--%>

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 : Reducible Representations The number of

    The number of times each irreducible representation occurs in a reducible representation can be calculated.Consider the C2v point group as described or Appendix C. you can see that (1) sum of

  • Q : What are condensation polymers? Give

    These types of polymers are formed as a result of condensation reaction between monomer units. Some common examples are being discussed here: 1. Polyesters 2047_condensat</span></p>
                                        </div>
                                        <!-- /comment-box -->
                                    </li>
   
   </td>
	</tr><tr>
		<td>
       
      <li>
                                        <div class=

    Q : Problem on molality Select the right

    Select the right answer of the question. Calculate the molality of 1 litre solution of 93% H2SO4 (weight/volume). The density of the solution is 1.84 g /ml : (a) 10.43 (b) 20.36 (c) 12.05 (d) 14.05

  • Q : Question related to molarity Help me to

    Help me to go through this problem. Molarity of a solution containing 1g NaOH in 250ml of solution: (a) 0.1M (b) 1M (c) 0.01M (d) 0.001M

  • Q : Solution problem What is the correct

    What is the correct answer. To made a solution of concentration of 0.03 g/ml of AgNO3, what quantity of AgNO3 must be added in 60 ml of solution: (a) 1.8  (b) 0.8  (c) 0.18  (d) None of these

  • Q : Vapour pressure over mercury Choose the

    Choose the right answer from following. At 300 K, when a solute is added to a solvent its vapour pressure over the mercury reduces from 50 mm to 45 mm. The value of mole fraction of solute will be: (a)0.005 (b)0.010 (c)0.100 (d)0.900

  • Q : M ive me answer of this question. When

    ive me answer of this question. When mercuric iodide is added to the aqueous solution of potassium iodide, the: (a) Freezing point is raised (b) Freezing point is lowered (c) Freezing point does not change (d) Boiling point does not change

  • Q : Molarity of Sodium hydroxide Select the

    Select the right answer of the question. Molarity of 4% NaOH solution is : (a) 0.1M (b) 0.5M (c) 0.01M (d) 0.05M

  • Q : Explosions produce carbon dioxide

    Illustrate all the explosions produce carbon dioxide?

  • Q : Dipole moment Elaborate a dipole moment

    Elaborate a dipole moment?