What does the maximum work of a reversible expansion mean?

[sgstudent]

What does it mean by maximum work of an isothermal, reversible expansion of a perfect gas from initial volume Vi to final volume Vf at temperature T? I understand that heat enters the system during this expansion to ensure that no internal energy is lost but it doesn't really help in anyway.

Also a second question is this.

When we are calculating the maximum work we use w=-nRTlnVf/Vi. But why can't we use w=-PextΔV? I mean the gas will still expand against the opposing pressure in the isothermal reversible expansion conditions so why do we have to use the first formula mentioned?

Thank you

 

[Chestermiller]

What does it mean by maximum work of an isothermal, reversible expansion of a perfect gas from initial volume Vi to final volume Vf at temperature T? I understand that heat enters the system during this expansion to ensure that no internal energy is lost but it doesn't really help in anyway.

The maximum work that the system can do on the surroundings in an isothermal expansion (i.e., with the boundary of system in contact with a constant temperature reservoir) of an ideal gas is obtained if the expansion is carried out reversibly. All irreversible isothermal expansions will result in less work being done on the surroundings.

Also a second question is this.

When we are calculating the maximum work we use w=-nRTlnVf/Vi. But why can't we use w=-PextΔV? I mean the gas will still expand against the opposing pressure in the isothermal reversible expansion conditions so why do we have to use the first formula mentioned?

The sign convention you have used here gives the work done by the surroundings on the system, rather than the work done by the system on the surroundings. Thus, with this sign convention, the reversible work is a minimum, rather than maximum.

Regarding the two equations you have written, please see my recent Physics Forums Insight article https://www.physicsforums.com/insights/understanding-entropy-2nd-law-thermodynamics/, in the section of the first law. If we use the sign convention that work done by the system on the surroundings is considered positive, then the equation dw = P_extdV always represents the work done, irrespective of whether the process is reversible or irreversible (in my Insight article, instead of P_ext, I call P_Int the pressure at the interface between the system and the surroundings). For a reversible path though, P_ext=P, where P is the (uniform) pressure throughout the system. In addition, for the reversible path, P represents the pressure calculated from the ideal gas law at the system temperature T (which is also uniform throughout the system for a reversible path). So, for the reversible isothermal path, $dw=P_{ext}dV=\frac{nRT}{V}dV$.

By writing w=P_extΔV, you are implicitly assuming that the external force per unit area P_ext is constant during the isothermal expansion. This is not possible for the case of an isothermal reversible expansion. For such an expansion, the external force per unit area must be continuously adjusted such that P_extV = P_initialV_initial at all locations along the path. That is, the pressure inside the system must be uniform and satisfy the ideal gas law. So, if the piston moves in such a way that external force per unit area is held constant during the gas expansion, the process path will be irreversible.

Chet

 

[sgstudent]

The maximum work that the system can do on the surroundings in an isothermal expansion (i.e., with the boundary of system in contact with a constant temperature reservoir) of an ideal gas is obtained if the expansion is carried out reversibly. All irreversible isothermal expansions will result in less work being done on the surroundings.

The sign convention you have used here gives the work done by the surroundings on the system, rather than the work done by the system on the surroundings. Thus, with this sign convention, the reversible work is a minimum, rather than maximum.

Regarding the two equations you have written, please see my recent Physics Forums Insight article https://www.physicsforums.com/insights/understanding-entropy-2nd-law-thermodynamics/, in the section of the first law. If we use the sign convention that work done by the system on the surroundings is considered positive, then the equation dw = P_extdV always represents the work done, irrespective of whether the process is reversible or irreversible (in my Insight article, instead of P_ext, I call P_Int the pressure at the interface between the system and the surroundings). For a reversible path though, P_ext=P, where P is the (uniform) pressure throughout the system. In addition, for the reversible path, P represents the pressure calculated from the ideal gas law at the system temperature T (which is also uniform throughout the system for a reversible path). So, for the reversible isothermal path, $dw=P_{ext}dV=\frac{nRT}{V}dV$.

