# Why Gibbs free energy is zero at equilibrium? – Importance Of Free Energy In Living System The term Gibbs energy is quite similar in the two systems. The difference is that in the previous graph, it is just shown as the sum of the energies of the forces that interact with a gas molecule. Gibbs energy refers to a particular type of dynamic energy, one that does not depend on the state of the gas but instead on the state of the atoms and molecules. In other words, Gibbs energy is related to the rate at which the atoms and molecules in one gas changes their atomic states. The second graph shows the average kinetic energy of the particles as a function of position. There is a “squeezed” state, where the molecules move slowly but with a high rate of kinetic energy, and a “open” state, where the molecules move at the same speed but with lower kinetic energy. In general, the Gibbs energy is the rate at which molecular spins and positions change. In the previous graphs, the kinetic energy was calculated as an average of the three energy states. The other two energy states are described below. I have left all of these energy states as an exercise for the reader. There are two other ways to calculate Gibbs energy. The simplest way is by dividing the total energy of the gas (the total energy of all the molecules) by a specific density (the number of individual gas molecules). The resulting number of free energy particles (GJ) is the inverse of the density of molecules per unit volume of gas.

Density = 1/3 (Kg/L), where K is the total mass of the gas mass. (L = cubic miles.)

The other approach is to use the volume for the gas to determine the fraction of free energy to be expressed in specific heat. (This is the approach taken by Gibbs) The energy density of 1/3 of the volume of the gas, is roughly equal to 12 kilojoules per cubic centimeter. (L = 3.54 moles per cubic centimeter) To calculate the specific heat of water,

Specific Heat = V h (molecules per cubic centimeter) / (molecules per volume) where

V h = Specific heat of water = 0.12 kJ/mole (Mol/m3)

V h = Specific Heat = 1/3 + 1/3 + 1/3

V h = (K + J)/(Mol/m3)

K = mass of water (kg)

J = amount of heat

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