is an entropy change, and the well-known formula for entropy change,

Thanks very much

 

[Red_CCF]

Sorry the figure is attached to this post.

 

[Chestermiller]

Is it reasonable to find an spatial average of σ_I over the piston face by using the dv/dx of the piston and not necessarily the local dv/dx of the gas (which likely leads to the use of CFD)?

Oops. Those dv/dx's should have been partial derivatives ∂v/∂x at constant time. The piston is regarded as a rigid body, so, the partial derivative of its velocity with respect to x, ∂v/∂x = 0.

So in engineering practice, it is easier/more practical to use P_ext (the pressure the piston exerts on the gas if the system is gas only, or surrounding onto piston if the system includes the piston) as opposed to P_I or σ_I that most of our discussion revolved around?

Excellent question.

If the process is reversible, then we need to take into account the pressure of the gas; it is uniform within the cylinder, and satisfies the equilibrium equation of state of the gas. The P_ext then has to be customized to make good on this.

If the process is irreversible, we don't know what the pressure of the gas at the interface is going to be, and all we can do is specify how P_ext is imposed as a function of time. We have been devoting most of our discussion to the irreversible case because you were curious about what was going on within the cylinder during the irreversible case, and how the force per unit area within the gas at the piston face comes about. The original questions centered around why the gas force per unit area at the piston face was not (a uniform) p throughout the cylinder. We noted that, in an irreversible expansion, the gas pressure varies with position along the cylinder. We also noted that viscous stresses contribute to the gas force per unit area at the piston face.

In both the reversible and irreversible cases, P_ext gives the work done by the system on the surroundings. If the piston is frictionless and massless, then σ_I=P_ext. And, if the piston is frictionless and massless, and the process is reversible, then P_ext=p.

In my example I took the gas as the system. If I am compressing a piston with gas-cylinder wall friction, work is needed to overcome wall shear stress. I am thinking the force balance to be P_{ext,piston on gas}A_piston = σ_IA_piston + $\tau$_gas-cylinderA_cylinder. I saw an action-reaction pair of $\tau$_gas-cylinder between the gas and cylinder. I also see σ_I of the gas onto the piston and P_ext as piston onto the gas. The problem is for the piston to move and overcome friction P_ext > σ_I, and the gas-piston forces onto each other are not equal; I believe that I have a conceptual error about how the forces interact.

No. You're right. You have a conceptual error. For the massless frictionless piston case that you are describing here, the force balance on the piston gives: P_{ext,piston on gas}A_piston = σ_IA_piston. The shear stress within the cylinder at the wall does not come into play in this equation (because it is not acting on the piston). But, the wall shear stress is included in the gas dynamics analysis. A simplified example of how this comes about is to look at the steady flow of an incompressible fluid in a pipe. In this case, the fluid velocity is not a function of axial position x. The force balance on the fluid between cross sections at x and x + Δx is:

$$(P(x)-P(x+Δx))A=πDΔxτ_{wall}$$

You can see that the pressure gradient along the pipe is determined by the shear stress at the wall. The gas dynamics case is more complicated, but this example at least shows how the wall shear stress comes into play.

This probably has the same answer my question above. If my force balance above was correct, had I used W =∫σ_IdV, then I would not have accounted for the extra energy used to overcome $\tau$_gas-cylinder and thus be incorrect?

True, but your force balance was not correct.

This is from Pippard's Classical Thermodynamics (figure attached) at the end of his development of the TdS equations

What he's saying (rather poorly and ambiguously) is: If we integrate pdV using p and V from the ideal gas law, we get a definite value for the integral, but it bears no relation to the work done in free expansion (which is zero). The integral of pdV only gives the correct value for the work W if the process is reversible.

However the wiki article on free expansion appear to agree with you:

The wiki article is clearer, but there is still a problem with what they say. If the gas is released through a stopcock, then there is a pressure drop across the stopcock, and this does not constitute free expansion. In order to have free expansion, there can't be any resistance to the gas expanding. There have been several threads in Physics Forums recently in which we have worked problems involving a high pressure gas passing through a valve into a lower pressure container until the pressures in the two containers equilibrate. Even if the initial pressure in the low pressure container is zero (pure vacuum), this is not free expansion.

