--%>

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 : Problem on bubble point The following

    The following mixture of hydrocarbons is obtained as one stream in a petroleum refinery.

    Q : Meaning of Molar solution Molar

    Molar solution signifies 1 mole of solute present/existed in: (i) 1000g of solvent (ii) 1 litre of solvent (iii) 1 litre of solution (iv) 1000g of solution

  • Q : Problem on Osmotic Pressure of solution

    The osmotic pressure of a 5% solution of cane sugar at 150oC  is (mol. wt. of cane sugar = 342)(a) 4 atm (b) 3.4 atm (c) 5.07 atm (d) 2.45 atmAnswer: (c) Π = (5 x 0.0821 x 1000 x 423)/(342 x 100) = 5.07 atm

  • Q : Advantages of doing your own chemistry

    What are the advantages of doing your own chemistry assignments? State your comment?

  • Q : Mole fraction of water Give me answer

    Give me answer of this question. A solution contains 25%H2O 25%C2H5OH , and 50% CH3 COOH by mass. The mole fraction of H2O would be: (a) 0.25 (b) 2.5 (c) 0.503 (d) 5.03.

  • Q : Question based on mole concept Help me

    Help me to solve this Question. The number of moles of SO2Cl2 in 13.5 gm is in is : (a) 0.1 (b) 0.2 (c) 0.3 (d) 0.4

  • Q : Nuclear Magnetic Resonance The nuclear

    The nuclear states produced by a magnetic field are studied in nuclear magnetic resonance spectroscopy.The frequency of the radiation that corresponds to the nuclear magnetic energy level spacings and the weakness of the radiation absorption that must be e

  • Q : Problem on molarity-normality-molality

    Can someone please help me in getting through this problem. The solution ofAl2(SO4)3 d = 1.253gm/m comprise 22% salt by weight. The molarity, normality and molality of the solution is: (1) 0.805 M, 4.83 N, 0.825 M (2)

  • Q : Direction of dipole moment expected

    Illustrate the direction of the dipole moment expected for hydrogen bromide?

  • Q : Dipole moment of chloro-octane Describe

    Describe the dipole moment of chloro-octane in brief?