Irreversible Quasistatic PV Work

[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!