Chet

 

[Red_CCF]

Oops. Those dv/dx's should have been partial derivatives ∂v/∂x at constant time. The piston is regarded as a rigid body, so, the partial derivative of its velocity with respect to x, ∂v/∂x = 0.

Since gas-piston interface moves mimics that of the piston, does this mean that ∂v/∂x = 0 for the layer of gas in contact with the piston on average, or is local turbulence such that on average ∂v/∂x deviates from 0 at all times?

With a stress transducer (or maybe two stress transducers, one to measure the normal stress and one to measure the shear stress).

Is this the same as a pressure transducer? For instance, is it possible to differentiate pressure stress and total stress?

What he's saying (rather poorly and ambiguously) is: If we integrate pdV using p and V from the ideal gas law, we get a definite value for the integral, but it bears no relation to the work done in free expansion (which is zero). The integral of pdV only gives the correct value for the work W if the process is reversible.

I'm a little confused by this, what's the significance of using P and V from ideal gas, is it that of the entire gas and not the interface? How does δW = pdV only apply for reversible processes? Also if q = 0, is he just saying that TdS = Tδσ (σ is the notation for entropy generation from irreversibilities).

Really? I'm very surprised, because, in freshman physics, you did a zillion problems where exactly this same type of situation prevailed, and you had absolutely no trouble with it at all. Suppose you have a rigid massless table top supported by cinder blocks at its four corners, and you put a big heavy sheet birthday cake on top of the table. The cake covers the entire surface of the table, and exerts a downward pressure of W/A uniformly over the entire table top, but the cinder blocks exert much higher upward pressures over their respective areas A' at the corners of the table: W/(4A'), where A >>4A'. In the portion of the table away from the corners, the pressure on the top of the table is W/A, while the pressure on the bottom of the table is zero.

Sorry to come back to this again. In the above example, between the table and cinder block, at every point of the contact the pressure they apply onto each other is W/(4A'). However, at the piston-gas interface, if P_ext,piston (perfectly uniform onto the gas) may not necessarily equal σ_I locally (but does over the piston face), this seems to imply a violation of Newton's Third Law on a local basis, as I should be able to find a small area where the forces on either side of the gas-piston interface are not equal.

Thank you very much

 

[Chestermiller]

Since gas-piston interface moves mimics that of the piston, does this mean that ∂v/∂x = 0 for the layer of gas in contact with the piston on average, or is local turbulence such that on average ∂v/∂x deviates from 0 at all times?

No, ∂v/∂x ≠ 0 for the layer of gas in contact with the piston. The only mechanical quantities that have to be continuous at the interface between the gas and the piston are

• the velocity
• the stress

The gas flow is not necessarily turbulent; it can also be laminar. But, ∂v/∂x positive means that the gas is expanding, and ∂v/∂x negative means that the gas is compressing.

Is this the same as a pressure transducer? For instance, is it possible to differentiate pressure stress and total stress?

A pressure transducer is the same thing as a normal-stress transducer; it measures the total stress perpendicular to the boundary, and we cannot resolve the pressure portion of the stress from the total normal stress.

To measure the shear stress on a boundary, one needs to use another type of transducer.

I'm a little confused by this, what's the significance of using P and V from ideal gas, is it that of the entire gas and not the interface?

It's neither. The P we are talking about here is what you calculate from P=nRT/V. As we've said before, the thermodynamic pressure of the gas is not even constant spatially within the cylinder, and, during an irreversible deformation, the thermodynamic pressure is not even equal to the total normal stress at any location.

How does δW = pdV only apply for reversible processes?

In a reversible process, the thermodynamic pressure of the gas within the cylinder actually is constant spatially, and the thermodynamic pressure is equal to the total normal stress at any location. Viscous stresses are not contributing at all. So the pressure at the piston face is the same as the pressure you calculate from the ideal gas law.

If the process is irreversible, the only way you can get the work is to use σ_IdV=P_extdV (assuming a massless, frictionless piston). In a reversible process, σ_IdV = (nRT/V)dV = P_extdV.

Also if q = 0, is he just saying that TdS = Tδs (s is the notation for entropy generation from irreversibilities).

