Chet Miller

physicsforums.com/threads/…

Please submit you question(s) at PHYSICSFORUMS.COM which offers a much more framework for discussions like this. I will be happy to work with you there. Thanks.

I would like.to move this to Physicsfotums.com if that is ok with you.

YES. THAT is one example of an Adiabatic irreversible compression. Is that the one you want to focus on?

Please specify a specific process we can focus on.

My point is that, for an irreversible expansion or compression, what we call $P_{int}$ is not equal to the pressure that one would calculate from the ideal gas law (or other applicable equation of state), except at the initial and final thermodynamic equilibrium states.. The same goes for what we call $P_{ext}$.

Suppose you choose the gas as your system. Then $P_{ext}$ is the magnitude of the force per unit area exerted by the inside face of the piston on the gas, and $P_{int}$ is the magnitude of the force per unit area exerted by the gas on the inside face of the piston. How are $P_{ext}$ and $P_{int}$ related?

@JánLalinský In thermodynamics, we divide the universe into two parts: the system and the surroundings. Do. you regard the piston as part of your system or part of its surroundings?

@hyportnex I don’t follow this at all. If a gas goes between these two equilibrium states, the work done by the gas must be $\int{pdV}$.

By your rationale, it is useless to study plane geometry since there is no such think as a perfect circle, a perfect square, a perfect rectangle, etc; boring with these involves additional assumptions that only complicate the discussion (of physics, for example). I have only one more question for you. In a rapid irreversible expansion or compression, is the "internal pressure' of the gas described by the ideal gas law, even though the gas is not at thermodynamic equilibrium?

The piston doesn't have infinite acceleration. Instead, the force per unit area exerted by the gas on the inside face of the piston will be equal to the external pressure exerted on the outside face of the piston. But, since the system is not at thermodynamic equilibrium, the force per unit area on the inside face of the piston is not equal to the value calculated from the ideal gas law. The ideal gas law applies only at thermodynamc equilibrium.l. Are you familiar with the rheological equation for a Newtonian fluid?

What if the pistol is massless and frictionless?

If you can move the piston (day using a control system with feedback (such that the force per unit area on the inside face of the piston is the same as initially throughout the change) and a very large heat flux is imposed (such the the process is irreversible), then $\Delta H$ will equal Q (provided the final state is a thermodynamic equilibrium state). Of course, it is very difficult to control the force on the inside face of the piston in this throughout an irreversible process like this.

I agree with what you say.

Sowa P_{wzt}$ represent the force per unit area of the inside face of the piston on the gas. Or does it represent the force per unit area exerted on the outside face of the piston, or something else? If you do a force balance on the piston using Newton's 2nd law of motion for an irreversible expansion or compression, what do you get?

My head is spinning from reading these discussions. H = U +PV is independent of any process or whether a process is reversible or irreversible. The parameters in this equation refer to a thermodynamic equilibrium state. You speak of $P_{ext}$ and $P_{int}$, but you are imprecise as to what these represent. Does $P_{int}$ represent the force per unit area of the gas on the insidede face of the piston?

Wait for my response.

How about we solve this assuming that the gas undergoes adiabatic reversible expansion in each chamber and see where it leads. If you want to do this, go to PhysicsForums.com, register, and start a conversation with user chestermiller.

If the initial and final states are thermodynamic equilibrium states, that is not enough. For a process to be reversible, the processs must be at thermodynamic equilibrium at all intermediate states. In other words, the process must consist of a continuous sequence of thermodynamic equilibrium states (over the entire process path). No spontaneous changes, involving non-equilibrium states, are allowed.

See Transport Phenomena, Bird et Al, homework problem 11.D.1

Do you think that the ideal gas law is valid during rapid irreversible expansions or compressions, or is it only valid at thermodynamic equilibrium??

@BobD What you are describing would really be equivalent to controlling it extremely.

The only way to make the barrier move such that the process is reversible is if one exerts external force to control the motion of the barrier. If the barrier is free to move, but comes to rest in the final state, there will be generation of entropy with each of the two compartments, and the reversible equations will not apply.

@BobD You can, of course, do it by doing work on the partition. But you can't if it is free to move.

@BobD Both gases undergo irreversible changes. If only one undergoes reversible, the is no basis for deciding which one.

