--%>

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 : Vapour pressure of benzene Give me

    Give me answer of this question. The vapour pressure of benzene at a certain temperature is 640mm of Hg. A non-volatile and non-electrolyte solid weighing 2.175g is added to 39.08g of benzene. The vapour pressure of the solution is 600,mm of Hg . What is the mo

  • Q : What are the chemical properties of

    Haloalkanes are extremely reactive category of aliphatic compounds. Their reactivity is due to the presence of polar carbon-halogen bond in their mole

  • Q : Procedure to judge that organic

    Describe briefly the procedure to judge that the given organic compound is pure or not?

  • Q : Vander Waals forces Wax is an example

    Wax is an example of: (a) Ionic crystal  (b) Covalent crystal  (c) Metallic crystal  (d) Molecular crystalAnswer: (d) Iodine crystals are molecular crystals, in which constituent particles are molecules having inter particle

  • Q : Dipole attractions for london dispersion

    Illustrate how are dipole attractions London dispersion forces and hydrogen bonding similar?

  • Q : Degree of dissociation The degree of

    The degree of dissociation of Ca(No3)2 in a dilute aqueous solution containing 14g of the salt per 200g of water 100oc is 70 percent. If the vapor pressure of water at 100oc is 760 cm. Calculate the vapor pr

  • Q : P block why BiF3 is ionic whereas other

    why BiF3 is ionic whereas other trihalides are covalent in nature?

  • 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 : Dipole moment of chloro-octane Describe

    Describe the dipole moment of chloro-octane in brief?

  • Q : Explain methods for industrial

    The important methods for the preparation of alcohol on large-scale are given below: