Why is Qir < than Qr? Carnot cycle and change in G

[Abhishek Jain]

I am learning physics on khan academy and they do a proof to show that delta G for a reversible reaction is negative and how for a irreversible reaction it is positive. However in the proof, they assume that the heat put in by the isotherm is less for an irreversible reaction compared with a reversible reaction. They state the the heat generated by friction that is produced reduces heat required to be added to the system. However, my intuition is that it would require more heat because it would require more work to reach the same height as the reversible piston due to friction. Wouldn't it need to do more work to counteract the work done by friction? Yes the work done by friction is generating heat which will require less heat to reach T1. However, since it requires more work to move the piston due to friction, wouldn't Qir+Qf>Qr, since Wir>Wr to reach same position? The only way for Qir to be less than Qr would be for Qf to account for the work done by friction + some more to make Qir smaller. However, since it is an isotherm delta U = 0, so work = q. so Wf = Qf? What am I missing here? Please help. This is a foundational concept and I feel it is necessary for me to understand this to understand this topic fully. The link to the video is below.

https://www.khanacademy.org/science...us-gibbs-free-energy-spontaneity-relationship

 

[Chestermiller]

I am learning physics on khan academy and they do a proof to show that delta G for a reversible reaction is negative and how for a irreversible reaction it is positive. However in the proof, they assume that the heat put in by the isotherm is less for an irreversible reaction compared with a reversible reaction.

I have a few comments about this before addressing the friction issue.

1. They are not talking about $\Delta G$ for a reversible chemical reaction compared to $\Delta G$ for an irreversible chemical reaction. They are talking about an reversible gas expansion compared to an irreversible gas expansion (with no chemical reaction) in a cylinder between the same initial and final equilibrium states, both of which are at the same temperature.

2. There are many technical mistakes in this video, and you should not pay any attention to the video. It starts out OK in the discussion the isothermal reversible expansion, but, from that point on, it goes farther and farther astray from fundamental thermodynamic reality. In the end, it incorrectly determines the $\Delta G$ between the initial and final states of the system.

3. The addition of friction between the piston and cylinder for the irreversible case is a "red herring" that did not need to be introduced to describe the difference between an irreversible expansion and a reversible expansion. Moreover, the effect of friction is extremely tricky to analyze in this problem, as you have already learned. So I am going to first analyze the expansions quantitatively without friction to illustrate what is happening. After that, I'll proceed to analyzing the friction effect.

The present analysis assumes that the cylinder is in a vacuum so that external force on the top side of the piston is due only to the weights (gravel or larger weights) on the top of the piston. The piston itself is assumed frictionless, with a cross sectional area of A, and a weight W. The mass of weights on top of the piston in the initial state are w, and, in the final state, these weights have been entirely removed. The gas in the cylinder is an ideal gas (n moles).

INITIAL STATE:
Temperature T
Pressure $P_1$
Volume $V_1$
$P_1A=(W+w)$

FINAL STATE:
Temperature T
Pressure $P_2$
Volume $V_2$
$P_2A=W$

REVERSIBLE PATH:
For the reversible path, the weights w are removed gradually, so that the system is always close to thermodynamic equilibrium at the constant temperature T. So, for this path, we have: $$\Delta U=0$$and $$Q=Work=\int_{V_1}^{V_2}{PdV}=\int_{V_1}^{V_2}{\frac{nRT}{V}dV}=nRT\ln{(V_2/V_1)}$$But, for an ideal gas, $$\frac{V_2}{V_1}=\frac{P_1}{P_2}=\frac{(W+w)}{W}$$So,$$Q=Work=nRT\ln{\left(\frac{W+w}{W}\right)}$$And the change in entropy of the system is:$$\Delta S=\frac{Q}{T}=nR\ln{\left(\frac{W+w}{W}\right)}$$The change in Gibbs Free energy of the system is then:$$\Delta G=\Delta U+\Delta (PV)-T\Delta S=-nRT\ln{\left(\frac{W+w}{W}\right)}$$The change in entropy of the constant temperature reservoir is -Q/T, so the change in entropy of the system plus the reservoir is zero.

For this reversible change, the granular weights are removed from the piston very gradually as the piston rises, so that some of the weights are removed at the original elevation of the piston, while others are removed at higher elevations, until the piston reaches its final elevation. So the total work done is equal to the increase in potential energy of the piston $W(V_2-V_1)/A$, plus the increases in potential energy of the particles comprising the weight w, which are distributed over elevations between the original height of the piston and the final height. So the work done by the gas on the piston and weights is significantly more than just that required to raise the piston to its final elevation.

IRREVERSIBLE PATH:
For the irreversible path, the entire weight of granules w is removed suddenly from the piston at time zero. As a result of this, the piston spontaneously accelerates upward because of the unbalanced net force caused by the removal of the granule weights; and the piston even overshoots its final equilibrium elevation. But eventually, because of viscous damping by the gas within the cylinder, the oscillations of the piston about the final equilibrium elevation decrease, and the piston ultimately comes to rest at its new elevation. In this case, the potential energy of the granules that were removed at the initial elevation of the piston do not change; only the potential energy of the piston itself changes. So, in the irreversible case, the amount of work done by the gas is less than in the reversible case, and is given by: $$Work=W(V_2-V_1)/A=P_2(V_2-V_1)=P_2V_2(1-V_1/V_2)=nRT(1-P_2/P_1)=nRT\left(1-\frac{W}{W+w}\right)=nRT\left(\frac{w}{W+w}\right)$$In order for the work to be less, the average force of the gas on the piston (averaged over the displacement variations of the piston) must be less than in the reversible case. This lower average force is made possible by viscous stresses in the gas which act in the direction opposite to the movement of the piston.

Since the change in internal energy $\Delta U$ between the initial and final equilibrium states of the gas is again zero, the heat added from the reservoir is equal to the work done by the gas on the piston:$$Q=Work=nRT\left(\frac{w}{W+w}\right)$$So, in the irreversible case, the expanding gas does less work on its surroundings (the piston) and the amount of heat absorbed by the gas from the reservoir is also less.

In the irreversible case, since the reservoir is ideal, its change in entropy is $-Q/T=-nR\left(\frac{w}{W+w}\right)$. So, in the irreversible case, since the change in entropy of the system is unchanged, the change in entropy of the system plus surroundings is > 0. This is the result of entropy generation within the system as a result of viscous dissipation of mechanical energy.

SUMMARY:
So you can see that, even without friction, the irreversible gas expansion does less work on the surroundings than in the reversible case, and the amount of heat transferred from the reservoir to the gas is also less. For more discussion of how viscous dissipation contributes to reducing the amount of expansion work that a gas can do in an irreversible process, see my Physics Forums Insights article in https://www.physicsforums.com/insights/reversible-vs-irreversible-gas-compressionexpansion-work/

I will stop here an give you a chance to digest what I have said and to ask questions. If you wish, I will address the friction issue after that.

