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

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SUMMARY

The Helmholtz Free Energy (HFE) change during constant temperature and volume processes can be analyzed using the differential form dA = -SdT - PdV. In scenarios where temperature (T) and volume (V) are held constant, the change in HFE is influenced by non-PV work, such as electrical work in electro-chemical reactions. A practical example involves a rigid container with nitrogen gas in one compartment and vacuum in another, where removing the partition allows the system to re-equilibrate, resulting in a change in HFE despite constant T and V. The discussion emphasizes that the change in HFE can occur due to variations in entropy (S) and the number of moles of reactants and products.

PREREQUISITES
  • Understanding of Helmholtz Free Energy (HFE) and its differential form.
  • Familiarity with thermodynamic concepts such as entropy (S) and pressure (P).
  • Knowledge of electro-chemical reactions and non-PV work.
  • Basic principles of gas behavior, particularly ideal gases.
NEXT STEPS
  • Study the implications of Helmholtz Free Energy in non-mechanical work scenarios.
  • Explore the relationship between entropy changes and Helmholtz Free Energy in thermodynamic processes.
  • Investigate the effects of free expansion on ideal gases and its impact on thermodynamic properties.
  • Learn about the role of chemical potentials in determining changes in Helmholtz Free Energy.
USEFUL FOR

Students and professionals in thermodynamics, chemical engineering, and physical chemistry, particularly those interested in the applications of Helmholtz Free Energy in non-PV work scenarios.

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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