Maximum non - PV Work at Constant Temperature and Pressure and Reversi

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Discussion Overview

The discussion revolves around the extraction of maximum non-PV work at constant temperature and pressure, particularly in the context of reversible and irreversible chemical reactions, such as those occurring in galvanic cells. Participants explore the implications of thermodynamic equilibrium on work extraction and the relationship between Gibbs energy and non-PV work.

Discussion Character

  • Debate/contested
  • Conceptual clarification
  • Technical explanation

Main Points Raised

  • Some participants propose that maximum non-PV work can be extracted when processes are carried out reversibly, equating this work to the change in Gibbs energy of the system.
  • Others argue that if a chemical reaction is at equilibrium, the Gibbs energy is zero, raising questions about how non-PV work can be extracted in such a state.
  • It is suggested that reversible processes can still do work while maintaining thermodynamic equilibrium at every stage, as illustrated by the reversible isothermal expansion of an ideal gas.
  • Some participants question whether carrying a reaction reversibly necessarily implies that the system is at chemical equilibrium throughout the process.
  • A later reply discusses the operation of galvanic cells, noting that applying an external potential allows for gradual work extraction, even when the reaction is nearly at equilibrium.
  • There is a comparison made between the operation of galvanic cells and isothermal gas expansion, emphasizing the balance of external and internal pressures or potentials.
  • Some participants highlight a trade-off between efficiency and power in practical applications, noting that processes operating close to reversible conditions may achieve high efficiency but produce minimal power.

Areas of Agreement / Disagreement

Participants express differing views on the relationship between reversible processes, equilibrium, and non-PV work extraction. There is no consensus on whether reversible reactions can be at equilibrium while still allowing for non-PV work to be done.

Contextual Notes

Participants note the complexities involved in defining equilibrium and the assumptions about Gibbs energy changes during reversible processes. The discussion highlights the limitations of applying theoretical models to practical scenarios.

Dario56
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TL;DR
What does it mean to carry out chemical reaction reversibly or irreversibly?
When we talk about systems at constant temperature and pressure, maximum amount of non-PV work can be extracted if process is carried reversibly and in that case it is equal to change in Gibbs energy of the system (decrease in Gibbs energy if system does non - PV work, A.K.A work is extracted from system).

When we have chemical reaction in the system (for example electrochemical process in galvanic cell) than Gibbs energy of the system changes because of that reaction.

Basically, what does it mean to carry chemical reaction reversibly or irreversibly? Does carrying reaction reversibly mean that system is at chemical equilibrium where reaction Gibbs energy equals zero? If so, how can we extract any non - PV work (in our case non - PV work is electrical work since we are talking about galvanic cell) if system/reaction is already at equilibrium?

Electromotive force of cell is zero at equilibrium and such cell can't be used to power anything (electric motor for example).
 
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When you carry out any process reversibly, the system is at thermodynamic equilibrium at every stage of the process, and yet reversible work is done on the surroundings. In the case of reversible isothermal expansion of an ideal gas against a piston imposing sequentially decreasing pressure, for example, do you think that no reversible work is done (because the gas is at thermodynamic equilibrium over the sequence of states).
 
Chestermiller said:
When you carry out any process reversibly, the system is at thermodynamic equilibrium at every stage of the process, and yet reversible work is done on the surroundings. In the case of reversible isothermal expansion of an ideal gas against a piston imposing sequentially decreasing pressure, for example, do you think that no reversible work is done (because the gas is at thermodynamic equilibrium over the sequence of states).
To answer your question, no. However, there is a difference between PV and non-PV work. If carrying reaction reversibly at constant p and T means that system is at chemical equilibrium during process than reaction Gibbs energy is zero. We know that maximum non-PV work can be extracted if reaction is carried reversibly and in that case is equal to decrease/change of Gibbs energy of system. But, if so since system is at chemical equilibrium at all times and change in Gibbs energy of system is zero and thus non-PV work is zero. Maybe, carrying reaction reversibly doesn't mean system is at chemical equilibrium at all times during change. If we take for example galvanic cell, how can we carry the reaction reversibly?
 
Dario56 said:
To answer your question, no. However, there is a difference between PV and non-PV work. If carrying reaction reversibly at constant p and T means that system is at chemical equilibrium during process than reaction Gibbs energy is zero. We know that maximum non-PV work can be extracted if reaction is carried reversibly and in that case is equal to decrease/change of Gibbs energy of system. But, if so since system is at chemical equilibrium at all times and change in Gibbs energy of system is zero and thus non-PV work is zero. Maybe, carrying reaction reversibly doesn't mean system is at chemical equilibrium at all times during change. If we take for example galvanic cell, how can we carry the reaction reversibly?
The case of the Galvanic cell is very similar to the case of isothermal gas expansion, although, at first glance, it doesn't look that way. In the case of the galvanic cell, you apply an external electrode potential to balance the potential of the cell. Then you allow the cell to do work very gradually by allowing one electron at a time to flow through the external circuit and drive whatever motor (or whatever) it drives. The reaction is always very nearly at equilibrium at the imposed potential, and reactants are being consumed as products are being allowed to form (but the concentrations of the reactants and product ions in the cells are very large, so they are not changing significantly) while a small amount of charge is flowing. In short, the work done is equal to the cell potential times the amount of charge that is allowed to flow. This can be evaluated per mole of charge.
 
Chestermiller said:
The case of the Galvanic cell is very similar to the case of isothermal gas expansion, although, at first glance, it doesn't look that way. In the case of the galvanic cell, you apply an external electrode potential to balance the potential of the cell. Then you allow the cell to do work very gradually by allowing one electron at a time to flow through the external circuit and drive whatever motor (or whatever) it drives. The reaction is always very nearly at equilibrium at the imposed potential, and reactants are being consumed as products are being allowed to form (but the concentrations of the reactants and product ions in the cells are very large, so they are not changing significantly) while a small amount of charge is flowing. In short, the work done is equal to the cell potential times the amount of charge that is allowed to flow. This can be evaluated per mole of charge.
Yes, applying outer voltage to balance cell voltage is analogous to balancing external and internal pressure on piston during state change. And changing outer voltage in very small steps (infinitesimally for perfectly reversible process) is analogous to small changes in external pressure. So, galvanic cell working perfectly reversibly can't give any current or that current is very small if real process is carried close to reversible. Such cell would be very efficient (with efficiency close to thermodynamic limit), but would develop almost no power in the same way if engines had state changes close to reversible they would develop almost no power, but would be working very close to Carnot efficiency. When we compare real/practical processes with reversible processes we see there is a trade off between efficiency and power. Even though processes operating close to reversible would be maximally efficient, they would be useless for practical/engineering applications as they create almost no power.
 
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