 

[Abhishek Jain]

So in essence, because the reversible system is doing some work pushing the piston W + some amount of the granules w (changing as you remove granules) up throughout the process, while the irreversible is just pushing the piston W up through the whole process, the work done by irreversible is less. Am I understanding that correctly?

Since S is a state variable, though the irreversible system took a different route to get from P₁V₁ to P₂V₂, Delta S for the is the same for both the reversible and irreversible expansion. However, since less work is required for irreversible system, the amount of heat required to put into the system from the surroundings is less. So delta s of system + Surroundings for irreversible is greater than 0. Am I understanding that part correctly?

Finally, above you said that video incorrectly determines Delta G between initial and final states of both systems. What would be the correct way to interpret it?

Thank you for all your help. This question had been eating at me for weeks until I found this site!

 

[Chestermiller]

So in essence, because the reversible system is doing some work pushing the piston W + some amount of the granules w (changing as you remove granules) up throughout the process, while the irreversible is just pushing the piston W up through the whole process, the work done by irreversible is less. Am I understanding that correctly?

Yes, but not exactly. I only used the weights as an example. If you had manually applied the same force over time that the piston and weights are applying in the reversible and irreversible cases, then the gas would do the same amount of work on you. And the amount of work done by the gas in the reversible case would be greater than in the irreversible case.

The key feature of an irreversible expansion is viscous stresses. When the gas is deforming rapidly, it exhibits stress/deformation behavior different from that which prevails when it is deforming very gradually. In a rapid expansion, the force exerted by the gas on the piston face depends not only on the amount of deformation (i.e., the volume change) but also on the time rate of change of deformation. This latter (viscous) effect reduces that force that the gas would be exerting on the piston face at a given gas volume compared to the reversible case where the rate of deformation is very low. The viscous stresses in a gas are proportional to the local rate of deformation, and, in an expansion, they are negative (tensile).

Since S is a state variable, though the irreversible system took a different route to get from P₁V₁ to P₂V₂, Delta S for the is the same for both the reversible and irreversible expansion. However, since less work is required for irreversible system, the amount of heat required to put into the system from the surroundings is less. So delta s of system + Surroundings for irreversible is greater than 0. Am I understanding that part correctly?

Absolutely.

Finally, above you said that video incorrectly determines Delta G between initial and final states of both systems. What would be the correct way to interpret it?

The correct way of interpreting $\Delta G$ in this scenario is that it is minus the maximum amount of work that the gas can do on its surroundings over all the possible constant-reservoir-temperature reversible and irreversible paths between the initial and final equilibrium states of the system. This is the same as minus the reversible work over all reversible paths.

 

[Abhishek Jain]

Thanks for the help. I was so lost before this. :)

 

[DoubtExpert]

I have a few comments about this before addressing the friction issue.

1. They are not talking about $\Delta G$ for a reversible chemical reaction compared to $\Delta G$ for an irreversible chemical reaction. They are talking about an reversible gas expansion compared to an irreversible gas expansion (with no chemical reaction) in a cylinder between the same initial and final equilibrium states, both of which are at the same temperature.

2. There are many technical mistakes in this video, and you should not pay any attention to the video. It starts out OK in the discussion the isothermal reversible expansion, but, from that point on, it goes farther and farther astray from fundamental thermodynamic reality. In the end, it incorrectly determines the $\Delta G$ between the initial and final states of the system.

3. The addition of friction between the piston and cylinder for the irreversible case is a "red herring" that did not need to be introduced to describe the difference between an irreversible expansion and a reversible expansion. Moreover, the effect of friction is extremely tricky to analyze in this problem, as you have already learned. So I am going to first analyze the expansions quantitatively without friction to illustrate what is happening. After that, I'll proceed to analyzing the friction effect.

The present analysis assumes that the cylinder is in a vacuum so that external force on the top side of the piston is due only to the weights (gravel or larger weights) on the top of the piston. The piston itself is assumed frictionless, with a cross sectional area of A, and a weight W. The mass of weights on top of the piston in the initial state are w, and, in the final state, these weights have been entirely removed. The gas in the cylinder is an ideal gas (n moles).

INITIAL STATE:
Temperature T
Pressure $P_1$
Volume $V_1$
$P_1A=(W+w)$

FINAL STATE:
Temperature T
Pressure $P_2$
Volume $V_2$
$P_2A=W$

REVERSIBLE PATH:
For the reversible path, the weights w are removed gradually, so that the system is always close to thermodynamic equilibrium at the constant temperature T. So, for this path, we have: $$\Delta U=0$$and $$Q=Work=\int_{V_1}^{V_2}{PdV}=\int_{V_1}^{V_2}{\frac{nRT}{V}dV}=nRT\ln{(V_2/V_1)}$$But, for an ideal gas, $$\frac{V_2}{V_1}=\frac{P_1}{P_2}=\frac{(W+w)}{W}$$So,$$Q=Work=nRT\ln{\left(\frac{W+w}{W}\right)}$$And the change in entropy of the system is:$$\Delta S=\frac{Q}{T}=nR\ln{\left(\frac{W+w}{W}\right)}$$The change in Gibbs Free energy of the system is then:$$\Delta G=\Delta U+\Delta (PV)-T\Delta S=-nRT\ln{\left(\frac{W+w}{W}\right)}$$The change in entropy of the constant temperature reservoir is -Q/T, so the change in entropy of the system plus the reservoir is zero.

For this reversible change, the granular weights are removed from the piston very gradually as the piston rises, so that some of the weights are removed at the original elevation of the piston, while others are removed at higher elevations, until the piston reaches its final elevation. So the total work done is equal to the increase in potential energy of the piston $W(V_2-V_1)/A$, plus the increases in potential energy of the particles comprising the weight w, which are distributed over elevations between the original height of the piston and the final height. So the work done by the gas on the piston and weights is significantly more than just that required to raise the piston to its final elevation.

IRREVERSIBLE PATH:
For the irreversible path, the entire weight of granules w is removed suddenly from the piston at time zero. As a result of this, the piston spontaneously accelerates upward because of the unbalanced net force caused by the removal of the granule weights; and the piston even overshoots its final equilibrium elevation. But eventually, because of viscous damping by the gas within the cylinder, the oscillations of the piston about the final equilibrium elevation decrease, and the piston ultimately comes to rest at its new elevation. In this case, the potential energy of the granules that were removed at the initial elevation of the piston do not change; only the potential energy of the piston itself changes. So, in the irreversible case, the amount of work done by the gas is less than in the reversible case, and is given by: $$Work=W(V_2-V_1)/A=P_2(V_2-V_1)=P_2V_2(1-V_1/V_2)=nRT(1-P_2/P_1)=nRT\left(1-\frac{W}{W+w}\right)=nRT\left(\frac{w}{W+w}\right)$$In order for the work to be less, the average force of the gas on the piston (averaged over the displacement variations of the piston) must be less than in the reversible case. This lower average force is made possible by viscous stresses in the gas which act in the direction opposite to the movement of the piston.

