Concise Physical Chemistry

c02 JWBS043-Rogers September 13, 2010 11:23 Printer Name: Yet to Come

PROBLEMS AND EXERCISES 29

Solution 2.1 Expanding the van der Waals equation and collecting terms, we get

(

p+

a

V^2

)

(V−b)=RT

pV−pb+

a

V

−

ba

V^2

=RT

pV^3 −pbV^2 +aV−ba=RT V^2

pV^3 −(pb+RT)V^2 +aV−ba= 0

Comment: Nonideal Behavior—Another View We know that molecules are not

dimensionless points. Figure 2.8 shows that Boyle’s law, the lower curve, is not

obeyed above about 200 bars pressure for nitrogen. Positive deviation from Boyle’s

law is a high-pressure phenomenon shown by all real gases because all real molecules

occupy a nonzero volume. The real sample volume is larger than it “ought to be” on

the basis of Boyle’s law. Molecules, which may be thought of roughly as hard spheres,

are bumping into each other and refusing to invade each other’s space. The volume

in which the particles can move is the total volume minus the volume actually taken

up by the particles. As higher pressures are imposed, the particle volume becomes a

larger proportion of the total volume; hence the deviation from the Boyle’s law curve

is larger. The aggregate of molecular volume, as distinct from total volume, for all

molecules in the sample is theexcluded volume.

V (dm^3 )

0.0 0.1 0.2 0.3 0.4 0.5

p

(bar

)

0

100

200

300

400

500

600

T = 273.15 K

FIGURE 2.8 Boyle’s law plot for an ideal gas (lower curve) and for nitrogen (upper curve).