--%>

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 : Mole fraction in vapours Choose the

    Choose the right answer from following. If two substances A and B have P0A P0B= 1:2 and have mole fraction in solution 1 : 2 then mole fraction of A in vapours: (a) 0.33 (b) 0.25 (c) 0.52 (d) 0.2

  • Q : Dissolving Group IV Carbonate Explain

    Explain how dissolving the Group IV carbonate precipitate with 6M CH3COOH, followed by the addition of extra acetic acid.

  • Q : Question associated to vapour pressure

    Choose the right answer from following. The vapour pressure lowering caused by the addition of 100 g of sucrose(molecular mass = 342) to 1000 g of water if the vapour pressure of pure water at 25degree C is 23.8 mm Hg: (a)1.25 mm Hg (b) 0.125 mm Hg (c) 1.15 mm H

  • Q : Problem on equilibrium constant Ethanol

    Ethanol is manufactured from carbon monoxide and hydrogen at 600 K and 20 bars according to the reaction2 C0(g) + 4 H2(g) ↔ C2H5OH(g) + H2O (g)The feed stream contains 60 mol% H2, 20 m

  • Q : Question based on normality Provide

    Provide solution of this question. A 5 molar solution of H2SO4 is diluted from 1 litre to 10 litres. What is the normality of the solution : (a) 0.25 N (b) 1 N (c) 2 N (d) 7 N

  • Q : M ive me answer of this question. When

    ive me answer of this question. When mercuric iodide is added to the aqueous solution of potassium iodide, the: (a) Freezing point is raised (b) Freezing point is lowered (c) Freezing point does not change (d) Boiling point does not change

  • 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 : Volume of solution containing solute

    What volume of solution contains 0.1 mole of the solute: (a) 100ml (b) 125ml  (c) 500ml (d) 62.5ml Choose the right answer from above.

  • 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 : Explain various chemicals associated

    During processing of food, several chemicals are added to it to augment its shelf life and to make it more attractive as well. Main types of food addi