Since the change in internal energy $\Delta U$ between the initial and final equilibrium states of the gas is again zero, the heat added from the reservoir is equal to the work done by the gas on the piston:$$Q=Work=nRT\left(\frac{w}{W+w}\right)$$So, in the irreversible case, the expanding gas does less work on its surroundings (the piston) and the amount of heat absorbed by the gas from the reservoir is also less.

In the irreversible case, since the reservoir is ideal, its change in entropy is $-Q/T=-nR\left(\frac{w}{W+w}\right)$. So, in the irreversible case, since the change in entropy of the system is unchanged, the change in entropy of the system plus surroundings is > 0. This is the result of entropy generation within the system as a result of viscous dissipation of mechanical energy.

SUMMARY:
So you can see that, even without friction, the irreversible gas expansion does less work on the surroundings than in the reversible case, and the amount of heat transferred from the reservoir to the gas is also less. For more discussion of how viscous dissipation contributes to reducing the amount of expansion work that a gas can do in an irreversible process, see my Physics Forums Insights article in https://www.physicsforums.com/insights/reversible-vs-irreversible-gas-compressionexpansion-work/

I will stop here an give you a chance to digest what I have said and to ask questions. If you wish, I will address the friction issue after that.

Why is the change in entropy of the reservoir takes as -Q/T? isn't this applicable only when the process is reversible? Why do you say that the reservoir is ideal?

 

[Chestermiller]

Why is the change in entropy of the reservoir takes as -Q/T? isn't this applicable only when the process is reversible? Why do you say that the reservoir is ideal?

In the thermodynamics literature, the term "reservoir" is used synonymously with "ideal reservoir." Are you asking: "what kind of thermal properties does a reservoir need to have to approach ideal behavior?"

 

[DoubtExpert]

In the thermodynamics literature, the term "reservoir" is used synonymously with "ideal reservoir." Are you asking: "what kind of thermal properties does a reservoir need to have to approach ideal behavior?"

Yes...What are the thermal properties? And one more thing - Why do you take the the change in entropy of a reservoir as ${{ - Q} \over T}$ ? According to the definition of entropy, $\Delta S = {{{q_{rev}}} \over T}$ , where $q_{rev}$ is the heat added to the system in a reversible process. How can you use this expression for finding change in entropy of the reservoir in an $irreversible\;process$?

 

[Chestermiller]

Yes...What are the thermal properties? And one more thing - Why do you take the the change in entropy of a reservoir as ${{ - Q} \over T}$ ? According to the definition of entropy, $\Delta S = {{{q_{rev}}} \over T}$ , where $q_{rev}$ is the heat added to the system in a reversible process. How can you use this expression for finding change in entropy of the reservoir in an $irreversible\;process$?

I can help you work out all of this. First let me state my understanding of an ideal reservoir to make sure we are starting from the same place.

An ideal reservoir is an entity that engages in heat transfer with a system, and which has the following characteristics:
1. For any finite amount of heat transferred between the system and the reservoir (in a reversible or irreversible process), the average temperature of the reservoir does not change significantly
2. Irrespective of the actual process path, the reservoir maintains its temperature constant at the interface with the system throughout.

Are you in agreement with this?

 

[DoubtExpert]

I can help you work out all of this. First let me state my understanding of an ideal reservoir to make sure we are starting from the same place.

An ideal reservoir is an entity that engages in heat transfer with a system, and which has the following characteristics:
1. For any finite amount of heat transferred between the system and the reservoir (in a reversible or irreversible process), the average temperature of the reservoir does not change significantly
2. Irrespective of the actual process path, the reservoir maintains its temperature constant at the interface with the system throughout.

Are you in agreement with this?

Yes...Agreed!

 

[Chestermiller]

Yes...Agreed!

Excellent. Now, what do you think the thermal properties of a real reservoir would have to be like in order to approach this kind of ideal behavior? I am thinking in terms of the reservoir heat capacity, its thermal conductivity, the mass of the reservoir, etc. Please see if you can get some ideas on how to reason this out. After you get back with me, I will present a specific example to help quantify these ideas.

Chet

 

[DoubtExpert]

Excellent. Now, what do you think the thermal properties of a real reservoir would have to be like in order to approach this kind of ideal behavior? I am thinking in terms of the reservoir heat capacity, its thermal conductivity, the mass of the reservoir, etc. Please see if you can get some ideas on how to reason this out. After you get back with me, I will present a specific example to help quantify these ideas.

Chet

So a real reservoir would have a low heat capacity i.e any amount of heat transferred would alter its temperature. The interface between the system and the reservoir wouldn't remain at a constant temperature.

 

[Chestermiller]

So a real reservoir would have a low heat capacity i.e any amount of heat transferred would alter its temperature.

That's a different way of saying it, but I know what you mean. So, if we have a real reservoir at temperature $T_i$ before the process, and a final thermodynamic equilibrium temperature $T_f$ after the process, and if the total amount of heat transferred to the reservoir from the system was Q during the process, then $$MC(T_f-T_i)=Q\tag{1}$$where M is the mass of material in the reservoir and C is its specific heat capacity. In addition, the change in entropy of our finite reservoir from its initial state to its final state would be: $$\Delta S=MC\ln{(T_f/T_i)}\tag{2}$$If we solve Eqn. 1 for $T_f$ and substitute the result into Eqn. 2, we obtain: $$\Delta S=MC\ln{\left(\frac{T_i+\frac{Q}{MC}}{T_i}\right)}=MC\ln{\left(1+\frac{Q}{MCT_i}\right)}\tag{3}$$For our real reservoir, the second term in parenthesis becomes small compared to unity if MC becomes very large. In the limit of very large MC, Eqn. 3 approaches:$$\Delta S=\frac{Q}{T}\tag{4}$$ where, in this limit, $T_f\rightarrow T_i=T$. So, for a real reservoir to approach ideal reservoir behavior, the mass times the specific heat capacity of the reservoir material must be very large.

The interface between the system and the reservoir wouldn't remain at a constant temperature.

What other physical property of the reservoir material would have to be large in order for the interface temperature of the real reservoir to be essentially constant throughout the process?

 

[DoubtExpert]

So, if we have a real reservoir at temperature $T_i$ before the process,

Is this process a reversible or an irreversible process?

In addition, the change in entropy of our finite reservoir from its initial state to its final state would be: $$\Delta S=MC\ln{(T_f/T_i)}\tag{2}$$

If this is an irreversible process, how is this expression for change in entropy true? As far as I know there is no direct expression for calculating entropy for an irreversible process. (This what I read in my textbook and I am not too sure about this. And this is what I meant from my original question.)

What other physical property of the reservoir material would have to be large in order for the interface temperature of the real reservoir to be essentially constant throughout the process?

Okay...I got now why high heat capacity would lead a real reservoir to ideal behaviour. Is the other physical property thermal conductivity?

