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

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1. Oct 5, 2016

### 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 beacause 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.

2. Oct 6, 2016

### Staff: Mentor

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.

3. Oct 6, 2016

### 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!

4. Oct 7, 2016

### Staff: Mentor

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).
Absolutely.
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.

5. Oct 7, 2016

### Abhishek Jain

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

6. Apr 21, 2017

### DoubtExpert

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?

7. Apr 21, 2017

### Staff: Mentor

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?"

8. Apr 22, 2017

### DoubtExpert

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 $irreversible\;process$?

9. Apr 22, 2017

### Staff: Mentor

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?

10. Apr 22, 2017

### DoubtExpert

Yes...Agreed!

11. Apr 22, 2017

### Staff: Mentor

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

12. Apr 23, 2017

### DoubtExpert

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.

13. Apr 23, 2017

### Staff: Mentor

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.

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?

14. Apr 24, 2017

### DoubtExpert

Is this process a reversible or an irreversible process?

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.)

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

15. Apr 24, 2017

### Staff: Mentor

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.

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.

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.

16. May 4, 2017

### DoubtExpert

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)

17. May 4, 2017

### Staff: Mentor

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.

Last edited: May 4, 2017
18. May 5, 2017

### DoubtExpert

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?

19. May 7, 2017

### DoubtExpert

Sir, could you answer these questions?

20. May 7, 2017

### Staff: Mentor

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.
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.
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).