Yes, but if you use those adiabatic reversible equations for both gases, you will not be able to satisfy the aero change in internal energy condition. Tr it and see what you get.

@JohnRennie If the barrier is moving at even on the order of 10% of the speed of sound, there will still be a siighnificant rate of deformation of the gas. on both sides of the barrier, and significant viscous dissipation within the gas on both sides. see Transport Phenomena by Bird et al, Homework problem 11.D.1 Entopy generation balance. In addition, if you assume that the process is adiabatic and reversible on both sides of the barrier in this particular problem, you will not be able to satisfy the condition that the internal energy of the combined system does not change,

Even with a massless barrier, the barrier would not travel with the speed of sound. But compression- and expansion waves adjacent to the barrier would travel at local sonic velocity. The important thing is that, within these expansion and compression zones, the gas is deforming very rapidly, and viscous dissipation of mechanical energy into internal energy is occurring. Even in the case of a piston with significant mass, the process would not be reversible since, eventually the kinetic energy of the piston would be dissipated by viscous effects.

You first. You made a statement. Please explain your rationale.

Why would the partition have to move at speed comparable to the speed of sound for the process to be irreversible?

I could solve this, but I feel that it's too much work for a problem you just dreamed up.

Realistically, you need to consider viscous drag and the proximity of the wall in scenario 2. The ball acts as a valve, and slows the water flow. But, unless viscous drag in the region between the walll and ball is included, you will not predict the lower velocity..

In scenario 2, neglecting drag by the sphere and the tube wall, the velocity is determined by the Bernoulli equation.

I don't understand this question. I thought the focus was on the velocity of the sphere.

I think that, if it can not move horizontally or vertically, its velocity is zero.

In scenario 2, it is held stationary in the vertical direction, but can move in the horizontal direction? Why are you omitting viscous interactions with the walll/. It seems that these can be very important. won't the ball clog the exit hole if it is allowed to move horizontal?

Google "Falling Ball Viscometry" for part 1. Even though your ball is rising (due to buoyancy), the same equations apply. For experiment 2, I don't understand the geometry.

But the gas does have viscous stresses in addition to the isotropic (pressure) component of stress. And these do damp the motion of the piston. Do You think that the rapid expanding gas satisfies the ideal gas law, even if, at equilibrium it does?

Eventualy, viscous stresses within the gas will cause the kinetic energy of the piston to dissipate, and the piston will stop moving. When this happens, the change in internal energy U from the initial state to the final state in the process you described will be zero: $$\Delta U=0$$

Do you think that the piston continues moving forever, or, even with a frictionless piston, does it eventually come to rest with zero KE?

Are you talking about your responses to Phillip Wood's answer?

No. In your car example, there is viscous drag also. I suggest you consult a fluid mechanics or transport phenomena text and read over the sections on Newton's law of viscosity extended to 3D deformations, in which the stress tensor is a linear function of the velocity gradient tensor.

As the gas volume oscillates with time, there are velocity gradients within the gas. This gives rise to viscous stresses and viscous dissipation. Are you aware of the mechanical constitutive equation for a Newtonian viscous fluid (like a gas).

You can't ignore viscous damping. It will be present and will eventually bring the piston to a stop. At that point, the work done by the gas will exactly equal the work done externally on the surroundings, and the system will be at equilibrium.

In other words, you are saying that there is no damping mechanism if the piston is non-conductive. But there actually is a damping mechanism that you are overlooking. the piston will come to rest even if the piston is frictionless.

Why would heat conduction after the mechanical response? And if it does not conduct heat, then you are asserting that the oscillation will continue forever?

Do you think the oscillation will go on forever,, even if the piston is frictionless?

If the initial and final states are thermodynamic equilibrium states, you certainly can consider the piston as part of the overall system or, alternately, part of the overall surroundings. If I do a force balance on the piston using Newton's 2nd law, I get $F_g-P_{ext}A=\frac{m}{A}\frac{d^2V}{dt^2}$ where $F_g$ is the force that the gas exerts on the inside face of the piston, $P_{ext}$ is the external pressure on the outside face of the piston, m is the mass of the piston, A is the area of the piston, and V is the volume of gas. Do you think the piston willl keep moving or will it stop?

Is the piston part of your system or part of the surroundings?

If the piston has mass, the piston does not exert equal and opposite force on system and surroundings.