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

• Dario56
In summary, when discussing systems at constant temperature and pressure, the maximum amount of non-PV work that can be extracted is equal to the change in Gibbs energy of the system. This is only possible if the process is carried out reversibly, meaning that the system is at thermodynamic equilibrium at every stage of the process. In the case of a chemical reaction in the system, the Gibbs energy changes due to the reaction. Carrying out a reaction reversibly means that the system is at chemical equilibrium during the process, resulting in a zero change in Gibbs energy and therefore no non-PV work. However, in the case of a galvanic cell, the reaction can be carried out reversibly by applying an external electrode potential and allowing the reaction
Dario56
TL;DR Summary
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.

Chestermiller

## 1. What is maximum non-PV work at constant temperature and pressure?

Maximum non-PV work at constant temperature and pressure refers to the maximum amount of work that can be extracted from a system without any change in the system's volume or pressure. This type of work is also known as reversible work and is achieved when the system undergoes a reversible process.

## 2. How is maximum non-PV work calculated?

The maximum non-PV work at constant temperature and pressure can be calculated using the formula W = -nRTln(V2/V1), where W is the work, n is the number of moles of gas, R is the gas constant, T is the temperature, and V1 and V2 are the initial and final volumes of the gas, respectively.

## 3. What is the significance of maximum non-PV work?

The maximum non-PV work is significant because it represents the maximum amount of useful work that can be obtained from a system. It is a measure of the system's energy and its ability to do work. It is also a key concept in thermodynamics and is used to determine the efficiency of various processes.

## 4. How does maximum non-PV work relate to reversibility?

Maximum non-PV work is only achievable in reversible processes, where the system can be returned to its initial state without any energy loss. In irreversible processes, some energy is lost as heat, resulting in a decrease in the maximum non-PV work that can be obtained from the system.

## 5. Can maximum non-PV work be negative?

Yes, maximum non-PV work can be negative. This occurs when the final volume of the system is greater than the initial volume, resulting in a negative value for ln(V2/V1). This indicates that the work done by the system is less than the work done on the system, and the process is not reversible.

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