 

[Chestermiller]

Is this process a reversible or an irreversible process?

Once we specify the initial and final states of a system (or, in this case, a reservoir being treated as a system), the details of the process that got it from its initial state to its final state are irrelevant. The change in thermodynamic functions like entropy and internal energy are intrinsic physical properties of the material comprising the system, and depend only on the two end states.

If this is an irreversible process, how is this expression for change in entropy true? As far as I know there is no direct expression for calculating entropy for an irreversible process. (This what I read in my textbook and I am not too sure about this. And this is what I meant from my original question.)

I sympathize with your confusion. This material is taught very poorly in almost all thermo courses and textbooks. There actually is a direct way to calculate the entropy change for an irreversible process. In an effort to deal with this gap in understanding for members of Physics Forums studying thermodynamics, I wrote the following Cookbook describing how to determine the entropy change of the system and the surroundings for an irreversible process: https://www.physicsforums.com/insights/grandpa-chets-entropy-recipe/
The Cookbook includes a number of examples to illustrate specifically how the methodology is applied.

Okay...I got now why high heat capacity would lead a real reservoir to ideal behaviour. Is the other physical property thermal conductivity?

Yes! The thermal conductivity of the reservoir has to be very high so that the difference between the boundary temperature and the bulk temperature becomes vanishingly small.

 

[DoubtExpert]

In an effort to deal with this gap in understanding for members of Physics Forums studying thermodynamics, I wrote the following Cookbook describing how to determine the entropy change of the system and the surroundings for an irreversible process: https://www.physicsforums.com/insights/grandpa-chets-entropy-recipe/
The Cookbook includes a number of examples to illustrate specifically how the methodology is applied.

Thank you for providing the CookBook. It cleared the confusion of entropy calculation for an irreversible process.

I still fail to understand certain terms in thermodynamics.

I am not comfortable with the term irreversible. When do you call a process irreversible?
Consider an ideal gas in a container with a massless, frictionless piston with some big rocks kept above it. Let the system be in contact with an ideal reservoir of temperature T. Now I remove some rocks above the piston and the gas rapidly expands due to the drop in pressure. Then I place the rocks back on the piston and thus the original pressure is restored. Since the temperature remained constant, the gas must compress and return to its original state with the same initial P,V and T. So, haven't I conducted a reversible process?

(I know this process mentioned above is irreversible because it has not been done quasistatically but I want to know the exact point to what led me to irreversibility during this process)

 

[Chestermiller]

Thank you for providing the CookBook. It cleared the confusion of entropy calculation for an irreversible process.

I still fail to understand certain terms in thermodynamics.

I am not comfortable with the term irreversible. When do you call a process irreversible?
Consider an ideal gas in a container with a massless, frictionless piston with some big rocks kept above it. Let the system be in contact with an ideal reservoir of temperature T. Now I remove some rocks above the piston and the gas rapidly expands due to the drop in pressure. Then I place the rocks back on the piston and thus the original pressure is restored. Since the temperature remained constant, the gas must compress and return to its original state with the same initial P,V and T. So, haven't I conducted a reversible process?

(I know this process mentioned above is irreversible because it has not been done quasistatically but I want to know the exact point to what led me to irreversibility during this process)

These are excellent questions.

Let me start out by defining an Internally Reversible Process. An internally reversible process is one for which your system passes through a continuous sequence of thermodynamic equilibrium states. Such a process is, of course, quasistatic. The process you described above is not internally reversible. For it to be internally reversible, you would have to decrease the load of rocks (granules) very gradually, with each individual granule having infinitesimal mass.

A Fully Reversible Process is one in which both the system and the surroundings (treated as a separate system) undergo matching internally reversible processes. In such a process, both the system and the surroundings can be returned to their original states (by following the reverse path) without significantly affecting anything else (either thermally or mechanically) in the universe.

An alternate test for an internally reversible process is if $\Delta S$ for the system is equal than the integral of $dq/T_B$, where $T_B$ is the temperature at the boundary between the system and the surroundings during the process. If it is greater than the integral, then the process is irreversible. This is the Clausius inequality.

An alternate test for a Fully Reversible Process is to determine the sum of the entropy changes for the system and the surroundings. If it is zero, then the process is Fully Reversible. If it is greater than zero, then the process is not Fully Reversible. If it is negative, then you made a mistake.

 

[DoubtExpert]

These are excellent questions.

Let me start out by defining an Internally Reversible Process. An internally reversible process is one for which your system passes through a continuous sequence of thermodynamic equilibrium states. Such a process is, of course, quasistatic. The process you described above is not internally reversible. For it to be internally reversible, you would have to decrease the load of rocks (granules) very gradually, with each individual granule having infinitesimal mass.

A Fully Reversible Process is one in which both the system and the surroundings (treated as a separate system) undergo matching internally reversible processes. In such a process, both the system and the surroundings can be returned to their original states (by following the reverse path) without significantly affecting anything else (either thermally or mechanically) in the universe.

An alternate test for an internally reversible process is if $\Delta S$ for the system is equal than the integral of $dq/T_B$, where $T_B$ is the temperature at the boundary between the system and the surroundings during the process. If it is greater than the integral, then the process is irreversible. This is the Clausius inequality.

An alternate test for a Fully Reversible Process is to determine the sum of the entropy changes for the system and the surroundings. If it is zero, then the process is Fully Reversible. If it is greater than zero, then the process is not Fully Reversible. If it is negative, then you made a mistake.

Why is it necessary to conduct a reversible process in thermodynamics?
How is the process of removing one big rock different from removing infinitesimal granules at a time? What difference does it make?
Could you give an example of how you would conduct a fully reversible process?

 

[DoubtExpert]

These are excellent questions.

Let me start out by defining an Internally Reversible Process. An internally reversible process is one for which your system passes through a continuous sequence of thermodynamic equilibrium states. Such a process is, of course, quasistatic. The process you described above is not internally reversible. For it to be internally reversible, you would have to decrease the load of rocks (granules) very gradually, with each individual granule having infinitesimal mass.

A Fully Reversible Process is one in which both the system and the surroundings (treated as a separate system) undergo matching internally reversible processes. In such a process, both the system and the surroundings can be returned to their original states (by following the reverse path) without significantly affecting anything else (either thermally or mechanically) in the universe.

An alternate test for an internally reversible process is if $\Delta S$ for the system is equal than the integral of $dq/T_B$, where $T_B$ is the temperature at the boundary between the system and the surroundings during the process. If it is greater than the integral, then the process is irreversible. This is the Clausius inequality.

An alternate test for a Fully Reversible Process is to determine the sum of the entropy changes for the system and the surroundings. If it is zero, then the process is Fully Reversible. If it is greater than zero, then the process is not Fully Reversible. If it is negative, then you made a mistake.

Sir, could you answer these questions?

 

[Chestermiller]

Why is it necessary to conduct a reversible process in thermodynamics?

