--%>

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 normality Help me to solve

    Help me to solve this problem. 0.5 M of H2AO4 is diluted from 1 lire to 10 litre, normality of resulting solution is : (a)1 N (b) 0.1 N (c)10 N (d)11 N

  • Q : What is laser and explain its working?

    Laser action relies on a non-Boltzmann population inversion formed by the absorption of radiation and vibrational deactivation that forms a long lived excited electronic state. An excited state molecule can move to a lower energy state or return to the

  • Q : Q what is the basicity of primary

    what is the basicity of primary secondary and tertiary amines in chlorobenzene

  • Q : Diffusion Molecular View When the

    When the diffusion process is treated as the movement of particles through a solvent the diffusion coefficient can be related to the effective size of diffusing particles and the viscosity of the medium.To see how the experimental coefficients can be treat

  • Q : P block bif3 is ionic while other

    bif3 is ionic while other trihalides are covalent in nature

  • Q : Vapour pressure of volatile substance

    Provide solution of this question. According to Raoult's law the relative lowering of vapour pressure of a solution of volatile substance is equal to: (a) Mole fraction of the solvent (b) Mole fraction of the solute (c) Weight percentage of a solute (d) Weight perc

  • 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 : Question on Raoults law Give me answer

    Give me answer of this question. For a dilute solution, Raoult's law states that: (a) The lowering of vapour pressure is equal to mole fraction of solute (b) The relative lowering of vapour pressure is equal to mole fraction of solute (c) The relative lowering of v

  • 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 : Basic concepts Determination of correct

    Determination of correct mol. Mass from Roult's law is applicable to :