--%>

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 : PH of an Alkyl Halide Briefly state the

    Briefly state the pH of an Alkyl Halide?

  • 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 : Molarity 20mol of hcl solution requires

    20mol of hcl solution requires 19.85ml of 0.01 M NAOH solution for complete neutralisation. the molarity of hcl solution

  • Q : Describe chemical properties of amines.

    Like ammonia, primary, secondary and tertiary amines have a single pair of electrons on N atom. Hence chemical behavior of amines is similar to ammonia. Amines are basic in nature, and in most of the reactions they act as nucleophiles.      1. Reaction wi

  • Q : What are emulsions?Describe its

    Emulsions are colloidal solutions in which disperse phase as well as dispersion medium is both liquids. Emulsions can be broadly classified into two types: (i) Oil in water (O/W type) emulsions: in this type of emulsions, oil acts disperse phase and water acts

  • 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 : What are lattices and unit cells? The

    The repeating, atomic level structure of a crystal can be represented by a lattice and by the repeating unit of the lattice, the unit cell.It was apparent very early in the study of crystals that the shapes of crystals stem from an ordered array of smaller

  • Q : Problem on Adiabatic expansion

    Calculate the change in entropy for the system for each of the following cases. Explain the sign that you obtain by a physical argument a) A gas undergoes a reversible, adiabatic expansion from an initial state at 500 K, 1 MPa, and

  • Q : Problems related to entropy change A)

    A) Two compartments each of 1 m3 capacity are joined by a valve and insulated from the surroundings and from one another. One compartment has saturated steam at 683.6 kPa and the other contains steam at the same temperature but at a pressure of 101.3 kPa. T

  • Q : What is Spectroscopy? This is a very

    This is a very important aspect of Physical Chemistry in which knowledge of the size, shape, rigidity and electronic structure of molecules deduced from the experimental methods treated here goes hand in hand with the theoretical approaches of chemical reactions. Spec