This kind-of assumes that entropy (and, rate of change of entropy) can be evaluated locally at all points within a system experiencing an irreversible process. One then integrates the above equation over the volume of the system to get the total rate of entropy generation. There are many in Physics Forums that have taken issue with this assumption in the past, saying that the entropy of a system can only be determined when the system is in an equilibrium state. On the other hand, I happen to subscribe to and accept this assumption because it has been found to give the correct overall change in entropy between the initial and final states when the rate of entropy generation is expressed by a particular equation.

Sorry to come back to this again. In the above example, between the table and cinder block, at every point of the contact the pressure they apply onto each other is W/(4A'). However, at the piston-gas interface, if P_ext,piston (perfectly uniform onto the gas) may not necessarily equal σ_I locally (but does over the piston face), this seems to imply a violation of Newton's Third Law on a local basis, as I should be able to find a small area where the forces on either side of the gas-piston interface are not equal.

This is an incorrect application of Newton's third law. The correct rendering of Newton's third law is: On the gas-side of the piston, the stress that the gas exerts on the piston face σ_I is equal and opposite to the stress that the gas-side of the piston face exerts on the gas at each an every point on the piston face. If the piston is massless and frictionless, the force exerted on the non-gas-side of the piston is equal in magnitude and opposite in direction to the stress σ_I integrated over the gas-side of the piston face. The latter is not Newton's third law, but rather the result of a force balance applied to the piston.

Chet

 

[Red_CCF]

It's neither. The P we are talking about here is what you calculate from P=nRT/V. As we've said before, the thermodynamic pressure of the gas is not even constant spatially within the cylinder, and, during an irreversible deformation, the thermodynamic pressure is not even equal to the total normal stress at any location.

For irreversible processes (like the free expansion process), what would the P calculated from ideal gas law represent; is it some averaged pressure term for the whole system? My current impression is that this P term is more specific than what we are discussing, which is a general stress term σ_I specific to the system interface?

Below is from the same paragraph from Pippard as I posted earlier

It is only for a reversible change, however, that q = TdS and w = -pdV; neither of these equalities hold for an irreversible change - q ≠ TdS and w ≠ -pdV - but dU = TdS - pdV is still valid; if q = TdS - ε then w = -pdV + ε. This point is illustrated ... [rest of the paragraph on the free expansion example posted earlier]

Do you happen to know what he means by the bolded and why it is true?

In a reversible process, the thermodynamic pressure of the gas within the cylinder actually is constant spatially, and the thermodynamic pressure is equal to the total normal stress at any location. Viscous stresses are not contributing at all. So the pressure at the piston face is the same as the pressure you calculate from the ideal gas law.

For the TdS equation dU = TdS - pdV, I understand T to be the temperature at the system boundary where heat transfer is occurring regardless of the process, but does p = σ_I for irreversible processes (if not what pressure are we taking for p)? If so can σ_I be considered a state property?

If the process is irreversible, the only way you can get the work is to use σ_IdV=P_extdV (assuming a massless, frictionless piston). In a reversible process, σ_IdV = (nRT/V)dV = P_extdV.

Is the bolded also true for quasistatic irreversible process?

Thank you very much

 

[Chestermiller]

For irreversible processes (like the free expansion process), what would the P calculated from ideal gas law represent; is it some averaged pressure term for the whole system? My current impression is that this P term is more specific than what we are discussing, which is a general stress term σ_I specific to the system interface?

In an irreversible process, the pressure and the temperature are both functions of position (at a given time), there is probably not just one unique average temperature and average pressure for the system that also satisfy the ideal gas law. Even if we mathematically define average pressure and temperature in such a way that the ideal gas law is satisfied, it is unlikely that these values will have any physical relevance, and they couldn't be used to correctly calculate the amount of work that is done. So, what good is that.

Below is from the same paragraph from Pippard as I posted earlier

It is only for a reversible change, however, that q = TdS and w = -pdV; neither of these equalities hold for an irreversible change - q ≠ TdS and w ≠ -pdV - but dU = TdS - pdV is still valid; if q = TdS - ε then w = -pdV + ε. This point is illustrated ... [rest of the paragraph on the free expansion example posted earlier]
Do you happen to know what he means by the bolded and why it is true?

dU = TdS - pdV describes the changes in U, S, and V between two differentially separated equilibrium states. For an irreversible process between these same two differentially separated equilibrium states, the equation q = TdS - ε suggests that the heat transferred along the irreversible path is less than the heat transferred along the reversible path; but this is equivalent to the Clausius inequality only if T for the irreversible path is taken as the temperature at the interface for the reversible path. However, then T_IdS in not really the heat transferred over the reversible path. I'm a little uncomfortable with this interpretation. I would have preferred that they had written: q = T_IdS - ε. Maybe they felt that, for differentially separated equilibrium states, the difference between T for the reversible path and T_I for the irreversible path would be insignificant.