A reversible process represents the limiting behavior of real processes. It tells us the maximum amount of work that can be done and the maximum amount of heat that can be transferred, or the maximum temperature change to expect. It is also important in chemical thermodynamics, for deriving equations for the equilibrium constants for chemical reactions.

How is the process of removing one big rock different from removing infinitesimal granules at a time? What difference does it make?

The amount of work will be different. The equation for the work is $W=\int{P_{ext}dV}$ where $P_{ext}$ is the force exerted by the piston on the gas at the piston face. In big rock situation, $P_{ext}$ will drop to a lower value suddenly, while in the infinitesimal granules case, $P_{ext}$ will decrease gradually with V.

Could you give an example of how you would conduct a fully reversible process?

If I wanted to do a fully adiabatic reversible expansion of a gas, I would have the granules sitting on top of a piston, and, as I removed them (say by pushing each of them, in turn, horizontally sideways), I would slide them onto shelves at the elevations they had when they left the piston. Then, for the reverse path, I would add one new infinitesimal granule at the top to get things started, and then gradually slide them off their shelves onto the piston at their various elevations until the piston was at its original elevation again. In the end, the granules, the piston, and the gas would be in the exact same state at which they started (except for the infinitesimal granule added at the top, and the additional infinitesimal granule remaining on the piston when it reached its bottom location again).

 

[DoubtExpert]

It tells us the maximum amount of work that can be done and the maximum amount of heat that can be transferred, or the maximum temperature change to expect.

Could you give me an instance where I would want to know the maximum amount of work that can be obtained or the maximum amount of heat that can be transferred?

The process you described above is not internally reversible. For it to be internally reversible, you would have to decrease the load of rocks (granules) very gradually, with each individual granule having infinitesimal mass.

Just by decreasing the load of rocks (granules) very gradually, how does it become internally reversible? (why is quasistatic necessary for reversibility?)
In the process that I had mentioned, by removing that big rock and adding it back, I eventually returned to the original state. Then how come its not reversible?

If I wanted to do a fully adiabatic reversible expansion of a gas, I would have the granules sitting on top of a piston, and, as I removed them (say by pushing each of them, in turn, horizontally sideways), I would slide them onto shelves at the elevations they had when they left the piston.

Instead of placing the granules onto the shelves, can't I just collect them in my hands as I removed them and place them back subsequently so there is no addition or remaining of any granule? I hope I am not disturbing the environment in the process.

 

[Chestermiller]

Could you give me an instance where I would want to know the maximum amount of work that can be obtained or the maximum amount of heat that can be transferred?

I am trying to design an engine to continually do a certain amount of repetitious work. How do I design it so it is most efficient and requires the least amount of fuel?

Just by decreasing the load of rocks (granules) very gradually, how does it become internally reversible? (why is quasistatic necessary for reversibility?)

In the reversible case, more work is done on the surroundings than if you decrease the load suddenly. Plus, if you stop removing granules at any time, the system is immediately at thermodynamic equilibrium. After you suddenly remove the large rock, the process continues spontaneously until it finally equilibrates. So, in the reversible case, the system passes through a continuous sequence of thermodynamic equilibrium states, while, in the rapid removal case, only the final state of the system is one of thermodynamic equilibrium.

In the process that I had mentioned, by removing that big rock and adding it back, I eventually returned to the original state.

No you didn't. The final volume and the final temperature were not the same.

Then how come its not reversible?

Like I said, the work will be different, and the final temperatures and pressures will be different.

What you need to do is solve a focus problem to examine the difference. Do you want me to define a gas expansion problem with (a) a sudden removal of rock and (b) a gradual removal of granules so you can solve it, and make the comparison yourself?

Instead of placing the granules onto the shelves, can't I just collect them in my hands as I removed them and place them back subsequently so there is no addition or remaining of any granule? I hope I am not disturbing the environment in the process.

You are disturbing the environment. Your body is part of the environment, and by collecting the granules and moving them to different elevations, your body has done work on them. And, your body is not a perfectly reversible engine. By sliding them onto and off of frictionless shelves at the various elevations, your body has done negligible work.

 

[DoubtExpert]

I always try to understand any concept intuitively. So I would appreciate if you answer that way.
Reversible, literally means you are able to bring it back to its original state i.e from where it started.

In the reversible case, more work is done on the surroundings than if you decrease the load suddenly.

I don't find any connection between - doing more work on the surroundings and thus be able to get the system back to its original state.

So, in the reversible case, the system passes through a continuous sequence of thermodynamic equilibrium states, while, in the rapid removal case, only the final state of the system is one of thermodynamic equilibrium.

Now, when a system is in thermodynamic equilibrium, what I understand is that the macrostates are uniform throughout the system (homogenous). If a system has uniform macrostates at each and every moment, how does that help it to attain back the previous state (reversible)?

No you didn't. The final volume and the final temperature were not the same.

How is that possible? I kept the system in contact with an ideal reservoir throughout the process. So the temperature of the system should be constant. In the end, the final pressure was same as the initial pressure i.e the weight of the rock. #P# and #T# are the same as before, so should #V#. So the system has attained the same state as before.

By sliding them onto and off of frictionless shelves at the various elevations, your body has done negligible work.

Then, in the process that you mentioned, how were you sliding them onto shelves at various elevations, if not manually?

 

[Chestermiller]

I always try to understand any concept intuitively. So I would appreciate if you answer that way.
Reversible, literally means you are able to bring it back to its original state i.e from where it started.

You need to be able to bring both the system and the surroundings back to their initial state.

Now, when a system is in thermodynamic equilibrium, what I understand is that the macrostates are uniform throughout the system (homogenous). If a system has uniform macrostates at each and every moment, how does that help it to attain back the previous state (reversible)?[/qoute]
Like I said, it must be both the system and the surroundings. If the system passes through a continuous sequence of thermodynamic equilibrium states, that qualifies the system to be able to be part of a reversible process. But the surroundings must also pass through a matching/corresponding set of thermodynamic equilibrium states for the change to be fully reversible.

How is that possible? I kept the system in contact with an ideal reservoir throughout the process. So the temperature of the system should be constant. In the end, the final pressure was same as the initial pressure i.e the weight of the rock. #P# and #T# are the same as before, so should #V#. So the system has attained the same state as before.

I was thinking of an adiabatic process. For an isothermal process, the amount of heat and the amount of work for irreversible and reversible paths will be different.

Then, in the process that you mentioned, how were you sliding them onto shelves at various elevations, if not manually?

It takes negligible work on your part to slide them off frictionless piston onto frictionless shelf.

I see that you are not willing to take up my challenge and actually solve a problem for a reversible and irreversible path to compare. Once you do that you will understand the differences. I have in mind two problems: 1. Expansion in contact with a constant temperature reservoir and 2. Adiabatic expansion.

