Understanding reversible and irreversible process

[rkmurtyp]

Hello,
It is said that any process accompanied by dissipating effects is an irreversible process.For instance, a piston cylinder arrangement undergoing an expansion of a gas with heat lost as friction between surfaces is an irreversible process.Here the system does work on the surroundings with heat being rejected to the surroundings as a result of friction.The process can be summed up as:

Work transfer for the system= -W
Work transfer to the surroundings=+W

Heat lost from the system=-Q
Heat gained by the surroundings=+Q

Then logically the process could be reversed at any stage if external work is done by the surroundings on the system equal to W and heat lost as friction be added to the system so that both system and surroundings are restored to their initial conditions.So why it's still called an irreversible process?

The irreversible process involving expansion of a gas with friction is more complicated than necessary. A simpler process to understand the concept of reversible and irreversible processes would be to consider either expansion with pressure of the surroundings lower than the pressure of the system OR compression of a gas with friction present. After understanding these you can certainly consider expansion with friction.

Let us take the expansion experiment.

Here we are conducting an irrversible expansion process (pressure of the surroundings lower than the pressure of the system) that takes the system from an equilibrium state A to another equilibrium state B, by say, an isothermal process (keeping the system in thermal contact with a heat reservoir (HR) at the temperature, T, the temperature of the system in state A, T_A (ofcourse it is also the temperature of the system in state B, T_B i.e.T_A =T_B = T)). The work W done by the gas is less than the reversible work W_rev (the maximum work) it would have done in going from A to B by a reversible process. The system absobs Q (<Q_rev) units of heat from the reservoir.

Thus the changes in the surroundings (when the system goes from state A to state B are: A mass in the surroundings is found at a higher level (this is a way of saying the system did work on the surroundings) and a HR at temperature T, loses Q (<Q_rev) units of heat.

Now let us take the system to its original state by a reversible isothermal process. This needs minimum expenditure of work of W_rev units of work (which is more than W). The HR at temperature T receives Q_rev units (>Q units it lost). Thus when the system reaches its original state we find a mass in the surroundings at a level lower than it was in the beginning of the process and the HR gained (Q_rev - Q) units of heat (=W_rev - W).

The net result of this cyclic process is transformation of (W_rev - W) units of work into (Q_rev -Q) units of heat.

Now we can consider the compression with friction involved.

We keep the system in thermal contact with HR at T and apply pressure P_ext (>P_sym)to overcome friction. (Note that, you can employ a higher pressure to compress even if there was no friction). The isothermal work done on the system W is greater than W_rev (the minimum work needed to take the system fromB to A). Heat Q (=W) units is supplied to the HR at T.

Now according to Kelvin's postulate of the second law of thermodynamics, it is impossible to absorb Q units of heat from HR at T and convert it into W (=Q) units of work in a cyclic process leaving no other changes elsewhere.

The best we can do to take the system from state A to state B is to employ a reversible isothermal expansion process. This aborbs Q_rev (<Q) units of heat and delivers W_rev (<W) units of work (raises a mass to a level lower than its original level). The net result at the end of the cycle is: the system reached its original state and, in the surroundings a mass is found at a lower level and HR gained an equivalent amount of heat.

To understand the concept of irreversibility in thermodyanmics we don't need the presence of friction and its dissipative effects. It is in dynamics that friction leads to dissipative effects and loss of efficiency of processes.
In thermodyanmics, we have this situation: Once an irreversible process occurs, it is impossible to reverse it with no other changes left elsewhere - even employing the most efficient revrsible prcess.

 

[rkmurtyp]

Hello,
It is said that any process accompanied by dissipating effects is an irreversible process.For instance, a piston cylinder arrangement undergoing an expansion of a gas with heat lost as friction between surfaces is an irreversible process.Here the system does work on the surroundings with heat being rejected to the surroundings as a result of friction.The process can be summed up as:

Work transfer for the system= -W
Work transfer to the surroundings=+W

Heat lost from the system=-Q
Heat gained by the surroundings=+Q

Then logically the process could be reversed at any stage if external work is done by the surroundings on the system equal to W and heat lost as friction be added to the system so that both system and surroundings are restored to their initial conditions.So why it's still called an irreversible process?

The irreversible process involving expansion of a gas with friction is more complicated than necessary. A simpler process to understand the concept of reversible and irreversible processes would be to consider either expansion with pressure of the surroundings lower than the pressure of the system OR compression of a gas with friction present. After understanding these you can certainly consider expansion with friction.

