--%>

Relationship between free energy and pressure

The free energy of a gas depends on the pressure that confines the gas.


The standard free energies of formation, like those allow predictions to be made of the possibility of a reaction at 25°C for each reagent at 1-bar pressure. For these free-energy data to be of more general use, a means must be available for calculating free energies at other pressures and temperatures.

To start, we form a complete and detailed description for changes in free energy. From the defining equations G = H - TS and H = U + PV we obtain 

dG = dU + P dV + V dP - T dS - S dT

This expression has redundancies in it and can be simplified. The state of the system is determined when the temperature and the pressure, or one of these and one of the properties of the system, are fixed. Changes in any two of these variables determined the change in the state of the system. It follows that the change in any property of the system can be expressed in terms of changes in any two of these variables.

First, we deal with an "ordinary" process in which no mechanical energy other than P dV energy is evolved. In this case P dV = dUmech. Second, we imagine that the states of the system that we are considering can be connected by a reversible process. For such a process dS + dStherm = dS + dUtherm/T = 0, or T dS = -dUtherm. With these stipulation becomes,

dG = dU + dUmech + V dP + dUtherm - S dT

the first law sets the combination of the three U terms to zero, and we have

dG = V dP - S dT

we have arrived at an expression for changes in the free energy in the terms of changes in just two state-determining variables.

Now think of the free energy G as being a property of the system and, therefore, dependent on the state of the system. If this state is specified by  the temperature and the pressure, we can write the general total differential

dG = (∂G/∂P)T dp + (∂G/∂T)P dT

Comparison with equation lets us make the identifications

(∂G/∂P)T = V


And 

(∂G/∂P)P = -S


These results show how the free energy property changes when, separately, the pressure or the temperature is changed.

Notice that we arrived at these results by considering a special type of process. But since G is a property of the system, it will change by a certain amount when the pressure or temperature is changed, for any type of process.

We deal with the dependence of free energy on temperature and now we follow up on the expression obtained for the pressure dependence.

Liquids and solids have small molar volumes compared with gases. For many purposes the pressure dependence of the free energy of liquids and solids can be neglected.

For gases the dependence of free energy on pressure is appreciable and important. For an ideal gas, P and V are related by the ideal gas law, and the integration can be performed to give the free-energy change when the pressure is changed from P1 to P2 at constant temperature. Thus

G2 - G1 = ∫V dP = nRT ∫P2P1 dP/P = nRT In P2/P1

Of particular interest is the extent to which the free energy changes from its standard state value when the pressure changes from 1 bar. If state 1 is the standard state, then

P1 = 1 bar and G1 = G° 

P2 = P bar and G2 = G

With this notation for states 1 and 2 it can be we written for 1 mol as

G - G° = RT In P/1 bar

Or G = G° + RT In P [T const, P in bar, and 1 mol of an ideal gas]    

   Related Questions in Chemistry

  • Q : How to calculate solutions molar

    The contribution of an electrolyte, or an ion electrolyte, is reported as the molar of a conductance. The definition of the molar conductance is based on the following conductivity cell in which the electrodes are 1 m apart and of sufficient area that th

  • Q : Einsteins mass energy relation In

    In Einstein’s mass energy relation e = mc2 for what is c employed or why is light needed for the reactions. As the reactions are with the help of neutrons?

  • Q : HCl is polar or non-polar Can you

    Can you please illustrate that HCl is polar or non-polar? Briefly illustrate it.

  • Q : Problem related to molarity Provide

    Provide solution of this question. Increasing the temperature of an aqueous solution will cause: (a) Decrease in molality (b) Decrease in molarity (c) Decrease in mole fraction (d) Decrease in % w/w

  • Q : Molarity of Barium hydroxide 25 ml of a

    25 ml of a solution of barium hydroxide on titration with 0.1 molar solution of the hydrochloric acid provide a litre value of 35 ml. The molarity of barium hydroxide solution will be: (i) 0.07 (ii) 0.14 (iii) 0.28 (iv) 0.35

  • Q : Decanormal and decinormal solution

    Provide solution of this question.10N/and 1/10N solution is called: (a) Decinormal and decanormal solution (b) Normal and decinormal solution (c) Normal and decanormal solution (d) Decanormal and decinormal solution

  • Q : Problem on moles of solution The number

    The number of moles of a solute in its solution is 20 and total no. of moles are 80. The mole fraction of solute wil be: (a) 2.5 (b) 0.25 (c) 1 (d) 0.75

  • Q : What is electrolysis? Explain with

    Passage of a current through a solution can produce an electrolysis reaction.Much additional information on the properties of the ions in an aqueous solution can be obtained from studies of the passage of a direct current (dc) through a cell containing a s

  • Q : Question based on relative lowering of

    Give me answer of this question. When a non-volatile solute is dissolved in a solvent, the relative lowering of vapour pressure is equal to: (a) Mole fraction of solute (b) Mole fraction of solvent (c) Concentration of the solute in grams per litre

  • Q : Some basic concepts of chemistry an

    an atom of an element is 10.1 times heavier than the mass of a carbon atom.What is its mass in amu?