By writing w=P_extΔV, you are implicitly assuming that the external force per unit area P_ext is constant during the isothermal expansion. This is not possible for the case of an isothermal reversible expansion. For such an expansion, the external force per unit area must be continuously adjusted such that P_extV = P_initialV_initial at all locations along the path. That is, the pressure inside the system must be uniform and satisfy the ideal gas law. So, if the piston moves in such a way that external force per unit area is held constant during the gas expansion, the process path will be irreversible.

Chet

Hi Chestermiller thanks for the detailed reply :)

What does it mean by "All irreversible isothermal expansions will result in less work being done on the surroundings." And why would the work done by maximum if it were reversible?

I was taught that saying -w or w=-P_extΔV meant that work was done by the system to the surroundings. Would that be a terminology error?

I read the post you wrote. So in that isothermal expansion since the Pext continuously changes I cannot simply use PextΔV to get my work done. Instead I have to use nRTlnVf/Vi. Would this be correct?

Relating to the above question my lecturer gave us these tutorial questions.
(b) Calculate the work done when 1000 mol of nitrogen gas is expanded reversibly and isothermally at a temperature of 300 K from 1 to 3 m3 .
(c) Compare and explain your answer obtained in (b) to what you would have calculated from the perfect gas equation.

So for part B i simply used the nRTlnVf/Vi equation to get the work done. However for part C I'm unsure on how to calculate the work done using the perfect gas equation. I initially thought the method I proposed where I obtained the Pext from the final volume using P=nRT/V where V=3m^3 but that would be assuming that that expansion was irreversible.

Thanks again

 

[Chestermiller]

Hi Chestermiller thanks for the detailed reply :)

What does it mean by "All irreversible isothermal expansions will result in less work being done on the surroundings." And why would the work done by maximum if it were reversible?

It means that all irreversible isothermal expansions result in less work being done on the surroundings than the corresponding reversible expansion between the same two initial and final thermodynamic equilibrium states. The "why" should be answered from the results of your homework example.

I was taught that saying -w or w=-P_extΔV meant that work was done by the system to the surroundings. Would that be a terminology error?

I assume you are using the sign convention that $ΔU=q+w$, where w represents the work done by the surroundings on the system. With this sign convention, $w=-\int P_{ext}dV$, and the work done by the system on the surroundings is $+\int P_{ext}dV$. The equation $w=-P_{ext}ΔV$ gives the work done by this surroundings on the system in the specific situation where $P_{ext}$ is constant during the deformation. So it is never equal to the work done by the system on the surroundings.

I read the post you wrote. So in that isothermal expansion since the Pext continuously changes I cannot simply use PextΔV to get my work done. Instead I have to use nRTlnVf/Vi. Would this be correct?

Yes, for the reversible case.

Relating to the above question my lecturer gave us these tutorial questions.
(b) Calculate the work done when 1000 mol of nitrogen gas is expanded reversibly and isothermally at a temperature of 300 K from 1 to 3 m3 .
(c) Compare and explain your answer obtained in (b) to what you would have calculated from the perfect gas equation.

So for part B i simply used the nRTlnVf/Vi equation to get the work done. However for part C I'm unsure on how to calculate the work done using the perfect gas equation. I initially thought the method I proposed where I obtained the Pext from the final volume using P=nRT/V where V=3m^3 but that would be assuming that that expansion was irreversible.

Thanks again

My guess is that there is a typo in the specification of part (b). All your confusion would be cleared up if the word reversible were replaced by the word irreversible. That would mean that your analysis for part (c) would apply perfectly to part (b), and your analysis for part (b) would apply perfectly to part (c). In any event, if this change were made, you analysis would be spot on correct.

Chet

 

[Chestermiller]

You may want to check out the following thread: https://www.physicsforums.com/threa...-and-heat-capacity.796304/page-3#post-5094618

In this thread, Worryingchem and I have been working on the problem of isothermal irreversible expansion of an ideal gas, starting at post #15 and going through post #35. After completing that problem, we have been working on adiabatic irreverislbe expansion.

Chet

 

[sgstudent]

It means that all irreversible isothermal expansions result in less work being done on the surroundings than the corresponding reversible expansion between the same two initial and final thermodynamic equilibrium states. The "why" should be answered from the results of your homework example.

