Adiabatic or Isobaric process?

  • #51
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No I am not calculating integral pdx, I am just using mechanics consideration. And even this mechanical work is to be assumed to be done quasi statically. We use work energy theorem of Mechanics to conclude that the energy of the system on which work has been done must increase its energy when the final equilibrium state is reached. To conclude the change in internal energy we will have to consider another reversible process of heating quasi statically to achieve the final state with the help of a reservoir of infinite thermal capacity passing through series of equlibrium states of increasing temperature.
I have no understanding of what you are trying to say here, but I can tell you as one of Physics Forums experts on thermodynamics that I agree totally with what Philip Kosck and and Terry Bing have said. If what you are saying disagrees with them, then I too disagree with you.
 
  • #52
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I have no understanding of what you are trying to say here, but I can tell you as one of Physics Forums experts on thermodynamics that I agree totally with what Philip Kosck and and Terry Bing have said. If what you are saying disagrees with them, then I too disagree with you.
It is too much conditional Sir. After all I had used your.earlier post, to write what I wanted, where in you said that we can assume that external pressure can be assumed to be held constant at 3P, then we can calculate the external work done. My submission is that this kind of quasi statical integration is also present in the definition of potential energy in mechanics. Rather the work energy theorem it self talks about reversible work. I think it is not true that these considerations just pop up in thermodynamics alone but they are present in mechanics too. Having known that force varies as 1/r^2 our assertion that potential energy will be proportional to 1/r, does involve the idea of reversible work done.
 
  • #53
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It is too much conditional Sir. After all I had used your.earlier post, to write what I wanted, where in you said that we can assume that external pressure can be assumed to be held constant at 3P, then we can calculate the external work done.
My submission is that this kind of quasi statical integration is also present in the definition of potential energy in mechanics.
The compression process that occurs as a result of increasing the external pressure on the gas from P to 3P and then holding the pressure constant is neither quasi static nor reversible. It is a highly irreversible process, resulting in significant dissipation of mechanical energy to internal energy as a result of viscous friction. What does the work energy theorem say about processes in which friction is involved to dissipate mechanical energy?
 
  • #54
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The compression process that occurs as a result of increasing the external pressure on the gas from P to 3P and then holding the pressure constant is neither quasi static nor reversible. It is a highly irreversible process, resulting in significant dissipation of mechanical energy to internal energy as a result of viscous friction. What does the work energy theorem say about processes in which friction is involved to dissipate mechanical energy?
For that work energy theorem says nothing, as work energy theorem is originally conceptualized for a point particle change in its KE and PE. In the problem posed by OP it was not written that P was made 3P in a gradual manner. It was written P was suddenly made 3P it is mentioned in the problem that the process is isobaric and finally it reaches an equilibrium state. Whether it is stated or not we know the process is irreversible and adiabatic. It is only externally isobaric. P as a parameter of the gas is just not defined during the process. So we need to use the information that it is ideal gas and guess the final state and then connect these by reversible processes to evaluate the final equilibrium state values using the tenets of thermodynamics.

Do you not now stand for your statement on this thread that external work done will be Pext*(V2 - V1)

There are totally three things I would like to say
1. What ever I have written on this thread nowhere you have said I am wrong. Now you say you do not understand what I am saying but if I disagree with others then you also disagree with me.

2. Idea of quasi static process is present in the definition of PE of a point particle too in Mechanics

3. There are two may I say modern definitions of "work", one related with with 'Force' and other related with energy. Former is historical and well known. The latter is work is that transfer of energy from one system to another where temperature difference is not directly involved. I do not know whether it is proper to quote books on this thread but as an indulgent student let me state here these two definitions were given first in mechanics chapter and then second in thermodynamics chapter in second edition of Halliday and Resnick Physics. But in the fifth edition of the same book with more authors both definitions are given in the chapter on mechanics it self.
 
  • #55
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For that work energy theorem says nothing, as work energy theorem is originally conceptualized for a point particle change in its KE and PE. In the problem posed by OP it was not written that P was made 3P in a gradual manner. It was written P was suddenly made 3P it is mentioned in the problem that the process is isobaric and finally it reaches an equilibrium state. Whether it is stated or not we know the process is irreversible and adiabatic. It is only externally isobaric. P as a parameter of the gas is just not defined during the process. So we need to use the information that it is ideal gas and guess the final state and then connect these by reversible processes to evaluate the final equilibrium state values using the tenets of thermodynamics.
We don't need to guess anything. In this problem, during the deformation, 3P is the pressure of the gas at the piston face, and, from this we have enough information to precisely establish the final equilibrium state. We can precisely calculate the work done by the gas on the piston for this irreversible process, and we know that there is no heat transfer, so, from the first law of thermodynamics, we know the relationship between the work and the change in internal energy. We don't need to connect the initial and final states by a reversible process to evaluate the final equilibrium state; we already know the final thermodynamic equilibrium state for the irreversible process by applying the first law in conjunction with the ideal gas law.
Do you not now stand for your statement on this thread that external work done will be Pext*(V2 - V1)
Yes, I stand by this. This is work done by the gas on the piston face (which constitutes its surroundings), and it is also minus the work done by the piston face on the gas. If you want to call it external work, that's your prerogative. But it is really the work done by the gas on its surroundings.
 
