--%>

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 : Explain group 15 elements. The various

    The various elements

  • Q : How to test a gas to see if it was

    Write a short note to describe how to test a gas to see if it was hydrogen or not?

  • Q : Avogadro's hypothesis Law Principle

    Avogadro's hypothesis Law Principle- Berzelius, a chemist tried

  • Q : Strength of dilute acid of Sulfuric acid

    Select the right answer of the question.10ml of conc.H2SO4 (18 molar) is diluted to 1 litre. The approximate strength of dilute acid could be: (a)0.18 N (b)0.09 N (c) 0.36 N (d)1800 N

  • Q : Theory of one dimensional motion For

    For motion in one dimension, the distribution of the molecules over quantum states, speeds, and energies can be deduced.Here we show that the energy of a macroscopic gas sample can be described on the basis of our knowledge of the quantum states allowed to

  • Q : Problem on colligative properties

    Choose the right answer from following. The magnitude of colligative properties in all colloidal dispersions is : (a) Lowerthan solution (b)Higher than solution(c) Both (d) None

  • Q : Define Bond Energies - Bond Charges

    Energy changes in some chemical reactions can be used to deduce the energies of chemical bonds. Our understanding of the molecular basis of thermodynamic properties is extended when we ask why the enthalpy change for a reaction is what it is. We deduce,

  • Q : Finding strength of HCL solution Can

    Can someone please help me in getting through this problem. 1.0 gm of pure calcium carbonate was found to require 50 ml of dilute  HCL for complete reaction. The strength of the HCL  solution is given by: (a) 4 N  (b) 2 N  (c) 0.4 N  (d) 0.2 N

  • Q : Problem on decinormal Select the right

    Select the right answer of the question. How much water is required to dilute 10 ml of 10 N hydrochloric acid to make it exactly decinormal (0.1 N): (a) 990 ml (b) 1000 ml (c) 1010 ml (d) 100 ml

  • Q : Problem on making solutions The weight

    The weight of pure NaOH needed to made 250cm3 of 0.1 N solution is: (a) 4g  (b) 1g  (c) 2g  (d) 10g Choose the right answer from above.