I assume you are using the sign convention that $ΔU=q+w$, where w represents the work done by the surroundings on the system. With this sign convention, $w=-\int P_{ext}dV$, and the work done by the system on the surroundings is $+\int P_{ext}dV$. The equation $w=-P_{ext}ΔV$ gives the work done by this surroundings on the system in the specific situation where $P_{ext}$ is constant during the deformation. So it is never equal to the work done by the system on the surroundings.

Yes, for the reversible case.

My guess is that there is a typo in the specification of part (b). All your confusion would be cleared up if the word reversible were replaced by the word irreversible. That would mean that your analysis for part (c) would apply perfectly to part (b), and your analysis for part (b) would apply perfectly to part (c). In any event, if this change were made, you analysis would be spot on correct.

Chet

Hmm my notes uses ΔU=q-w maybe that is where the problem arises from.

Perhaps part b of the question wants us to use the nRTlnVf/Vi while part c wants us to assume it is not a reversible expansion? So the answer in part b is larger as it is the maximum expansion work since it is done in an isothermal reversible condition while in part c it is in an isothermal irreversible condition?

But I still don't really understand why the isothermal irreversible condition would result in a smaller expansion work done. In an irreversible change what is the difference?

 

[Chestermiller]

Hmm my notes uses ΔU=q-w maybe that is where the problem arises from.

With that sign convention, the work done by the system on the surroundings is dw = +P_extdV.

Perhaps part b of the question wants us to use the nRTlnVf/Vi while part c wants us to assume it is not a reversible expansion? So the answer in part b is larger as it is the maximum expansion work since it is done in an isothermal reversible condition while in part c it is in an isothermal irreversible condition?

Well to do part c, you would still need to use the ideal gas law to get the final state, so maybe that is what they are referring to.

But I still don't really understand why the isothermal irreversible condition would result in a smaller expansion work done. In an irreversible change what is the difference?

In the irreversible case, there would be viscous dissipation of mechanical energy as a result of the rapid deformation, and the heat generation from this viscous dissipation would limit the amount of heat that you could take in from the surroundings while at the same time maintaining the temperature constant. This would limit the amount of work, since only the amount of heat taken in from the surroundings determines the amount of work done in an isothermal expansion.

Chet

 

[sgstudent]

With that sign convention, the work done by the system on the surroundings is dw = +P_extdV.

Well to do part c, you would still need to use the ideal gas law to get the final state, so maybe that is what they are referring to.

In the irreversible case, there would be viscous dissipation of mechanical energy as a result of the rapid deformation, and the heat generation from this viscous dissipation would limit the amount of heat that you could take in from the surroundings while at the same time maintaining the temperature constant. This would limit the amount of work, since only the amount of heat taken in from the surroundings determines the amount of work done in an isothermal expansion.

Chet

Ohh would it be correct to say that in an irreversible expansion heat will be evolved so this reduces the amount of heat entering the system. So the expansion work done is limited? Just to make sure I totally understood what you said.

But why would there be a viscous dissipation of mechanical energy causing the heat to be generated?

Thanks again

 

[Chestermiller]

Ohh would it be correct to say that in an irreversible expansion heat will be evolved so this reduces the amount of heat entering the system. So the expansion work done is limited? Just to make sure I totally understood what you said.

Yes.

But why would there be a viscous dissipation of mechanical energy causing the heat to be generated?

What you are really asking here is "what are some of the details of how viscous stresses come into play during a transient irreversible expansion of an ideal gas," correct?

Chet

 

[sgstudent]

Yes.

What you are really asking here is "what are some of the details of how viscous stresses come into play during a transient irreversible expansion of an ideal gas," correct?

Chet

Yes. I'm not sure what the difference between a reversible and irreversible expansion. I read your article and thread and I learned that in a reversible change the Pint=Pext. The way you put it was "the temperature and pressure throughout the system along the entire reversible process path are completely uniform spatially." I understood the example in the thread you gave (I hope so) whereby you remove a mass from the piston causing the Pext and hence Pint to decrease and we solve for the work done in that manner.

But I don't really understand how an irreversible expansion scenario would look like.

Thanks

 

[Chestermiller]

Yes. I'm not sure what the difference between a reversible and irreversible expansion. I read your article and thread and I learned that in a reversible change the Pint=Pext.

