--%>

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 : Question on colligative property Choose

    Choose the right answer from following. Which of the following is a colligative property: (a) Osmotic pressure (b) Boiling point (c) Vapour pressure (d) Freezing point

  • Q : Organic structure of cetearyl alcohol

    Show the organic structure of cetearyl alcohol and state what the organic family is? Briefly state it.

  • Q : Problem on Neutralization What weight

    What weight of hydrated oxalic acid should be added for complete neutralisation of 100 ml of 0.2N - NaOH solution? (a) 0.45 g  (b)0.90 g  (c) 1.08 g  (d) 1.26 g      Answer

  • Q : Modes of concentration Which of the

    Which of the given modes of expressing concentration is fully independent of temperature: (1) Molarity (2) Molality (3) Formality (4) Normality Choose the right answer from above.

  • Q : Problem on normality Help me to solve

    Help me to solve this problem. 0.5 M of H2AO4 is diluted from 1 lire to 10 litre, normality of resulting solution is : (a)1 N (b) 0.1 N (c)10 N (d)11 N

  • Q : Explain the mechanism of Enzyme

    A mechanism for enzyme-catalyzed reactions that leads to the typical rate equation for these reactions can be described.A variety of rate equations are required to portray the rates of enzymes catalyzed reagents and physical conditions that are encountered

  • Q : Problem on vapour pressure Choose the

    Choose the right answer from following. If P and P are the vapour pressure of a solvent and its solution respectively N1 and N2 and are the mole fractions of the solvent and solute respectively, then correct relation is: (a) P= PoN1 (b) P= Po N2 (c)P0= N2 (d)

  • Q : Osmotic Pressure The O.P. (Osmotic

    The O.P. (Osmotic Pressure) of equimolar solution of Urea, BaCl2 and AlCl3, will be in the order:(a) AlCl3 > BaCl2 > Urea  (b) BaCl2 > AlCl3 > Urea  (c) Urea > BaCl2<

  • Q : Explosions produce carbon dioxide

    Illustrate all the explosions produce carbon dioxide?

  • Q : Problem on decinormal strength Can

    Can someone please help me in getting through this problem. How many grams of dibasic acid (having mol. wt. 200) must be present in 100ml  of its aqueous solution to provide decinormal strength: (i) 1g  (ii)2g  (iii) 10g  (iv) 20g<