--%>

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 - G= ∫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 : Problem on vapor-liquid equilibrium Two

    Two tanks which contain water are connected to each other through a valve. The initial conditions are as shown (at equilibrium): 683_tank question.jpg

  • Q : Molecular crystals Among the below

    Among the below shown which crystal will be soft and have low melting point: (a) Covalent  (b) Ionic  (c) Metallic  (d) MolecularAnswer: (d) Molecular crystals are soft and have low melting point.

  • Q : Group Cations Explain how dissolving

    Explain how dissolving the Group IV carbonate precipitate with 6M CH3COOH, followed by the addition of extra acetic acid, establishes a buffer with a pH of approximately

  • Q : Explain Phase Rule The relation between

    The relation between the number of phases, components and the degrees of freedom is known as the phase rule. One constituent systems: the identification of an area on a P-versus-T with one phase of a component system illustrates the two degrees of freedom that

  • Q : Problem based on molecular weight

    Select the right answer of the question. Molecular weight of urea is 60. A solution of urea containing 6g urea in one litre is : (a)1 molar (b)1.5 molar (c) 0.1 molar (d) 0.01 molar

  • Q : Atmospheric pressure Give me answer of

    Give me answer of this question. The atmospheric pressure is sum of the: (a) Pressure of the biomolecules (b) Vapour pressure of atmospheric constituents (c) Vapour pressure of chemicals and vapour pressure of volatile (d) Pressure created on to atmospheric molecules

  • Q : Problem on thermodynamic equilibrium In

    In the manufacture of sulphuric acid by the contact process, S02 is oxidized to SO3 over a vanadium catalyst: The reactor is adiabatic and operates at atmospheric pressure. The gases enter the reactor at 410&d

  • Q : What are emulsions?Describe its

    Emulsions are colloidal solutions in which disperse phase as well as dispersion medium is both liquids. Emulsions can be broadly classified into two types: (i) Oil in water (O/W type) emulsions: in this type of emulsions, oil acts disperse phase and water acts

  • Q : Depression in the freezing point When

    When 0.01 mole of sugar is dissolved in 100g of a solvent, the depression in freezing point is 0.40o. When 0.03 mole of glucose is dissolved in 50g of the same solvent, depression in the freezing point will be:(a) 0.60o  (b) 0.80o

  • Q : Normality of acetic acid Give me answer

    Give me answer of this question. The normality of 10% (weight/volume) acetic acid is: (a)1 N (b)10 N (c)1.7 N (d) 0.83 N