Let us take the expansion experiment.

Here we are conducting an irrversible expansion process (pressure of the surroundings lower than the pressure of the system) that takes the system from an equilibrium state A to another equilibrium state B, by say, an isothermal process (keeping the system in thermal contact with a heat reservoir (HR) at the temperature, T, the temperature of the system in state A, T_A (ofcourse it is also the temperature of the system in state B, T_B i.e.T_A =T_B = T)). The work W done by the gas is less than the reversible work W_rev (the maximum work) it would have done in going from A to B by a reversible process. The system absobs Q (<Q_rev) units of heat from the reservoir.

Thus the changes in the surroundings (when the system goes from state A to state B are: A mass in the surroundings is found at a higher level (this is a way of saying the system did work on the surroundings) and a HR at temperature T, loses Q (<Q_rev) units of heat.

Now let us take the system to its original state by a reversible isothermal process. This needs minimum expenditure of work of W_rev units of work (which is more than W). The HR at temperature T receives Q_rev units (>Q units it lost). Thus when the system reaches its original state we find a mass in the surroundings at a level lower than it was in the beginning of the process and the HR gained (Q_rev - Q) units of heat (=W_rev - W).

The net result of this cyclic process is transformation of (W_rev - W) units of work into (Q_rev -Q) units of heat.

Now we can consider the compression with friction involved.

We keep the system in thermal contact with HR at T and apply pressure P_ext (>P_sym)to overcome friction. (Note that, you can employ a higher pressure to compress even if there was no friction). The isothermal work done on the system W is greater than W_rev (the minimum work needed to take the system fromB to A). Heat Q (=W) units is supplied to the HR at T.

Now according to Kelvin's postulate of the second law of thermodynamics, it is impossible to absorb Q units of heat from HR at T and convert it into W (=Q) units of work in a cyclic process leaving no other changes elsewhere.

The best we can do to take the system from state A to state B is to employ a reversible isothermal expansion process. This aborbs Q_rev (<Q) units of heat and delivers W_rev (<W) units of work (raises a mass to a level lower than its original level). The net result at the end of the cycle is: the system reached its original state and, in the surroundings a mass is found at a lower level and HR gained an equivalent amount of heat.

To understand the concept of irreversibility in thermodyanmics we don't need the presence of friction and its dissipative effects. It is in dynamics that friction leads to dissipative effects and loss of efficiency of processes.
In thermodyanmics, we have this situation: Once an irreversible process occurs, it is impossible to reverse it with no other changes left elsewhere - even employing the most efficient revrsible prcess.

 

[Soumalya]

The irreversible process involving expansion of a gas with friction is more complicated than necessary. A simpler process to understand the concept of reversible and irreversible processes would be to consider either expansion with pressure of the surroundings lower than the pressure of the system OR compression of a gas with friction present. After understanding these you can certainly consider expansion with friction.

Let us take the expansion experiment.

Here we are conducting an irrversible expansion process (pressure of the surroundings lower than the pressure of the system) that takes the system from an equilibrium state A to another equilibrium state B, by say, an isothermal process (keeping the system in thermal contact with a heat reservoir (HR) at the temperature, T, the temperature of the system in state A, T_A (ofcourse it is also the temperature of the system in state B, T_B i.e.T_A =T_B = T)). The work W done by the gas is less than the reversible work W_rev (the maximum work) it would have done in going from A to B by a reversible process. The system absobs Q (<Q_rev) units of heat from the reservoir.

Thus the changes in the surroundings (when the system goes from state A to state B are: A mass in the surroundings is found at a higher level (this is a way of saying the system did work on the surroundings) and a HR at temperature T, loses Q (<Q_rev) units of heat.

Now let us take the system to its original state by a reversible isothermal process. This needs minimum expenditure of work of W_rev units of work (which is more than W). The HR at temperature T receives Q_rev units (>Q units it lost). Thus when the system reaches its original state we find a mass in the surroundings at a level lower than it was in the beginning of the process and the HR gained (Q_rev - Q) units of heat (=W_rev - W).

The net result of this cyclic process is transformation of (W_rev - W) units of work into (Q_rev -Q) units of heat.

Now we can consider the compression with friction involved.

We keep the system in thermal contact with HR at T and apply pressure P_ext (>P_sym)to overcome friction. (Note that, you can employ a higher pressure to compress even if there was no friction). The isothermal work done on the system W is greater than W_rev (the minimum work needed to take the system fromB to A). Heat Q (=W) units is supplied to the HR at T.

