
At equilibrium the change in Helmholtz free energy is:
(A) equal to zero
(B) less than zero
(C) greater than zero
(D) both A and C
Answer
518.7k+ views
Hint: One of the given equations mentioned above will be your answer. As it is about Helmholtz free energy, we should first see what we really mean by this.
Complete answer:
Let us understand the basic concept of Helmholtz free energy first before observing the variations based on the conditional changes.
Helmholtz free energy-
It is a thermodynamic potential that measures the useful work obtainable from a closed thermodynamic system at a constant temperature and volume (i.e. isothermal and isochoric).
The Helmholtz free energy is useful for systems with constant volume. This is because when we hold volume constant, the negative of the change in Helmholtz free energy during the process is the maximum amount of work that the system can perform.
Also, at constant temperature the Helmholtz free energy is minimised at equilibrium. This is factually defined as,
F = U-TS
where,
F = Helmholtz free energy; (sometimes denoted as A)
U = internal energy of the system
T = absolute temperature of the surroundings
S = entropy of the system.
Now,
Starting with laws of thermodynamics,
$dU=\partial Q+\partial W,F=U-TS$ and $\partial Q\le TdS$
where,
$\partial W$ = work done by external forces on the system
$\partial Q$ = heat transferred from the environment to the system
Thus, for any process we have,
$d\left( F+TS \right)=dF+TdS+SdT\le TdS+\partial W$
1. If the process is reversible-
$dF+SdT\le \partial W$
2. If process is irreversible; such that the external work done is zero and also the process is isothermal-
$dF\le 0$
And most of the processes are irreversible in nature.
As we can see, the change in Helmholtz free energy is going less than zero at equilibrium when we consider the concept of constant temperature.
Thus, at equilibrium the change in Helmholtz free energy is less than zero.
Therefore, option (B) is correct.
Note:
Do note that in contrast to Helmholtz free energy, Gibbs free energy (free enthalpy) is commonly used as a measure of thermodynamic potential when it’s concerned with constant pressure and temperature.
Complete answer:
Let us understand the basic concept of Helmholtz free energy first before observing the variations based on the conditional changes.
Helmholtz free energy-
It is a thermodynamic potential that measures the useful work obtainable from a closed thermodynamic system at a constant temperature and volume (i.e. isothermal and isochoric).
The Helmholtz free energy is useful for systems with constant volume. This is because when we hold volume constant, the negative of the change in Helmholtz free energy during the process is the maximum amount of work that the system can perform.
Also, at constant temperature the Helmholtz free energy is minimised at equilibrium. This is factually defined as,
F = U-TS
where,
F = Helmholtz free energy; (sometimes denoted as A)
U = internal energy of the system
T = absolute temperature of the surroundings
S = entropy of the system.
Now,
Starting with laws of thermodynamics,
$dU=\partial Q+\partial W,F=U-TS$ and $\partial Q\le TdS$
where,
$\partial W$ = work done by external forces on the system
$\partial Q$ = heat transferred from the environment to the system
Thus, for any process we have,
$d\left( F+TS \right)=dF+TdS+SdT\le TdS+\partial W$
1. If the process is reversible-
$dF+SdT\le \partial W$
2. If process is irreversible; such that the external work done is zero and also the process is isothermal-
$dF\le 0$
And most of the processes are irreversible in nature.
As we can see, the change in Helmholtz free energy is going less than zero at equilibrium when we consider the concept of constant temperature.
Thus, at equilibrium the change in Helmholtz free energy is less than zero.
Therefore, option (B) is correct.
Note:
Do note that in contrast to Helmholtz free energy, Gibbs free energy (free enthalpy) is commonly used as a measure of thermodynamic potential when it’s concerned with constant pressure and temperature.
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