To get a better idea of the comparison between reversible and irreversible expansions and compressions of ideal gases, see my Physics Forums Insights article in https://www.physicsforums.com/insights/reversible-vs-irreversible-gas-compressionexpansion-work/

 

[DoubtExpert]

You need to be able to bring both the system and the surroundings back to their initial state.I was thinking of an adiabatic process. For an isothermal process, the amount of heat and the amount of work for irreversible and reversible paths will be different.

It takes negligible work on your part to slide them off frictionless piston onto frictionless shelf.

I see that you are not willing to take up my challenge and actually solve a problem for a reversible and irreversible path to compare. Once you do that you will understand the differences. I have in mind two problems: 1. Expansion in contact with a constant temperature reservoir and 2. Adiabatic expansion.

To get a better idea of the comparison between reversible and irreversible expansions and compressions of ideal gases, see my Physics Forums Insights article in https://www.physicsforums.com/insights/reversible-vs-irreversible-gas-compressionexpansion-work/

Forgive me, I am willing to solve a problem. Give me one problem and I will try to solve it.

 

[Chestermiller]

Forgive me, I am willing to solve a problem. Give me one problem and I will try to solve it.

I have a vertical cylinder with an ideal gas inside of volume $V_0$ and temperature $T_0$. The bottom half of the cylinder is sitting in a constant temperature bath, also at $T_0$; the top half of the cylinder is insulated. There is a frictionless piston of mass m sitting on top of the gas, and a pile of rock granules of mass M sitting on top of the piston. The top of the cylinder connects to a vacuum chamber, so that the top of the piston, the rock, and the part of the cylinder above the piston are all exposed to a perfect vacuum at all times.

Does this description of the physical system make sense to you so far?

I will be posing questions to you along the way that require you to do some analysis to come up with the answers. In this way, we will work our way through to the end.

 

[DoubtExpert]

I have a vertical cylinder with an ideal gas inside of volume $V_0$ and temperature $T_0$. The bottom half of the cylinder is sitting in a constant temperature bath, also at $T_0$; the top half of the cylinder is insulated. There is a frictionless piston of mass m sitting on top of the gas, and a pile of rock granules of mass M sitting on top of the piston. The top of the cylinder connects to a vacuum chamber, so that the top of the piston, the rock, and the part of the cylinder above the piston are all exposed to a perfect vacuum at all times.
View attachment 203290Does this description of the physical system make sense to you so far?

I will be posing questions to you along the way that require you to do some analysis to come up with the answers. In this way, we will work our way through to the end.

Nice Question!

I have a silly question though. Is this situation possible in the real world? (maybe close to this)
If not why do we think of such hypothetical situation?

Other than this, it made good sense to me. You can proceed to ask questions.

 

[Chestermiller]

Nice Question!

I have a silly question though. Is this situation possible in the real world? (maybe close to this)
If not why do we think of such hypothetical situation?

Situations like this do occur in the real world, but, of course, there are additional complexities. When you are trying to learn something new, you start out simple, and then build on your knowledge. You don't go to the most complex problem from the outset, and then find that you get nowhere with it. I still need to see some demonstration on your part that you have mastered even the simple concepts. So, regarding our problem, here are some questions:

If the system in the figure is in thermodynamic equilibrium to begin with, what is the force F that the gas is exerting on the lower face of the piston (in terms of m and M)?

If the area of the piston is A, what is the pressure that the gas is exerting on the lower face of the piston?

What is the number of moles of gas in the cylinder?

We are first considering the situation where, at time zero, the pile of granules is suddenly removed from the top of the piston, and the gas experiences a rapid spontaneous expansion, at the end of which the system attains a new (final) thermodynamic equilibrium state.

In the final thermodynamic equilibrium state, what is the the force F that the gas is exerting on the lower face of the piston (in terms of m and M)?

What is pressure that the gas is exerting on the lower face of the piston?

What is the volume of the gas at the end of the expansion?

For the period of time during the expansion, if F(t) represents the force that the gas is exerting on the lower face of the piston at time t, write down a Newton's second law force balance for the motion of the piston.

 

[DoubtExpert]

Assuming that the situation described is on the surface of the $Earth$,

If the system in the figure is in thermodynamic equilibrium to begin with, what is the force F that the gas is exerting on the lower face of the piston (in terms of m and M)?

$$Force\;({F_0}) = \left( {M + m} \right)g$$

If the area of the piston is A, what is the pressure that the gas is exerting on the lower face of the piston?

$$Pressure\;({P_0}) = {{\left( {M + m} \right)g} \over A}$$

What is the number of moles of gas in the cylinder?

$$No\;of{\kern 1pt} Moles\;(n) = {{\left( {M + m} \right)g{V_0}} \over {AR{T_0}}}$$

In the final thermodynamic equilibrium state, what is the the force F that the gas is exerting on the lower face of the piston (in terms of m and M)?

$$Force\;({F_1}) = mg$$

What is pressure that the gas is exerting on the lower face of the piston?

$$Pressure\;({P_1}) = {{mg} \over A}$$

What is the volume of the gas at the end of the expansion?

$$Final\;Volume\;({V_1}) = \left( {{{M + m} \over m}} \right){V_0}$$

For the period of time during the expansion, if F(t) represents the force that the gas is exerting on the lower face of the piston at time t, write down a Newton's second law force balance for the motion of the piston.

Let $a(t)$ represent the $acceleration$ of the piston at time $t$. Then,$$\eqalign{
& F(t) - mg = ma(t) \cr
& F(t) = m(g + a(t)) \cr} $$

 

[Chestermiller]

Assuming that the situation described is on the surface of the $Earth$,
$$Force\;({F_0}) = \left( {M + m} \right)g$$
$$Pressure\;({P_0}) = {{\left( {M + m} \right)g} \over A}$$
$$No\;of{\kern 1pt} Moles\;(n) = {{\left( {M + m} \right)g{V_0}} \over {AR{T_0}}}$$
$$Force\;({F_1}) = mg$$
$$Pressure\;({P_1}) = {{mg} \over A}$$
$$Final\;Volume\;({V_1}) = \left( {{{M + m} \over m}} \right){V_0}$$
Let $a(t)$ represent the $acceleration$ of the piston at time $t$. Then,$$\eqalign{
& F(t) - mg = ma(t) \cr
& F(t) = m(g + a(t)) \cr} $$

Excellent!

Now, before we continue, I'm going to give you a little more information about reversible and irreversible processes, so that you can better understand what is happening to the gas in each case.

