Helmholtz free energy Definition and 9 Discussions

In thermodynamics, the Helmholtz free energy (or Helmholtz energy) is a thermodynamic potential that measures the useful work obtainable from a closed thermodynamic system at a constant temperature (isothermal). The change in the Helmholtz energy during a process is equal to the maximum amount of work that the system can perform in a thermodynamic process in which temperature is held constant. At constant temperature, the Helmholtz free energy is minimized at equilibrium.
In contrast, the Gibbs free energy or free enthalpy is most commonly used as a measure of thermodynamic potential (especially in chemistry) when it is convenient for applications that occur at constant pressure. For example, in explosives research Helmholtz free energy is often used, since explosive reactions by their nature induce pressure changes. It is also frequently used to define fundamental equations of state of pure substances.
The concept of free energy was developed by Hermann von Helmholtz, a German physicist, and first presented in 1882 in a lecture called "On the thermodynamics of chemical processes". From the German word Arbeit (work), the International Union of Pure and Applied Chemistry (IUPAC) recommends the symbol A and the name Helmholtz energy. In physics, the symbol F is also used in reference to free energy or Helmholtz function.

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  1. hagopbul

    I Another question from Ashcroft and Mermin: Fermi-Dirac Distribution

    Good Day : i reached the page 40 of Ashcroft Mermin book and after the equation 2.38 there is this expression of E(a,N) which is equal to Helmoltez Free energy F = U - TS , how this two terms F , E are related ? anyone can provide adequate explanation , and few useful references Best...
  2. Daria

    Liquid-solid transition

    Hello! I'm struggling with this exercise for three days already. Would appreciate any help. Thanks in advance. This is my thoughts... So a number density ρ=N/V. As I understand N doesn’t change through transition, but V changes The free energy F=U-TS or dF=-PdV-SdT As the liquid coexists...
  3. P

    Entropy and the Helmholtz Free Energy of a Mass-Piston System

    Attempt at a Solution: Heat Absorbed By The System By the first law of thermodynamics, dU = dQ + dW The system is of fixed volume and therefore mechanically isolated. dW = 0 Therefore dQ = dU The change of energy of the system equals the change of energy of the gas plus the change of energy...
  4. D

    Equilibrium volume of two differential van der Waal gases

    Homework Statement Two ideal van der Waals fluids are contained in a cylinder, separated by an internal moveable piston. There is one mole of each fluid, and the two fluids have the same values of the van der Waals constants b and c; the respective values of the van der Waals constant ''a'' are...
  5. W

    Helmholtz and Gibbs free energy for an adiabatic process

    Homework Statement Calculate changes in A and G of one mole of an ideal gas that undergoes the following processes respectively. 1. adiabatic expansion from (T1, P1) to (T2, P2) 2. isobaric expansion from (P, V1, T1) to (P, V2, T2) (if it is not isothermal) 3. isochoric expansion from (V, P1...
  6. arpon

    Problem on Thermodynamics

    Homework Statement Derive the equation ##U=-T(\frac{\partial A}{\partial T})_V## where ##U## is the internal energy, ##T## is the temperature, ##A## is the Helmholtz function. Reference: Heat and Thermodynamics, Zemansky, Dittman, Page 272, Problem 10.4 (a) Homework Equations ##dA=-PdV-SdT##...
  7. J

    Magnetic moment of paramagnetic crystal

    Hello, I've been having some trouble with a paramagnetism problem from my Statistical Mechanics class textbook (F. Mandl, Statistical Physics, 2nd edition, p. 25). The problem is as follows 1. Homework Statement 2. Homework Equations 1. The temperature parameter \displaystyle{ \beta =...
  8. A

    Helmholtz free energy in the canonical ensemble

    Hello everybody :D My question is: given the distribution of the canonical ensemble, how do we get the helmoltz free energy? I think we can't use A = U-TS because we don't know how to write S. So what's the solution? Thanks
  9. SalfordPhysics

    Derive Internal Energy from Thermodynamic Identity

    Homework Statement For a single molecule, derive the internal energy U = 3/2kBT In terms of the partition function Z, F = -kBTlnZ Where Z = V(aT)3/2 Homework Equations Thermodynamic identity: δF = -SδT - pδV p = kBT/V S = kB[ln(Z) + 3/2] The Attempt at a Solution U = F + TS δU = δF +...