For the TdS equation dU = TdS - pdV, I understand T to be the temperature at the system boundary where heat transfer is occurring regardless of the process, but does p = σ_I for irreversible processes (if not what pressure are we taking for p)? If so can σ_I be considered a state property?

As I indicated in my blog, dq/T_I for an irreversible path is less than dS between two differentially separated equilibrium states. Also, σ_I is definitely not a state property, since, for an irreversible path, it includes the effects of viscous stresses.

If the process is irreversible, the only way you can get the work is to use σIdV=PextdV (assuming a massless, frictionless piston). In a reversible process, σIdV = (nRT/V)dV = PextdV.

Is the bolded also true for quasistatic irreversible process?

In a quasitatic irreversible process, it is reasonable to assume that the viscous stresses are neglibible, and the pressure is uniform within the system. But this doesn't mean that the temperature is uniform. If the temperature is not uniform, and you know the temperature distribution, you can still calculate the pressure at the interface and use that to calculated dW = pdV. However, it will not be equal to (nRT/V)dV, and you would have to do calculations to take into account the variations of both temperature and molar density within the cylinder, under the constraint that the total number of moles is constant. But this is just the kind of thing you are trying to avoid (detailed analysis of the variations within the system) when you use the first law.

Chet

 

[Red_CCF]

In an irreversible process, the pressure and the temperature are both functions of position (at a given time), there is probably not just one unique average temperature and average pressure for the system that also satisfy the ideal gas law. Even if we mathematically define average pressure and temperature in such a way that the ideal gas law is satisfied, it is unlikely that these values will have any physical relevance, and they couldn't be used to correctly calculate the amount of work that is done. So, what good is that.

Is the point of the ideal gas law pressure/temperature to describe the state of the whole system during at any point in process with one pressure and temperature (only possible if reversible), and if the process is irreversible, this becomes meaningless due to spatial variations of the properties in the system?

In Pippard's free expansion example where he states δW≠ PdV, what pressure is this/what does it "describe"?

dU = TdS - pdV describes the changes in U, S, and V between two differentially separated equilibrium states. For an irreversible process between these same two differentially separated equilibrium states, the equation q = TdS - ε suggests that the heat transferred along the irreversible path is less than the heat transferred along the reversible path; but this is equivalent to the Clausius inequality only if T for the irreversible path is taken as the temperature at the interface for the reversible path. However, then T_IdS in not really the heat transferred over the reversible path. I'm a little uncomfortable with this interpretation. I would have preferred that they had written: q = T_IdS - ε. Maybe they felt that, for differentially separated equilibrium states, the difference between T for the reversible path and T_I for the irreversible path would be insignificant.

For the bolded, if some irreversible process is performed, why must we use T_I of a reversible process and not the T_I of the actual (irreversible) process we are performing for equivalence to Clausius inequality? How come T_IdS for a reversible process does not equal δQ?

What dictates in the TdS equation that if q = TdS - ε then w = -pdV + ε? Is ε = T*dσ (entropy generation)?

Does σ_I have any role in the TdS equations if the process is irreversible or is it only the pressure component of the stress included? What do we use for the pressure term in TdS equations during irreversible processes?

In a quasitatic irreversible process, it is reasonable to assume that the viscous stresses are neglibible, and the pressure is uniform within the system. But this doesn't mean that the temperature is uniform. If the temperature is not uniform, and you know the temperature distribution, you can still calculate the pressure at the interface and use that to calculated dW = pdV. However, it will not be equal to (nRT/V)dV, and you would have to do calculations to take into account the variations of both temperature and molar density within the cylinder, under the constraint that the total number of moles is constant. But this is just the kind of thing you are trying to avoid (detailed analysis of the variations within the system) when you use the first law.