Actually, P_Int is the symbol I use in place of P_ext for both reversible and irreversible processes. I just wanted to give people a more clear-cut indication that what we are talking about is the force per unit area acting at the interface between our system and its surroundings.

The way you put it was "the temperature and pressure throughout the system along the entire reversible process path are completely uniform spatially." I understood the example in the thread you gave (I hope so) whereby you remove a mass from the piston causing the Pext and hence Pint to decrease and we solve for the work done in that manner.

But I don't really understand how an irreversible expansion scenario would look like.

Thanks

OK. I'm going to construct a crude model of a system undergoing an irreversible expansion, and we are then going to solve that model to see how that compares with what happens in a reversible expansion. This model is going to include viscous stresses, which are negligible in a reversible expansion. From the model setup and solution, you will be able to see why the effects of viscous stresses and their influence on the work done are negligible when the expansion is reversible.

As I said, the model is going to be very crude, and it will be only a first order approximation to what is happening. However, for an irreversible expansion, it will be much closer to reality than simply assuming that the force per unit area at the interface is equal to the thermodynamic pressure nRT/V (which is valid only for the reversible case). If we really wanted to solve the irreversible case "exactly," we would have to solve a complicated set of non-linear partial differential equations. However, for our purposes, this is not necessary, because our analysis will capture, at least qualitatively, the essence of the effects of the viscous stresses.

Our focus problem will be the irreversible isothermal expansion of an ideal gas in a cylinder fitted with a massless, frictionless piston. The initial volume will be V₀ and the initial pressure will be P₀. Part of the surface of the cylinder will be immersed in a constant temperature bath at temperature T, and the remainder of the cylinder (as well as the piston) will be insulated. At time t = 0, the temperature of the gas will be T, and we will also assume that the conductive heat transfer within the gas is sufficiently rapid that the temperature throughout the gas will be T for all times. (This is part of the approximation we are making to enable us to focus on the effects of the viscous stresses, and that would be relaxed if we went to the full set of non-linear PDEs. However, relaxing this approximation would not change qualitatively what we would find for the effects of viscous stresses on the work.)

At time t = 0, the force per unit area that we need to exert on the piston to hold it in place is P_ext = P₀. However, at time t = 0, we suddenly decrease the external force per unit area we exert on the piston to a lower value, p_ext (note the lower case p), and keep it at this lower value for all times (by whatever means necessary). This allows the gas to expand rapidly and irreversibly.

In the irreversible case, the force per unit area exerted by the gas on its side of the piston P* is determined by a combination of two factors:

1. The thermodynamic pressure P = nRT/V

2. Viscous stresses

The viscous stresses add to the thermodynamic pressure P to give the total force per unit area P*. So, unlike the reversible case where only the thermodynamic pressure is present, there is also a viscous stress contribution to the force per unit area on the piston.

Let's focus on the viscous stresses. When you study fluid mechanics, you will learn that, in this situation, the contribution of the viscous stresses to the total force per unit area exerted by the gas on the piston is given by:

Viscous stress = $-\frac{2μ}{V}\frac{dV}{dt}$

where μ is the viscosity of the gas. Note that the thermodynamic pressure depends on the volume of the gas, while the viscous stress is determined by the rate of change of the gas volume with respect to time. The quantity $\frac{1}{V}\frac{dV}{dt}$ in the relationship represents physically the axial rate of strain of the gas.

So, the total force per unit area P* exerted by the gas on the piston is approximated by the equation:
$$P^*=P-\frac{2μ}{V}\frac{dV}{dt}=\frac{nRT}{V}-\frac{2μ}{V}\frac{dV}{dt}=\frac{P_0V_0}{V}-\frac{2μ}{V}\frac{dV}{dt}$$
Now, since the piston is massless and frictionless, we must have that $p_{ext}=P^*$
Therefore, for all times t > 0, we have:
$$\frac{P_0V_0}{V}-\frac{2μ}{V}\frac{dV}{dt}=p_{ext}$$
This will be our starting equation for the remainder of the irreversible expansion analysis.

This is where I'll stop for now, and give you a chance to digest this and ask questions.

Chet