Excluded volume per mole

We first look into the excluded-volume effect, represented by the parameter b. Since the atoms are impenetrable, no atom can get closer than a distance a from another. (Note that the location of each atom is represented by the location of the center of that atom.) This implies that each atom has a volume of (4/3)7ra3 around its center that is not accessible to the center of any other atom this volume is the excluded volume associated with each atom. (Figure 13.9 is an illustration of this situation for two colloidal particles of different radii.) The actual excluded volume per atom, b (b, the excluded volume per mole, is equal to NAb, with Na the Avogadro s number) is, however, smaller than (4/3)7texcluded volume of an atom as calculated above may overlap with that of other atoms. Therefore, to obtain an expression for b, we need to multiply the above value by N (since there are N atoms in the volume), take half of it since otherwise we will be double counting the excluded volumes, and divide by A to get excluded volume per atom, that is,... 

Van der Waals reasoned that when using an ideal-gas-like treatment, the excluded volume should be subtracted from the actual volume of the container to give the free volume in which the molecules could undergo translational motion. As the excluded volume is proportional to the number of molecules (or moles) in the container, it is written as nb, where b is the excluded volume per mole, and the free volume is V — nb. Thus, according to van der Waals, the repulsive force modifies the ideal gas equation to... 

Because b, the excluded volume per mole (the actual size of the molecules) is much less than Vm, x = b/Vm is less than 1 and we can use the Taylor expansion11... 

We know that the constant b is the volume of the rigid molecules (dm3 mob1), which is the volume unavailable for the molecules to move in, and the excluded volume per mole = 4/3 so3 because the spherical diameter, o, is the closest that one molecule can approach another. Thus the constant, b, can be written in terms of molecular diameter so that (see also Section 3.4)... 

To convert from excluded volume per molecule to excluded volume per mole, we need to multiply by Avogadro s number, Nx-... 

Consider a gas sample consisting of molecules with radius r. (a) Determine the excluded volume defined by two molecules and (b) calculate the excluded volume per mole (b) for the gas. Compare the excluded volume per mole with the volume actually occupied by a mole of the molecules. 

Strategy (a) Use the formula for volume of a sphere to calculate the volume of a sphere of radius 2r. This is the excluded volume defined by two molecules, (b) Multiply the excluded volume per molecule by Avogadro s number to determine excluded volume per mole. 

Practice Problenn A Determine the excluded volume per mole and the total volume of the molecules in a mole for a gas consisting of molecules with radius 165 pm. [Note To obtain the volume in hters, we must express the radius in decimeters (dm).]... 

Think About It The excluded volume per mole is four times the volume actually occupied by a mole of molecules. 

The van der Waals equation and the ideal gas equation both have the form pressure factor X volume factor = nRT. The van der Waals equation uses a modified pressure factor, P + rp-afV, in place of P and a modified volume factor, V — nb, in place of V. In the modified volume factor, the term nb accounts for the volume of the molecules themselves. The parameter b is called the excluded volume per mole, and, to a rough approximation, it is the volume that one mole of gas occupies when it condenses to a liquid. Because the molecules are not point masses, the volume of the container must be no smaller than nb, and the volume available for molecular motion is — nb. As suggested in Figure 6-22(b), the volume available for molecular motion is quite small at high pressures. 

The constant b is the volume excluded by 1 mol of molecules and should therefore be close to V, the volume per mole in the liquid state, where molecules are essentially in contact with each other. For example, the density of liquid nitrogen is 0.808 g cm. One mole of N2 weighs 28.0 g, so... 

When two gas moleeules (assumed to be spherical) approach each other, the distance of elosest approaeh is the sum of their radii (2r). The volume around eaeh moleeule into whieh the een-ter of another molecule caimot penetrate is called the excluded volume. The effect of the excluded volume is to limit the fraction of the eontainer volume actually available for molecules to move about in a gas sample. Thus, the volume term in the van der Waals equation, V - nb, is the container volume V minus the correction for the exeluded volume, nb, where is the munber of moles of the gas and b is the excluded voliune per mole of the gas. 

As we begin to add structural features like atoms onto crystal lattice sites, we have to pay attention to the size of the features. How efficiently can we pack atoms onto a lattice We have already discussed two measures of the size of an atom or molecule. In Section 9.7, the van der Waals parameter b was related to the volume excluded per mole of molecules, so blNp is one measure of molecular size. In... 

Thus, b isa constant corresponding to the volume excluded per mole of molecules. The potential energy describing this situation is that of an impenetrable hard sphere. 

In the case where X is H20, that is, for the water-exchange reaction, the pressure-dependence of the rate has been measured and the volume of activation found to be +1.2 ml per mole. This result definitely excludes a predominantly SN2 mechanism, but it does not agree satisfactorily with an extreme 5N1 mechanism either. It is most consistent with a transition state in which the initial Co—OH2 bond is stretched quite far while formation of a new Co—bond is only beginning to occur, that is, a predominantly dissociative mechanism. 

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