REVERSIBLE PROCESSES
As we said earlier, in a reversible process, the system experiences a deformational and thermal path that is only slightly removed from a continuous sequence of thermodynamic equilibrium states. For a thermodynamic equilibrium state of a gas within a cylinder, the temperature, pressure, and specific volume (reciprocal of density) of the gas are uniform throughout the cylinder, and the compressive stresses (forces per unit area) within gas are the same in all directions; for an ideal gas:
$$\sigma_x=P=\frac{RT}{V}$$
$$\sigma_y=P=\frac{RT}{V}$$
$$\sigma_z=P=\frac{RT}{V}$$where the $\sigma$'s are the compressive stresses in the coordinate directions, T is the uniform temperature of the gas within the cylinder, p is the uniform pressure, and V is the total volume of the gas in the cylinder divided by the number of moles (i.e., the uniform specific volume)

IRREVERSIBLE PROCESSES
The mechanical behavior of a gas, in general, is more complicated than the (equilibrium) equation of state of the gas; it depends not only on the P-V-T behavior of the gas, but also on the rate at which the gas is being deformed. The latter is described by the spatial derivatives of the velocity of the gas. The more general equations for the compressive stresses in a gas that is not in thermodynamic equilibrium (derived in 3D from Newton's Law of viscosity) are given by:
$$\sigma_x=P-2\mu\frac{\partial u_x}{\partial x}=\frac{RT}{V}-2\mu\frac{\partial u_x}{\partial x}$$
$$\sigma_y=P-2\mu\frac{\partial u_y}{\partial y}=\frac{RT}{V}-2\mu\frac{\partial u_y}{\partial y}$$
$$\sigma_z=P-2\mu\frac{\partial u_x}{\partial x}=\frac{RT}{V}-2\mu\frac{\partial u_z}{\partial z}$$where the u's are the velocity components in the coordinate directions and $\mu$ is the viscosity of the gas. In these equations, we still call P the pressure of the gas, although this "thermodynamic pressure" is no longer the total force per unit area acting on every surface; the local pressure P just contributes to the force per unit area. Note also that, in an irreversible process, the temperature, pressure, and specific volume of the gas in these equations are no longer uniform spatially within the cylinder (e.g., the specific volume is no longer the total volume of the cylinder divided by the total number of moles); all three are functions of the spatial coordinates x, y, and z. Thus, T = T(x,y,z), P = P(x,y,z), and V = V(x,y,z). Note also that, when the velocity components of a gas become very small, as in a reversible deformation, the equations for the stresses reduce to those for a reversible process.

There are several reasons why the temperature, pressure, and specific volume are not uniform in an irreversible process. First of all, the gas has mass, so it has to be accelerated just like the piston is accelerated. But it is not solid like the piston. So, in a rapid deformation, it experiences a kind of "sloshing" effect, where different parts of the gas experience different velocities and accelerations. Secondly, heat transfer is occurring at a finite rate from the reservoir to the gas, so portions of the gas near the wall experience temperature changes at a different rate than portions of the gas closer to the axis. This also causes the viscosity of the gas (which is temperature dependent) to affect the compressive stresses in the gas non-uniformly. Finally, there can be small scale turbulence in the gas as it deforms, and this influences the effective value of the viscosity (which feeds back into the compressive stresses).

The bottom line is that, in an irreversible process, many complicated phenomena are occurring within the gas that affects its behavior in a difficult-to-quantify manner, and we can't simply say that the compressive stress acting on the piston face can be determined by merely using the ideal gas equation under the assumption that the pressure, temperature, and specific volume are uniform throughout.

Now, back to our problem. The equation you wrote for the force exerted by the gas on the piston face was as follows:
$$F(t) = m(g + a(t))$$where we now know that $F(t)=\sigma_zA=(\frac{RT}{V}-2\mu\frac{\partial u_z}{\partial z})A$, with the spatial derivative of vertical gas velocity evaluated at the piston face. Another way of writing your equation is:
$$F(t)=m(g+\frac{du}{dt})$$where u is the piston velocity. If we multiply this equation on both sides by u=dz/dt, we obtain:
$$F(t)\frac{dz}{dt}=m(g\frac{dz}{dt}+u\frac{du}{dt})$$
If I integrate the left hand side of this equation with respect to t, from time zero to time t, I get:
$$W(t)=\int_0^t{F(t')\frac{dz'}{dt'}dt'}=\int_{z_0}^{z(t)}{F(z')dz'}$$where W(t) is the work done by the gas on the piston from time zero (at which the pile of rocks was suddenly removed) to time t, z' and t' are dummy variables of integration, and $z_0$ is the elevation of the piston at time zero. What do you get when you integrate the right hand side of the equation with respect to t, and then combine the results?

 

[DoubtExpert]

Excellent!

Now, before we continue, I'm going to give you a little more information about reversible and irreversible processes, so that you can better understand what is happening to the gas in each case.

REVERSIBLE PROCESSES
As we said earlier, in a reversible process, the system experiences a deformational and thermal path that is only slightly removed from a continuous sequence of thermodynamic equilibrium states. For a thermodynamic equilibrium state of a gas within a cylinder, the temperature, pressure, and specific volume (reciprocal of density) of the gas are uniform throughout the cylinder, and the compressive stresses (forces per unit area) within gas are the same in all directions; for an ideal gas:
$$\sigma_x=P=\frac{RT}{V}$$
$$\sigma_y=P=\frac{RT}{V}$$
$$\sigma_z=P=\frac{RT}{V}$$where the $\sigma$'s are the compressive stresses in the coordinate directions, T is the uniform temperature of the gas within the cylinder, p is the uniform pressure, and V is the total volume of the gas in the cylinder divided by the number of moles (i.e., the uniform specific volume)

IRREVERSIBLE PROCESSES
The mechanical behavior of a gas, in general, is more complicated than the (equilibrium) equation of state of the gas; it depends not only on the P-V-T behavior of the gas, but also on the rate at which the gas is being deformed. The latter is described by the spatial derivatives of the velocity of the gas. The more general equations for the compressive stresses in a gas that is not in thermodynamic equilibrium (derived in 3D from Newton's Law of viscosity) are given by:
$$\sigma_x=P-2\mu\frac{\partial u_x}{\partial x}=\frac{RT}{V}-2\mu\frac{\partial u_x}{\partial x}$$
$$\sigma_y=P-2\mu\frac{\partial u_y}{\partial y}=\frac{RT}{V}-2\mu\frac{\partial u_y}{\partial y}$$
$$\sigma_z=P-2\mu\frac{\partial u_x}{\partial x}=\frac{RT}{V}-2\mu\frac{\partial u_z}{\partial z}$$where the u's are the velocity components in the coordinate directions and $\mu$ is the viscosity of the gas. In these equations, we still call P the pressure of the gas, although this "thermodynamic pressure" is no longer the total force per unit area acting on every surface; the local pressure P just contributes to the force per unit area. Note also that, in an irreversible process, the temperature, pressure, and specific volume of the gas in these equations are no longer uniform spatially within the cylinder (e.g., the specific volume is no longer the total volume of the cylinder divided by the total number of moles); all three are functions of the spatial coordinates x, y, and z. Thus, T = T(x,y,z), P = P(x,y,z), and V = V(x,y,z). Note also that, when the velocity components of a gas become very small, as in a reversible deformation, the equations for the stresses reduce to those for a reversible process.

