--%>

Non-ideal Gases Fugacity

The fugacity is a pressure like quantity that is used to treat the free energy of nonideal gases.

Now we begin the steps that allow us to relate free energy changes to the equilibrium constant of real, nonideal gases. The thermodynamic reaction (∂G/∂P) t = V is used with the ideal gas relation PV = RT, or V = RT/P, to obtain G = G° = R in P. it was this equation that led to the familiar equilibrium constant expression. If the ideal gas relation PV = RT is not satisfactory, some other quality equations, that of van der Waals, for example, could be used to express the pressure dependence of V. if that were done, the integration of (∂G/∂P)T = V would produce an awkward expression for the equilibrium constant. Thus a route that preserves the simple form of the equilibrium constant expression is preferable.

A satisfactory procedure is the introduction of a function called the fugacy ƒ.  This procedure insists on the free energy equation having the convenient form of the nonideal complications are hidden in the fugacy term. A number of manipulations are necessary; we begin with the thermodynamic equation for mol 1 of gas at constant temperature.

G2 - G1 = V dP

The quantity RT/P can be added to and subtracted from the integrand to give

G2 - G1 = [RT/P + (V - RT/P0] dP

= RT/P dP = (V - RT/P dP

= RTY in P2/P1 + (V - RT/P) dP


Thus the ratio f/P can be calculated at any temperature for which viral coefficient data are available and for any pressure in the range in which these data are applicable. If the real gas behavior is expressed by any other equation of state, the integration can be carried out graphically or with the help of a computer.

Fugacity and the law of corresponding states: for gases for which molar volume measurements have not been made and an equation of state is not available, the law of corresponding states can be used to estimate the fugacities at various reduced variables PR, VR and TR all gases follow the same imperfection and therefore the same nonideality. Furthermore, the variation of the compressibility factor Z with the reduced pressure has been represented for various values TR. These data are all that is necessary for the integration values of:

Z = PV/RT

From which we obtain:

V = RT/P × Z

With this relation eq. can be written as:

RT In ƒ/P = ∫PO (RT/P × Z - RT/P) dP = RT  ∫PO (Z - 1) dP/P

Or, In ƒ/P = 
 ∫PO (Z - 1) dP/P =  ∫PO (Z - 1) d PR/PR

The data of Z as a function of PR for a given value of TR then allow graphical integrations to be performed to give curves.

Example: estimate the fugacity of methane at 200 bar and 25°C, but use the correlation that is based on the law of corresponding states. The critical data give = 46.3 bar and T = 190.6 K for methane.

Solution: at 200 bar the reduced pressure is 200 bar/46.3 bar = 4.32. At 25°C the reduced temperature is 298.15/190.6 K = 1.56. From the value of ƒ/P is estimated at about 0.8, given ƒ = 160 bar.

   Related Questions in Chemistry

  • Q : What do you mean by the term medicine

    What do you mean by the term medicine dropper? Explain briefly?

  • Q : Composition of the vapour Choose the

    Choose the right answer from following. An ideal solution was obtained by mixing methanol and ethanol. If the partial vapour pressure of methanol and ethanol are 2.619KPa and 4.556KPa respectively, the composition of the vapour (in terms of mole fraction) will be: (

  • Q : Vapour pressure of a liquid Help me to

    Help me to go through this problem. The vapour pressure of a liquid depends on: (a) Temperature but not on volume (b) Volume but not on temperature (c) Temperature and volume (d) Neither on temperature nor on volume

  • Q : Finding Molarity of final mixture Can

    Can someone help me in finding out the right answer. 25ml of 3.0 MHNO3 are mixed with 75ml of 4.0 MHNO3. If the volumes are adding up the molarnity of the final mixture would be: (a) 3.25M (b) 4.0M (c) 3.75M (d) 3.50M

  • Q : Problem on Molar solution Can someone

    Can someone please help me in getting through this problem. 2.0 molar solution is acquired, when 0.5 mole solute is dissolved in: (i) 250 ml solvent (ii) 250 g solvent (iii) 250 ml solution (iv) 1000 ml solvent

  • Q : Molar mass what is the equation for

    what is the equation for calculating molar mass of non volatile solute

  • Q : Problem based on lowering in vapour

    Help me to solve this problem. An aqueous solution of glucose was prepared by dissolving 18 g of glucose in 90 g of water. The relative lowering in vapour pressure is: (a) 0.02 (b)1 (c) 20 (d)180

  • Q : Calculation of concentration of the

    Choose the right answer from following. 200ml of a solution contains 5.85 dissolved sodium chloride. The concentration of the solution will be(Na= 23: cl = 35.5 ) (a) 1 molar (b) 2 molar (c) 0.5 molar (d) 0.25 molar

  • Q : Determining maximum Osmotic pressure

    Which of the following would have the maximum osmotic pressure (assume that all salts are 90% dissociated): (a) Decimolar aluminium sulphate (b) Decimolar barium chloride (c) Decimolar sodium sulphate (d) A solution obtained by mix

  • Q : Problem on convection coefficient An

    An experiment to determine the convection coefficient associated with airflow over the surface of a thick stainless steel casting involves insertion of thermocouples in the casting at distances of 10 mm and 20 mm from the surface.  When the experiment was perform