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

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SUMMARY

The discussion centers on the extraction of maximum non-PV work at constant temperature and pressure, specifically in the context of galvanic cells and reversible processes. It is established that maximum non-PV work corresponds to the change in Gibbs energy, which is zero at chemical equilibrium, leading to the conclusion that non-PV work cannot be extracted when the system is at equilibrium. The conversation highlights the importance of applying external electrode potentials to maintain near-equilibrium conditions in galvanic cells, allowing for gradual work extraction. Ultimately, a balance between efficiency and power output is emphasized, noting that while reversible processes are efficient, they yield minimal power, making them impractical for engineering applications.

PREREQUISITES
  • Understanding of Gibbs energy and its role in thermodynamics
  • Knowledge of galvanic cells and electrochemical processes
  • Familiarity with concepts of reversible and irreversible processes
  • Basic principles of thermodynamic equilibrium
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  • Explore the principles of electrochemical potential in galvanic cells
  • Study the relationship between Gibbs energy and non-PV work extraction
  • Investigate the implications of reversible isothermal expansion in thermodynamics
  • Learn about the trade-offs between efficiency and power in engineering applications
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Students and professionals in chemistry, chemical engineering, and thermodynamics, particularly those interested in electrochemical systems and the practical applications of reversible processes.

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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