The Helmholtz free energy is a thermodynamic potential that measures the useful work of a closed system under constant temperature and volume conditions. Helmholtz free energy is denoted as F y is defined as the difference of the internal energy OR minus product of temperature T by entropy S:
F = U - T⋅S
Since it is energy, it is measured in Joules in the International System (SI), although other appropriate units can also be ergs (CGS), calories or electron volts (eV).
The negative variation of the Helmholtz energy during a process is equated with the maximum work that the system can do in an isochoric process, that is, at constant volume. When the volume is not kept constant, some of this work can be done on the environment.
In this case we refer to work in which the volume does not vary, such as electrical work: dW = Φdq, with Φ as the electric potential and q as the electric charge.
If the temperature is also constant, the Helmholtz energy is minimized when equilibrium is reached. For all this, the Helmholtz energy is particularly useful in constant volume processes. In this case you have:
- For a spontaneous process: ΔF < 0
- When the system is in equilibrium: ΔF = 0
- In a non-spontaneous process: ΔF> 0.
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As stated at the beginning, the Helmholtz energy is defined as "the internal energy U of the system, minus the product of the absolute temperature T of the system, by the entropy S of the system":
F = U - T⋅S
It is a function of temperature T and volume V. The steps to visualize this are as follows:
- Starting from the first law of thermodynamics, the internal energy U is related to the entropy S of the system and its volume V for reversible processes through the following differential relationship:
dU = dQ - dW = TdS - PdV
From this it follows that the internal energy U is a function of the variables S Y V, Thus:
U = U (S, V)
- Now we take the definition of F and it is derived:
dF = dU - d (TS) = dU - TdS - SdT
- Substituting there the differential expression obtained for dU in the first step, it remains:
dF = TdS - PoV - TdS - SdT = -SdT - PoV
- Finally it is concluded that F is a function of the temperature T and the volume V and can be expressed as:
F = F (T, V)
Helmholtz energy can be applied as a general criterion of spontaneity in isolated systems, but first it is convenient to specify some concepts:
- A system closed can exchange energy with the environment, but cannot exchange matter.
- Instead a isolated system does not exchange matter or energy with the environment.
- Finally a open system exchange matter and energy with the environment.
In reversible processes, the variation of internal energy is calculated as follows:
dU = TdS - PdV
Now suppose a process at constant volume (isochoric), in which the second term of the previous expression has zero contribution. It should also be remembered that according to the Clausius inequality:
dS ≥ dQ / T
Such inequality applies to an isolated thermodynamic system.
So for a process (reversible or not) in which the volume remains constant, the following is true:
T dS ≥ dU (at fixed volume)
Taking into account that:
dF = dU - T dS
We will have that in an isochoric process at constant temperature it is satisfied that: dF ≤ 0, as indicated at the beginning.
So the Helmholtz energy F is a decreasing quantity in a spontaneous process as long as it is an isolated system. F reaches its minimum and stable value when reversible equilibrium has been reached.
Calculate the variation of the Helmholtz free energy F for 2 moles of ideal gas at a temperature of 300K during an isothermal expansion that takes the system from an initial volume of 20 liters to a final volume of 40 liters.
Starting from the definition of F:
F = U - T S
Then a finite variation of F, called ΔF, will be:
ΔF = ΔU - T ΔS
Since the statement states that the temperature is constant: ΔT = 0. Now, in ideal gases the internal energy only depends on their absolute temperature, but since it is an isothermal process, then ΔU = 0 Y ΔF = - T ΔS. For ideal gases, the entropy change of an isothermal process is written like this:
ΔS = n.R.ln (Vtwo/ V1)
Applying this expression:
ΔS = 2 moles x 8.314 J / (K mol) x ln (40L / 20L) = 11.53 J / K
Finally, the change in Helmholtz energy is:
ΔF = - T ΔS = - 300K x 11.53 J / K = -3457.70 J.
Inside a cylinder there is a piston that divides it into two sections and on each side of the piston there are n moles of a monatomic ideal gas, as shown in the figure below.
The cylinder walls are good conductors of heat (diathermic) and are in contact with a reservoir of temperature Tor.
The initial volume of each of the cylinder sections are V1i and V2i, while its final volumes are V1f and V2f after quasi-static displacement. The piston is moved by means of a plunger that hermetically passes through the two cylinder caps.
It is asked to find:
a) The change in the internal energy of the gas and the work done by the system and
b) The variation of the Helmholtz energy.
Since the piston moves quasi-statically, the external force applied on the piston must balance the force due to the pressure difference in the two sections of the cylinder..
The job dW performed by external force Fext during an infinitesimal shift dx it is:
dW = - Fext dx = (P1 - Ptwo) a dx = P1 dV1 + Ptwo dVtwo
Where the relationship has been used dV1 = - dVtwo = a dx, being to the plunger area. On the other hand, the variation of the Helmholtz energy is:
dF = -SdT - PdV
Since the temperature does not change during the process, then dT = 0 Y dF = - PoV. Applying this expression to each section of the cylinder we have:
dW = P1 dV1 + Ptwo dVtwo = - dF1 - dFtwo
Being F1 Y Ftwo the Helmholtz energies in each of the chambers.
The finite work W can be calculated from the finite variation of the Helmholtz energy of each chamber:
W = -ΔF1 - ΔFtwo
To find the change in Helmholtz energy, the definition is used: F = U - T S. As in each chamber there is a monatomic ideal gas at constant temperature Tor, the internal energy does not change (ΔU = 0), so that: ΔF = - Tor ΔS. What's more:
ΔS = nR ln (VF/Saw)
That when substituting finally allows the work done to be:
W = -Tor nR ln (V1f / V1i) -To nR ln (V2f / V2i) = -ΔF1 -ΔFtwo
W = - To nR ln [(V1f ⋅ V1i) / (V2f .V2i)] = - ΔFtotal
Being ΔFtotal the total change in Helmholtz energy.
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