--%>

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 : Units of Measurement Unit of

    Unit of measurement- These are also some systems for units:      (1) C.G.S.

  • Q : Calculating total vapour pressure

    Select the right answer of the question. The vapour pressure of two liquids P and Q are 80 and 600 torr, respectively. The total vapour pressure of solution obtained by mixing 3 mole of P and 2 mole of Q would be: (a) 140 torr (b) 20 torr (c) 68 torr (d) 72 torr

  • Q : Molecular mass from Raoults law Provide

    Provide solution of this question. Determination of correct molecular mass from Raoult's law is applicable to: (a) An electrolyte in solution (b) A non-electrolyte in a dilute solution (c) A non-electrolyte in a concentrated solution (d) An electrolyte in a liquid so

  • 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 : Molecular Structure type The ionic

    The ionic radii of Rb+ and I- respectively are 1.46 Å and 2.16Å. The very most probable type of structure exhibited by it is: (a) CsCl type  (b) ZnS type  (c) Nacl type  (d) CaF2 type

    Q : Mole fraction of solute The mole

    The mole fraction of the solute in 1 molal aqueous solution is: (a) 0.027 (b) 0.036 (c) 0.018 (d) 0.009What is the correct answer.

  • Q : Formula of diesel Write a short note on

    Write a short note on the formula of diesel, petrol and also CNG?

  • Q : Help 1) Chromium(III) hydroxide is

    1) Chromium(III) hydroxide is highly insoluble in distilled water but dissolves readily in either acidic or basic solution. Briefly explain why the compound can dissolve in acidic or in basic but not in neutral solution. Write appropriate equations to support your answer. 2) Explain how dissolving t

  • Q : Molecular energies and speeds The

    The average translational kinetic energies and speeds of the molecules of a gas can be calculated.The result that the kinetic energy of 1 mol of the molecules of a gas is equal to 3/2 RT can be used to obtain numerical values for the

  • Q : What is Distillation Separation by

    Separation by distillation can be described with a boiling point diagram. The important process of distillation can now be investigated. From the boiling point diagram one can see that if a small amount of vapour were removed from a liquid of composit