--%>

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 : Water under pressure problem-henry law

    Can someone help me in going through this problem. The statement “When 0.003 moles of a gas are dissolved in 900 gm of water under a pressure of 1 atm, 0.006 moles will be dissolved under the pressure of 2 atm", signfies: (a)

  • Q : Hydrocarbons list and identify

    list and identify differences between the major classes of hydrocarbons

  • Q : Preparation of ammonium sulphate Select

    Select the right answer of the question. Essential quantity of ammonium sulphate taken for preparation of 1 molar solution in 2 litres is: (a)132gm (b)264gm (c) 198gm (d) 212gm

  • Q : Determining of normality of sodium

    Can someone please help me in getting through this problem. The normality of a solution of sodium hydroxide 100 ml of which includes 4 grams of NaOH is: (a) 0.1 (b) 40 (c) 1.0 (d) 0.4

  • Q : Einsteins mass energy relation In

    In Einstein’s mass energy relation e = mc2 for what is c employed or why is light needed for the reactions. As the reactions are with the help of neutrons?

  • Q : BASIC CHARACTER OF AMINES IN GAS PHASE,

    IN GAS PHASE, BASICITIES OF THE AMINES IS JUST OPPOSITE TO BASICITY OF AMINES IN AQEUOUS PHASE .. EXPLAIN

  • Q : Colligative property related question

    Select the right answer of the question. Which of the following is not a colligative property : (a) Osmotic pressure (b) Elevation in B.P (c) Vapour pressure (d) Depression in freezing point

  • Q : What are aliphatic amines and its

    In common system, the aliphatic amines are named by using prefix for alkyl group followed by the word amine.In case of mixed amines, the name of alkyl groups are arranged in alphabetical order. This is followed by the word amine. However, for simple secondary or tertiary amines anothe

  • Q : Molar mass of compound The freezing

    The freezing point of a solution having 4.8 g of a compound in 60 g of benzene is 4.48. Determine the molar mass of the compound (Kf = 5.1 Km-1) , (freezing point of  benzene = 5.5oC)          &n

  • Q : Colligative effect Choose the right

    Choose the right answer from following. Which one of the statements written below concerning properties of solutions, explain a colligative effect: (a) Boiling point of pure water decreases by the addition of ethano (b) Vapour pressure of pure water d