There are several reasons why the temperature, pressure, and specific volume are not uniform in an irreversible process. First of all, the gas has mass, so it has to be accelerated just like the piston is accelerated. But it is not solid like the piston. So, in a rapid deformation, it experiences a kind of "sloshing" effect, where different parts of the gas experience different velocities and accelerations. Secondly, heat transfer is occurring at a finite rate from the reservoir to the gas, so portions of the gas near the wall experience temperature changes at a different rate than portions of the gas closer to the axis. This also causes the viscosity of the gas (which is temperature dependent) to affect the compressive stresses in the gas non-uniformly. Finally, there can be small scale turbulence in the gas as it deforms, and this influences the effective value of the viscosity (which feeds back into the compressive stresses).

The bottom line is that, in an irreversible process, many complicated phenomena are occurring within the gas that affects its behavior in a difficult-to-quantify manner, and we can't simply say that the compressive stress acting on the piston face can be determined by merely using the ideal gas equation under the assumption that the pressure, temperature, and specific volume are uniform throughout.

Now, back to our problem. The equation you wrote for the force exerted by the gas on the piston face was as follows:
$$F(t) = m(g + a(t))$$where we now know that $F(t)=\sigma_zA=(\frac{RT}{V}-2\mu\frac{\partial u_z}{\partial z})A$, with the spatial derivative of vertical gas velocity evaluated at the piston face. Another way of writing your equation is:
$$F(t)=m(g+\frac{du}{dt})$$where u is the piston velocity. If we multiply this equation on both sides by u=dz/dt, we obtain:
$$F(t)\frac{dz}{dt}=m(g\frac{dz}{dt}+u\frac{du}{dt})$$
If I integrate the left hand side of this equation with respect to t, from time zero to time t, I get:
$$W(t)=\int_0^t{F(t')\frac{dz'}{dt'}dt'}=\int_{z_0}^{z(t)}{F(z')dz'}$$where W(t) is the work done by the gas on the piston from time zero (at which the pile of rocks was suddenly removed) to time t, z' and t' are dummy variables of integration, and $z_0$ is the elevation of the piston at time zero. What do you get when you integrate the right hand side of the equation with respect to t, and then combine the results?

I am trying my best to understand the information that you have given about reversible and irreversible processes.
I hope I got the difference between compressive stress and pressure right. Compressive stress is the force exerted by the gas on an area within the gas whereas pressure is the force exerted by the gas on an external object?
Forgive me for this- I am yet to learn spatial derivatives, differentials or integration in my college. I just know the basics of differentiation (and a little integration).

 

[Chestermiller]

I am trying my best to understand the information that you have given about reversible and irreversible processes.
I hope I got the difference between compressive stress and pressure right. Compressive stress is the force exerted by the gas on an area within the gas whereas pressure is the force exerted by the gas on an external object?

No. The term stress applies both to within the gas and on an external object. When you learned about pressure, it was called the force per unit area, but it was applied only to cases where the fluid was not deforming and, in these cases, it was the same in all directions. When a fluid is deforming, the mechanical behavior of the fluid is very different than when it is at equilibrium. The force per unit area is no longer the same on surfaces oriented in all directions, both internal and external. For a fluid that is deforming, we no longer call pressure the force per unit area; we now reserve the term "pressure" for the average force per unit area (averaged over all possible surface orientations). The remainder of the forces per unit area (which now depend on direction within the fluid) are "viscous stresses." These depend on the rate at which the fluid is deforming. In a deforming gas, the entity we now call the pressure is found to be very nearly equal to the 'thermodynamic pressure" calculated from the ideal gas law, but based on local values of the temperature and specific volume. The remaining contributions to the stress in different directions are related to the rate at which the gas is deforming.

 

[Chestermiller]

Forgive me for this- I am yet to learn spatial derivatives, differentials or integration in my college. I just know the basics of differentiation (and a little integration).

OK. Then I will provide the integration of the right hand side of the equation, and hope you are comfortable with it. So, the work done by the gas on the lower face of the piston (up to time t) is given by:
$$W(t)=mg[z(t)-z_0]+\frac{1}{2}mu^2(t)$$The first term on the right is the change in potential energy of the piston up to time t, and the second term is the change in its kinetic energy.

Now, my next question for you is "what do you think the frictionless piston motion will be like as time progresses?"
(a) It just keeps going higher and higher in elevation
(b) It overshoots the final equilibrium elevation, and then moves back downward again, oscillating back and forth about the equilibrium elevation forever (like as spring/mass system in simple harmonic motion)
(c) It begins oscillating as in (b), but the magnitude of the oscillation damps out, and vanishes when the piston finally settles at the equilibrium elevation (even though the piston is frictionless)

 

[DoubtExpert]

OK. Then I will provide the integration of the right hand side of the equation, and hope you are comfortable with it. So, the work done by the gas on the lower face of the piston (up to time t) is given by:
$$W(t)=mg[z(t)-z_0]+\frac{1}{2}mu^2(t)$$The first term on the right is the change in potential energy of the piston up to time t, and the second term is the change in its kinetic energy.

Now, my next question for you is "what do you think the frictionless piston motion will be like as time progresses?"
(a) It just keeps going higher and higher in elevation
(b) It overshoots the final equilibrium elevation, and then moves back downward again, oscillating back and forth about the equilibrium elevation forever (like as spring/mass system in simple harmonic motion)
(c) It begins oscillating as in (b), but the magnitude of the oscillation damps out, and vanishes when the piston finally settles at the equilibrium elevation (even though the piston is frictionless)

I think the right answer is Option (b). (Option (a) is not possible because of the pressure from the piston on the gas and Option (c) is not possible because there exist no agents that would assist in damping)
But I don't know why it undergoes a simple harmonic motion. Why do objects undergo a SHM? Why can't they settle in their new equilibrium position right away?

 

[Chestermiller]

I think the right answer is Option (b). (Option (a) is not possible because of the pressure from the piston on the gas and Option (c) is not possible because there exist no agents that would assist in damping)
But I don't know why it undergoes a simple harmonic motion. Why do objects undergo a SHM? Why can't they settle in their new equilibrium position right away?

Objects undergo SHM because there is nothing to drain the energy from the system. When the object passes through the equilibrium position, it has velocity so it overshoots. When it gets to an extreme position, its velocity is zero, but there is then a force acting on it from the spring. So, in the absence of an energy drain, the object keeps oscillating forever.

The actual correct answer to my question is Option (c), because there does exist an agent that would assist in damping. This is the viscous dissipation of mechanical energy to internal energy which occurs within the gas. The viscous forces within the gas act like a shock absorber in a a car, to dissipate mechanical energy. When you hit a bump, the car does not oscillate forever. The shock absorber causes the oscillation to damp out. In the case of a gas, both the springiness and the viscous dissipation are both present within the gas, rather than separately as in a car.

So, back to our problem. After an infinite amount of time, when our system achieves thermodynamic equilibrium and the piston has stabilized at its final equilibrium position, what is the equation for the amount of work that the gas has done on the piston?