--%>

Define Virial Equation

The constant of vander Waal's equation can be related to the coefficients of the virial equation. 

Vander Waal's equation provides a good overall description of the real gas PVT behaviour. Now let us focus on the description that this equation gives to the onset of nonideal behaviour. This stage is shown most clearly on displays of Z = PV/(RT) versus P. the first deviations from the ideal gas value of Z = 1 show up as straight line sections in Z-versus-P plots. These initial stages of non ideal behaviour are described by the simple virial expressions Z = 1 + BPP or Z = 1 + BV/V. we begin, therefore, by rearranging van der Waal's equation to a form that can be compared with the virial equations.

Multiplication of van der Waal's equation, in the form and for 1 mol by V/(RT) converts this equation to 

PV/RT = V/(V - b) - a/RTV = 1/(1 - b/V) - a/RTV


We develop an equation with the form of the virial equation with volume terms by recognizing that the 1/(1 - b/V) term can be expanded by using the binomial expansion (1 - x)-1 = 1 + x + x2 + .... If only the first three terms of the series are exhibited, we develop to

54_virial equation.png 

Comparison with virial equations, shows that van der Waals' equation implies the definition

BV = b - a/RT 

Experimental values for BV for neon are plotted, notice that the temperature dependence of this second virial coefficient is generally consistent with that suggested values of BV are negative at low temperatures where the second term dominates, and these values increase and become positive at higher temperatures where the first term dominates. The curve is drawn on the basis with a and b values adjusted to give a good fit to the experimental results. Fitting second virial coefficient data provides, as this example illustrates, another way for assigning values to van der Waals' a and b parameters.

Van der Waals' excluded volume and molecular diameters

The excluded volume b, introduced by van der Waals' as an empirical correction term, can be related to the size of the gas molecules. To do so, we assume the excluded volume is the result pairwise coming together of molecules. This assumption is justified when b values are obtained from second virial coefficient data. Fitting values for the empirical constants of van der Waals' equation. 

So that we need to deal with a single molecular size parameter, we treat molecules as spherical particles. The diameter of a molecule is d. the volume of a molecule is 4/3 ∏ (d/2)3.
The volume in which a pair of molecules cannot move because of each other's presence is indicated by the lightly shaded region. The radius of this excluded volume sphere is equal to the molecular diameter d. the volume excluded to the pair of molecules is 4/3∏d3. We thus obtain
Excluded volume per molecule = ½ (4/3 ∏d3)

= 4[4/3∏(d/2)3]


The expression in brackets is the volume of a molecule. Thus the excluded volume per molecule is 4 times the actual volume of the molecule.

Van der Waals' b term is the excluded volume per mole of molecules. Thus we have, with N representing Avogadro's number


B = 4 N [4/3∏ (d/2)3] = 4 N (volume of molecule)

Molecular size and Lennard-Jones Intermolecular attraction term based on second virial coefficient data:

Gas Excluded volume b, L mol-1 Molecular diam. D, pm εLJ, J × 10-21
He 0.021 255 0.14
Ne 0.026 274 0.49
Ar 0.050 341 1.68
Kr 0.058 358 2.49
Xe 0.084 405 3.11
H2 0.031 291 0.52
N2 0.061 364 1.28
O2 0.058 358 1.59
CH4 0.069 380 1.96
C(CH3)4 0.510 739 3.22

 

   Related Questions in Chemistry

  • Q : Thermodynamics I) Sulphur dioxide (SO2)

    I) Sulphur dioxide (SO2) with a volumetric flow rate 5000cm3/s at 1 bar and 1000C is mixed with a second SO2 stream flowing at 2500cm3/s at 2 bar and 200C. The process occurs at steady state. You may assume ideal gas behaviour. For SO2 take the heat capacity at constant pressure to be CP/R = 3.267

  • Q : Product of HCl Zn Illustrate  the

    Illustrate  the product of HCl Zn?

  • Q : Calculate PH value for a acetic acid 1.

    1. A solution of 0.100 M acetic acid is prepared. a) What is its pH value? b) If 20% of the initial acetic acid is converted to the acetate form by titration with NaOH, what is the resultant pH?

  • Q : Mole fraction in vapours Choose the

    Choose the right answer from following. If two substances A and B have P0A P0B= 1:2 and have mole fraction in solution 1 : 2 then mole fraction of A in vapours: (a) 0.33 (b) 0.25 (c) 0.52 (d) 0.2

  • Q : Molar concentration of Iron chloride

    Provide solution of this question. A certain aqueous solution of FeCl3 (formula mass =162) has a density of 1.1g/ml and contains 20.0% Fecl. Molar concentration of this solution is: (a) .028 (b) 0.163 (c) 1.27 (d) 1.47

  • Q : Unit of molality Select the right

    Select the right answer of the question. The unit of molality is: (a) Mole per litre (b) Mole per kilogram (c) Per mole per litre (d) Mole litre

  • Q : Group IV Cations Chromium(III)

    Chromium(III) hydroxide is highly insoluble in distilled water but dissolves readily in either acidic or basic solution. Briefly explain why the compound can dissolve in acidic or in basic but not in neutral solution. Write appropriate equations to support your answer.

  • Q : Chem Silicon has three naturally

    Silicon has three naturally occurring isotopes. 28Si, mass = 27.976927; 29Si, mass = 28.976495; 30Si, mass = 29.973770 and 3.10% abundance. What is the abundance of 28Si?

  • Q : Problem based on molarity Select the

    Select the right answer of the question. If 18 gm of glucose (C6H12O6) is present in 1000 gm of an aqueous solution of glucose, it is said to be: (a)1 molal (b)1.1 molal (c)0.5 molal (d)0.1 molal

  • Q : Value of molar solution Select the

    Select the right answer of the question. Molar solution contains: (a)1000g of solute (b)1000g of solvent (c)1 litre of solvent (d)1 litre of solution