--%>

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 : Polymers comparison of biodegradable

    comparison of biodegradable and non-biodegradable polymers

  • Q : What is Henry law constant and its

    1. The units of Henry Law constant are same as those of pressure, i.e. torr or h bar. 2. Different gases have dissimilar values of Henry law constant. The values of KH for some gases in water are given in tabl

  • Q : Theory of three dimensional motion

    Partition function; that the translational energy of 1 mol of molecules is 3/2 RT will come as no surprise. But the calculation of this result further illustrates the use of quantized states and the partition function to obtain macroscopic properties. The partition fu

  • Q : Symmetry Elements The symmetry of the

    The symmetry of the molecules can be described in terms of electrons of symmetry and the corresponding symmetry operations.Clearly some molecules, like H2O and CH4, are symmetric. Now w

  • Q : Mole fraction of solute The mole

    The mole fraction of the solute in 1 molal aqueous solution is: (a) 0.027 (b) 0.036 (c) 0.018 (d) 0.009What is the correct answer.

  • Q : Problem on vapor-liquid equilibrium Two

    Two tanks which contain water are connected to each other through a valve. The initial conditions are as shown (at equilibrium): 683_tank question.jpg

  • Q : Finding Normality Can someone please

    Can someone please help me in getting through this problem. Concentrated H2SO4 has a density of 1.98 gm/ml and is 98% H2SO4 by weight. The normality is: (a) 2 N  (b) 19.8 N  (c) 39.6 N  (d) 98

  • 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 : 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 : Alkaline medium The amount of KMnO 4

    The amount of KMnO4 required to prepare 100 ml of 0.1N solution in alkaline medium is: (a) 1.58 gm (b) 3.16 gm (c) 0.52 gm (d) 0.31 gmAnswer: (a) In alkaline medium KMnO4 act as ox