When you say that P_I ≠ nRT/V or RT/v, is the T and v here interface properties or something else?

Thanks very much

 

[Chestermiller]

Is the point of the ideal gas law pressure/temperature to describe the state of the whole system during at any point in process with one pressure and temperature (only possible if reversible), and if the process is irreversible, this becomes meaningless due to spatial variations of the properties in the system?

Yes. This is what I've been trying to say, but it seems like I haven't done a very good job.

In Pippard's free expansion example where he states δW≠ PdV, what pressure is this/what does it "describe"?

The discussion isn't very precise, but I think it's talking about nRT/V

For the bolded (but this is equivalent to the Clausius inequality only if T for the irreversible path is taken as the temperature at the interface for the reversible path. However, then TIdS in not really the heat transferred over the reversible path), if some irreversible process is performed, why must we use T_I of a reversible process and not the T_I of the actual (irreversible) process we are performing for equivalence to Clausius inequality?

The discussion has somehow gotten very convoluted with all the T's, T_I's, reversible's, and irreversible's. I didn't mean for you to interpret what I said in the way it turned out you did. Let me try to be more precise. For any irreversible path, we calculate the integral of dQ/T_I, where T_I is the interface temperature for that specific path. This integral is less than the ΔS between the initial and final equilibrium states.

How come T_IdS for a reversible process does not equal δQ?

It does. Again, I didn't explain things very well.

What dictates in the TdS equation that if q = TdS - ε then w = -pdV + ε? Is ε = T*dσ (entropy generation)?

I'm having trouble interpreting what Pippard is saying. I think what he's trying to say that q for the irreversible path is less than TdS for the reversible path and w for the irreversible path is greater than -pdV for the reversible path (apparently, in his notation, w is the work done by the surroundings on the system). I find it very confusing, and, I really don't like what he's done here. As far as ε = T*dσ, your guess is as good as mine.

Does σ_I have any role in the TdS equations if the process is irreversible or is it only the pressure component of the stress included? What do we use for the pressure term in TdS equations during irreversible processes?

The TdS equations only provide a constraint on the changes in the thermodynamic functions U, S, and V between two differentially separated equilibrium states. They can be applied to any finite reversible path (since for such paths, T and p are uniform and unambiguous), but they cannot be applied to finite irreversible paths.

When you say that P_I ≠ nRT/V or RT/v, is the T and v here interface properties or something else?

Thanks very much

Here's a crude example. Suppose that, during a quasistatic process, you somehow know how T is varying with x along the cylinder T = T(x), and you know the total number of moles of gas n within the cylinder. You know that the pressure is constant within the cylinder, but you don't know what its value is. Can you determine the pressure so that you can calculate the work at the interface? The answer is yes. From the ideal gas law, you know that the local molar density is:

$$ρ=\frac{p}{RT(x)}$$

So the number of moles of gas between x and x + dx is:

$$dn=A\frac{pdx}{RT(x)}$$

If we integrate this between x = 0 and x = L (the current distance between the piston and the dead end of the cylinder), we obtain:

$$n=pA\int_0^L{\frac{dx}{RT(x)}}$$

Therefore, the uniform pressure in the quasistatic process is given by:

$$p=\frac{nR}{A\int_0^L{\frac{dx}{T(x)}}}$$
This can be rewritten as:
$$pV=nR\overline{T}$$
where
$$\overline{T}=\frac{L}{\int_0^L{\frac{dx}{T(x)}}}$$

I hope this helps.

Chet

 

[Red_CCF]

The discussion isn't very precise, but I think it's talking about nRT/V

Okay, but in his free expansion example ideal gas law P isn't be uniform so such an expression would have little meaning?

The discussion has somehow gotten very convoluted with all the T's, T_I's, reversible's, and irreversible's. I didn't mean for you to interpret what I said in the way it turned out you did. Let me try to be more precise. For any irreversible path, we calculate the integral of dQ/T_I, where T_I is the interface temperature for that specific path. This integral is less than the ΔS between the initial and final equilibrium states.

It does. Again, I didn't explain things very well.

Is this equivalent to a re-statement of the Clausius inequality?

