How Does Helmholtz Free Energy Change During Constant Temperature and Volume?

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Discussion Overview

The discussion revolves around the Helmholtz free energy (HFE) and its behavior during processes at constant temperature and volume. Participants explore the implications of the differential form of HFE and its application to various scenarios, including free expansion and non-PV work.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant expresses confusion about how the Helmholtz free energy can change when both temperature and volume are held constant, referencing the differential form dA = -SdT - PdV.
  • Another participant corrects the earlier statement, suggesting that the natural variable should be temperature instead of entropy.
  • Some participants mention that the Helmholtz free energy is related to the capacity to perform non-mechanical work, which complicates the understanding of its change under constant conditions.
  • A participant introduces a scenario involving a rigid container with two compartments, questioning the change in Helmholtz free energy when the partition is removed.
  • There is a discussion about the nature of work done in the system, with references to electrical work and chemical potentials, indicating that changes in the number of moles may also play a role.
  • One participant suggests that in the case of free expansion, no work is done, and any change in Helmholtz free energy must be attributed to changes in entropy.

Areas of Agreement / Disagreement

Participants express varying levels of understanding and confusion regarding the implications of constant temperature and volume on Helmholtz free energy. There is no consensus on how to interpret the differential form or the scenarios presented, indicating multiple competing views and unresolved questions.

Contextual Notes

Some discussions involve assumptions about ideal gases and specific conditions that may not be universally applicable. The implications of non-PV work and changes in chemical potentials are also noted as areas of complexity.

Getterdog
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This is really basic,and I’m not seeing something obvious,but I’d appreciate help with this concept. In differential form dA= -tds-pdv. However s and v are the natural variables for this free energy and are held constant . As I understand it the Helmholtz free energy is the energy available to do work holding t and v constant. So according to the differential form ds and dv are 0 as these are held constant.,how can there be a change in the HFE during a process holding these constant.? Feeling really dumb.
 
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Sorry replace s with t as the natural variable,but so da=-sdt-Pdv,hold t and v constant.
 
It is referring to non pv work.
 
Now I’m more confused. Per wiki HFE is capacity todo non mechanical plus mechanical work. Still I don’t get da = -sdt-pdv if t and v are constant.
 
Getterdog said:
Now I’m more confused. Per wiki HFE is capacity todo non mechanical plus mechanical work. Still I don’t get da = -sdt-pdv if t and v are constant.
You left out the terms associated with the chemical potentials times the changes in the numbers of moles.
 
Well I’m simplifying,assuming no additional mass,,say just a fixed volume of nitrogen gas.
 
volume or density?
 
The Helmholtz free energy is the energy available to do work, more precisely, for a system in contact with a constant temperature heat reservoir at the same temperature as the initial temperature of the system, if the final temperature of the system is also at that temperature. In this case, the work is not necessarily P-V work, and other forms of work are also allowed. For example, if the system is also at constant volume, then electrical work can be performed, associated with electro-chemical reactions within the system. In this case, the numbers of moles of reactants and products change, even if the temperature and volume does not.
 
Here's a problem we can focus on to help us gain some understanding. We have a rigid container having 2 equal-volume compartments V, with a partition between the two compartments, and nitrogen in one compartment at pressure P, with vacuum in the other. The gas is initially at temperature T, and the container is in contact with an ideal constant-temperature reservoir at temperature T. We suddenly remove the partition and allow the system to re-equilibrate. To begin with, what is the change in Helmholtz free energy between the initial and final states of this system? How much work is done on the surroundings outside the container? How much heat is transferred?
 
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  • #10
Take for example a battery with dA=-SdT-VdP +UdQ where U is the electrical potential and Q charge. At constant T and P, dA=UdQ is the non-volume work.

Edit: Changed Current for Charge
 
Last edited:
  • #11
Chestermiller said:
Here's a problem we can focus on to help us gain some understanding. We have a rigid container having 2 equal-volume compartments V, with a partition between the two compartments, and nitrogen in one compartment at pressure P, with vacuum in the other. The gas is initially at temperature T, and the container is in contact with an ideal constant-temperature reservoir at temperature T. We suddenly remove the partition and allow the system to re-equilibrate. To begin with, what is the change in Helmholtz free energy between the initial and final states of this system? How much work is done on the surroundings outside the container? How much heat is transferred?
Well this is free expansion,and no work is done,and I’m assuming an idea gas,no change in temp, so no change in internal energy.. so a= u-tds so if there is a change in a it must be due to s?
 
  • #12
Getterdog said:
Well this is free expansion,and no work is done,and I’m assuming an idea gas,no change in temp, so no change in internal energy.. so a= u-tds so if there is a change in a it must be due to s?
Excellent. So the initial state is P,V, and T, and the final state is P/2, 2V, T. So do you know how to determine the change in A?
 

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