  • #56
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We don't need to guess anything. In this problem, during the deformation, 3P is the pressure of the gas at the piston face, and, from this we have enough information to precisely establish the final equilibrium state. We can precisely calculate the work done by the gas on the piston for this irreversible process, and we know that there is no heat transfer, so, from the first law of thermodynamics, we know the relationship between the work and the change in internal energy. We don't need to connect the initial and final states by a reversible process to evaluate the final equilibrium state; we already know the final thermodynamic equilibrium state for the irreversible process by applying the first law in conjunction with the ideal gas law.

I perfectly agree with this Sir. But if some one wants to know the work done in the irreversible process then we can connect by a reversible process and tell the work done in general especially in this case as the process is adiabatic.[/QUOTE]
 
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  • #57
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I perfectly agree with this Sir. But if some one wants to know the work done in the irreversible process then we can connect by a reversible process and tell the work done in general especially in this case as the process is adiabatic.
No Way. For this pair of initial and final states, there is no adiabatic reversible process that gives the same amount of work as for the present adiabatic irreversible path. The work for the present irreversible adiabatic process is ##3P(V_2-V_1)##. I challenge you to specify an adiabatic reversible path that even goes between these same two end states (temperatures, pressures, and volumes). Every reversible path will have to involve some transfer of heat.
 
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  • #58
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I agree again Sir, But Cp(T2-T1), is it not a reversible heat transfer? Or you say it is irreversible heat transfer that will take place in a real process? It is reversible heat transfer through reservoirs of infinite thermal capacity at temperatures increasing steadily from T1 to T2 in infinite infinitesimal steps.Then using the fact that internal energy is a state function this heat transfer must be equal to the work done in the actual irreversible process. What I wish to state Sir is whenever we use integration to calculate either work done by the system or heat supplied to the system we make use of the concept of a reversible process only.
 
  • #59
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I agree again Sir, But Cp(T2-T1), is it not a reversible heat transfer? Or you say it is irreversible heat transfer that will take place in a real process? It is reversible heat transfer through reservoirs of infinite thermal capacity at temperatures increasing steadily from T1 to T2 in infinite infinitesimal steps.Then using the fact that internal energy is a state function this heat transfer must be equal to the work done in the actual irreversible process. What I wish to state Sir is whenever we use integration to calculate either work done by the system or heat supplied to the system we make use of the concept of a reversible process only.
In other words whether in mechanics or thermodynamics the concept of quasi static and reversibility is required as a contrivance to let the 'mathematics' work!
 
  • #60
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Who says you can't find the change in entropy of your room? As @Philip Koeck correctly points out, all you need to know is the initial and final states, irrespective of whether the actual process from the initial state to the final state was irreversible. Didn't they teach you how to get the entropy change for an irreversible process?
I might have a way of solving the original problem via the entropy, but I'm not sure it's right. Anyway, the idea is this: The entropy of the surroundings is constant. If we imagine a reversible process between the same initial and final state, the entropy of the system is also constant (This might be be the problem.), since the total entropy has to be constant. Now we can use the Sackur-Tetrode equation to write down the entropy of the system for initial and final state and equate them. Then we also have the ideal gas law to eliminate the temperature in initial and final state. The result I get is V2 p23 = V1 p13. Does that make sense?
 
  • #61
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To address this, I'm going to take off my Member hat and put on my Physics Forums Mentor hat. Part of my responsibility as a Physics Forums Mentor is to make sure that the correct information is made available to the members, so that neophytes to a subject are not led astray. I don't mean to be rude or harsh, but I have to be frank with you. You seem very very off base on your understanding of thermodynamics, and I recommend for you a thorough review of the subject (at least the first and 2nd laws). The gaps in your understanding are too broad to cover within the format of a Physics Forums thread. So, after I respond to this post, I'm going to close the present thread.

I agree again Sir, But Cp(T2-T1), is it not a reversible heat transfer?
Generally, this equation does not describe heat transfer at all. It describes the change in enthalpy per mole ##\Delta h## of an ideal gas between two thermodynamic equilibrium states. For an irreversible process, it applies only to the two end states. For a reversible process, since all the intermediate states are also thermodynamic equilibrium states, it applies to all intermediate states as well
Or you say it is irreversible heat transfer that will take place in a real process?
Irreversible heat transfer will take place in a real process.
It is reversible heat transfer through reservoirs of infinite thermal capacity at temperatures increasing steadily from T1 to T2 in infinite infinitesimal steps.Then using the fact that internal energy is a state function this heat transfer must be equal to the work done in the actual irreversible process. What I wish to state Sir is whenever we use integration to calculate either work done by the system or heat supplied to the system we make use of the concept of a reversible process only.
In the process being discussed in this thread, the system is adiabatic, so there is no heat transfer taking place at all (that is the definition of adiabatic). And there are no reservoirs, and the zero heat transfer is not equal to the finite amount of work done.
 
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  • #62
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I might have a way of solving the original problem via the entropy, but I'm not sure it's right. Anyway, the idea is this: The entropy of the surroundings is constant. If we imagine a reversible process between the same initial and final state, the entropy of the system is also constant (This might be be the problem.), since the total entropy has to be constant. Now we can use the Sackur-Tetrode equation to write down the entropy of the system for initial and final state and equate them. Then we also have the ideal gas law to eliminate the temperature in initial and final state. The result I get is V2 p23 = V1 p13. Does that make sense?
This is an adiabatic irreversible process, so entropy is generated within the system, and the change in entropy of the system is positive. Therefore, any reversible process between the same two end states must involve heat transfer to the system. It is therefore impossible to devise an adiabatic reversible path between the two end states of an adiabatic irreversible process.
 
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