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 : Importance of organic chemistry

    Describe the importance of organic chemistry?

  • Q : What are electromotive force in

    The main objective of this particular aspect of Physical Chemistry is to examine the relation between free energies and the mechanical energy of electromotive force of electrochemical cells. The ionic components of aqueous solutions can be treated on the basis of the

  • Q : Question based on vapour pressure and

    Give me answer of this question. The vapour pressure of water at 20degreeC is 17.54 mm. When 20g of a non-ionic, substance is dissolved in 100g of water, the vapour pressure is lowered by 0.30 mm. What is the molecular weight of the substances: (a) 210.2 (b) 206.88

  • 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 : Problem based on normality Choose the

    Choose the right answer from following. NaClO solution reacts with H2SO3 as,. NaClO + H2SO3→NaCl+ H2SO4. A solution of NaClO utilized in the above reaction contained 15g of NaClO per litre. The

  • Q : Product of HCl Zn Illustrate  the

    Illustrate  the product of HCl Zn?

  • Q : Adiabatic compression A lean natural

    A lean natural gas is available at 18oC and 65 bars and must be compressed for economical pipeline transportation. The gas is first adiabatically compressed to 200 bars and then isobarically (i.e. at constant pressure) cooled to 25°C. The gas, which is

  • Q : Hydroxide is highly insoluble in

     : 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

  • Q : Formula of diesel Write a short note on

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

  • Q : Vapour pressure related question Help

    Help me to solve this question. Which of the following is incorrect: (a) Relative lowering of vapour pressure is independent (b)The vapour pressure is a colligative property (c)Vapour pressure of a solution is lower than the vapour pressure of the solvent (d)The

©TutorsGlobe All rights reserved 2022-2023.