The TdS equations only provide a constraint on the changes in the thermodynamic functions U, S, and V between two differentially separated equilibrium states. They can be applied to any finite reversible path (since for such paths, T and p are uniform and unambiguous), but they cannot be applied to finite irreversible paths.

I have seen many books say that the equation is valid for any process since every variable is a state function. Does this mean that, although the TdS equation is valid for any process, they can only be used to solve for reversible path only? Is there a limit on the number of reversible path between two equilibrium states?

Thank you very much

Here's a crude example. Suppose that, during a quasistatic process, you somehow know how T is varying with x along the cylinder T = T(x), and you know the total number of moles of gas n within the cylinder. You know that the pressure is constant within the cylinder, but you don't know what its value is. Can you determine the pressure so that you can calculate the work at the interface? The answer is yes. From the ideal gas law, you know that the local molar density is:

$$ρ=\frac{p}{RT(x)}$$

So the number of moles of gas between x and x + dx is:

$$dn=A\frac{pdx}{RT(x)}$$

If we integrate this between x = 0 and x = L (the current distance between the piston and the dead end of the cylinder), we obtain:

$$n=pA\int_0^L{\frac{dx}{RT(x)}}$$

Therefore, the uniform pressure in the quasistatic process is given by:

$$p=\frac{nR}{A\int_0^L{\frac{dx}{T(x)}}}$$
This can be rewritten as:
$$pV=nR\overline{T}$$
where
$$\overline{T}=\frac{L}{\int_0^L{\frac{dx}{T(x)}}}$$

I hope this helps.

Chet

For this example:

1. In the second last equation, would V/n be the average molar specific volume of the whole system?
2. Since p = RT/v, does this mean that the ratio T/v at any point in the gas is equal and so long temperature and specific volume are measured at the same point the pressure of the whole system can be deduced?

Thank you very much

 

[Chestermiller]

Okay, but in his free expansion example ideal gas law P isn't be uniform so such an expression would have little meaning?

Right!

Is this equivalent to a re-statement of the Clausius inequality?

Yes. You can tell that I have a lot of value for the Clausius inequality.

I have seen many books say that the equation is valid for any process since every variable is a state function. Does this mean that, although the TdS equation is valid for any process, they can only be used to solve for reversible path only? Is there a limit on the number of reversible path between two equilibrium states?

Let's suppose that you have two differentially separated equilibrium states of a system. You devise an irreversible path between these two equilibrium states that does not just involve small changes. If involves a very large excursion, with large values of the rate of heat flow and the rate of work being done. However, in the end, you wind up at a final equilibrium state that is only differentially separated from the initial equilibrium state. Even under these circumstances, the differential changes in U, S, and V are related by dU = TdS - PdV, where, because the change is differential, the initial and final temperatures and volumes are negligibly different from one another.

For this example:

1. In the second last equation, would V/n be the average molar specific volume of the whole system?

Yes.

2. Since p = RT/v, does this mean that the ratio T/v at any point in the gas is equal and so long temperature and specific volume are measured at the same point the pressure of the whole system can be deduced?

Yes. That's what this is all about. Of course, unless you do some heat transfer calculations, you are not going to know what T(x,t) is going to be, and you are not going to know the pressure. And, of course, this only applies to quasi static. Still, if you're willing to do the heat transfer calculations, you can get more out of the first law for such situations than just treating the contents of the cylinder as a black box. If you are willing to do gas dynamics calculations (which are pretty complicated), you can do the same thing for irreversible non-quasi-static cases. However, these are the kinds of things we are trying to avoid when we apply the first law to the overall macroscopic system.

Chet

 

[Red_CCF]

Right!

Just one last thing on this, in Pippard statement PdV ≠ 0 in free expansion, he is assuming the system can be represented by a uniform P=nRT/V while knowing that this isn't true?

Let's suppose that you have two differentially separated equilibrium states of a system. You devise an irreversible path between these two equilibrium states that does not just involve small changes. If involves a very large excursion, with large values of the rate of heat flow and the rate of work being done. However, in the end, you wind up at a final equilibrium state that is only differentially separated from the initial equilibrium state. Even under these circumstances, the differential changes in U, S, and V are related by dU = TdS - PdV, where, because the change is differential, the initial and final temperatures and volumes are negligibly different from one another.

