Irreversible Quasistatic PV Work

[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!