Now according to Kelvin's postulate of the second law of thermodynamics, it is impossible to absorb Q units of heat from HR at T and convert it into W (=Q) units of work in a cyclic process leaving no other changes elsewhere.

The best we can do to take the system from state A to state B is to employ a reversible isothermal expansion process. This aborbs Q_rev (<Q) units of heat and delivers W_rev (<W) units of work (raises a mass to a level lower than its original level). The net result at the end of the cycle is: the system reached its original state and, in the surroundings a mass is found at a lower level and HR gained an equivalent amount of heat.

To understand the concept of irreversibility in thermodyanmics we don't need the presence of friction and its dissipative effects. It is in dynamics that friction leads to dissipative effects and loss of efficiency of processes.
In thermodyanmics, we have this situation: Once an irreversible process occurs, it is impossible to reverse it with no other changes left elsewhere - even employing the most efficient revrsible prcess.

Very well explained

Thank You

 

[Chestermiller]

The irreversible process involving expansion of a gas with friction is more complicated than necessary. A simpler process to understand the concept of reversible and irreversible processes would be to consider either expansion with pressure of the surroundings lower than the pressure of the system OR compression of a gas with friction present. After understanding these you can certainly consider expansion with friction.

Let us take the expansion experiment.

Here we are conducting an irrversible expansion process (pressure of the surroundings lower than the pressure of the system) that takes the system from an equilibrium state A to another equilibrium state B, by say, an isothermal process (keeping the system in thermal contact with a heat reservoir (HR) at the temperature, T, the temperature of the system in state A, T_A (ofcourse it is also the temperature of the system in state B, T_B i.e.T_A =T_B = T)). The work W done by the gas is less than the reversible work W_rev (the maximum work) it would have done in going from A to B by a reversible process. The system absobs Q (<Q_rev) units of heat from the reservoir.

Thus the changes in the surroundings (when the system goes from state A to state B are: A mass in the surroundings is found at a higher level (this is a way of saying the system did work on the surroundings) and a HR at temperature T, loses Q (<Q_rev) units of heat.

Now let us take the system to its original state by a reversible isothermal process. This needs minimum expenditure of work of W_rev units of work (which is more than W). The HR at temperature T receives Q_rev units (>Q units it lost). Thus when the system reaches its original state we find a mass in the surroundings at a level lower than it was in the beginning of the process and the HR gained (Q_rev - Q) units of heat (=W_rev - W).

The net result of this cyclic process is transformation of (W_rev - W) units of work into (Q_rev -Q) units of heat.

Now we can consider the compression with friction involved.

We keep the system in thermal contact with HR at T and apply pressure P_ext (>P_sym)to overcome friction. (Note that, you can employ a higher pressure to compress even if there was no friction). The isothermal work done on the system W is greater than W_rev (the minimum work needed to take the system fromB to A). Heat Q (=W) units is supplied to the HR at T.

Now according to Kelvin's postulate of the second law of thermodynamics, it is impossible to absorb Q units of heat from HR at T and convert it into W (=Q) units of work in a cyclic process leaving no other changes elsewhere.

The best we can do to take the system from state A to state B is to employ a reversible isothermal expansion process. This aborbs Q_rev (<Q) units of heat and delivers W_rev (<W) units of work (raises a mass to a level lower than its original level). The net result at the end of the cycle is: the system reached its original state and, in the surroundings a mass is found at a lower level and HR gained an equivalent amount of heat.

To understand the concept of irreversibility in thermodyanmics we don't need the presence of friction and its dissipative effects. It is in dynamics that friction leads to dissipative effects and loss of efficiency of processes.
In thermodyanmics, we have this situation: Once an irreversible process occurs, it is impossible to reverse it with no other changes left elsewhere - even employing the most efficient revrsible prcess.

Hi rkmurtyp.

Thanks for the interesting analysis of reversibility/irreversibility for isothermal expansion without friction and for isothermal compression with friction. You may be interested in a thread that Red_CCF and I have been collaborating on quantifying the adiabatic quasi static compression and expansion of an ideal gas with friction between the piston and cylinder: https://www.physicsforums.com/showthread.php?t=750946
The first 12 posts of this thread focus on a fluid mechanics issue, but, starting with post #13, the thread switches to the thermo problem. We are currently focusing on determining the change in entropy, and confirming that it is greater than zero for this adiabatic irreversible process.

Chet

 

[rkmurtyp]

Yes, I will go to the thread you gave.