So dU = TdS - PdV is only valid for differential changes for any process but for irreversible processes we cannot integrate this equation between two non-differentially separated states? For reversible processes this equation would be valid because the process is at equilibrium at all times during the process and thus separated by an "infinite" number of differentially separated equilibrium states?

Yes. That's what this is all about. Of course, unless you do some heat transfer calculations, you are not going to know what T(x,t) is going to be, and you are not going to know the pressure. And, of course, this only applies to quasi static. Still, if you're willing to do the heat transfer calculations, you can get more out of the first law for such situations than just treating the contents of the cylinder as a black box. If you are willing to do gas dynamics calculations (which are pretty complicated), you can do the same thing for irreversible non-quasi-static cases. However, these are the kinds of things we are trying to avoid when we apply the first law to the overall macroscopic system.

Chet

On a general note, what are the gas dynamic equations?

Thank you very much

 

[Chestermiller]

Just one last thing on this, in Pippard statement PdV ≠ 0 in free expansion, he is assuming the system can be represented by a uniform P=nRT/V while knowing that this isn't true?

What he's saying (not too unambiguously) is that, if you try to use this equation for calculating the work in free expansion, you will get the wrong answer.

So dU = TdS - PdV is only valid for differential changes for any process but for irreversible processes we cannot integrate this equation between two non-differentially separated states? For reversible processes this equation would be valid because the process is at equilibrium at all times during the process and thus separated by an "infinite" number of differentially separated equilibrium states?

Yes.

On a general note, what are the gas dynamic equations?

See Bird, Stewart, and Lightfoot, Transport Phenomena, where they look at the problem of compressible flow through a nozzle.

Also, a while back, I wrote up a gas dynamics analysis of a problem involving an ideal gas within a cylinder that is separated into two compartments by a massless frictionless piston. The initial pressure in one of the compartments is higher than the other, and, at time zero, the piston is released. The analysis assumes zero viscosity for the gas, so the system oscillates back and forth forever. If you are interested in seeing this analysis, I can send it to you via email (it is in the form of a Word document). Just send me a message at my email address, and I'll email it back to you.

Chet

 

[rkmurtyp]

if the compression-expansion process is quasistatic but the process contain irreversibilities like friction which is converted into heat and absorbed by the system since it is insulated,

'Quasistatic' and friction don't go together! You can't have a quasistatic process in the presence of friction.

Therefore the process with friction is necessarily irreversible process. If and when you bring the system back to its original state (thereby completing a cycle) by a reversible process that consists of several steps the effect of irrevrsibility of the initial irreversible process ends up in changes in the surroundings such as lowering of a mass through certain height and supply of heat to a heat reservoir.

 

[Red_CCF]

if the compression-expansion process is quasistatic but the process contain irreversibilities like friction which is converted into heat and absorbed by the system since it is insulated,

'Quasistatic' and friction don't go together! You can't have a quasistatic process in the presence of friction.

Therefore the process with friction is necessarily irreversible process. If and when you bring the system back to its original state (thereby completing a cycle) by a reversible process that consists of several steps the effect of irrevrsibility of the initial irreversible process ends up in changes in the surroundings such as lowering of a mass through certain height and supply of heat to a heat reservoir.

How come quasistatic process cannot occur if friction is present? What if the process is slow enough such that any infinitessimal amount of heat generated from friction is added to the system, and the system is allowed to reach equilibrium before the process continues, I think it would possibly fulfill the definition of quasistatic process (passes through infinite number of equilibrium states) but yet would be irreversible?

Thank you very much

 

[rkmurtyp]

How come quasistatic process cannot occur if friction is present? What if the process is slow enough such that any infinitessimal amount of heat generated from friction is added to the system, and the system is allowed to reach equilibrium before the process continues, I think it would possibly fulfill the definition of quasistatic process (passes through infinite number of equilibrium states) but yet would be irreversible?

For a quasistatic (reversible) process we need to satisfy the condition P_ext= P_sym. When friction is present it is impossible to satisfy this condition.

The speed (or time rate of change) of a process (slow enough of fast enough etc) does not and should not enter thermodynamic arguments - time has no role to play in (equilibrium) thermodynamics.

A process that takes the system through a series of continuous set of equilibrium states has no chance to be out of equilibrium ever, consequently be irreversible ever!