Gas, Liquid, and Solid

5. Gas, Liquid, and Solid

When matter is compressed, its molecules exist closer to each other. As the intermo- lecular distance is reduced, the attractive force between adjacent molecules becomes large enough to reduce the molecular velocity. Through this process gas molecules slow down to a state at which the matter changes phase and becomes liquid. The atoms (that are part of mole- cules) in liquids can vibrate, and molecules can rotate around each other to assume any con- figuration as shown in Figure 22 . This rotational capability of the molecules disallows their placement at particular positions, and is a characteristic of a fluid. Liquid molecules contain negligible translational energy. The sum of their rotational and vibrational energies defines their warmth or “heat”. In general, the sum of the translational, rotational, and vibrational, en- ergies for fluids are comparable to the minimum potential energy with the consequence that fluids are mobile.

As liquids are compressed, the intermolecular distance l decreases further, and the net force on the molecules (i.e., the maximum attractive potential) declines to eventually become

negligibly small. Therefore, molecules cease to move around each other with the consequence that the rotational energy tends to zero, although the vibrational energy is still finite with the vibrations occurring about a fixed position l min (cf. Figure 23 ). The cessation of rotation

“glues” the molecules to definite positions as shown in Figure 23 , and at this fixed configura- tion matter becomes solid. As solids are compressed l < l min , although individual atoms con- tained in the various molecules vibrate, the intermolecular forces are repulsive. Upon stretch- ing solids, l > l min , and the intermolecular force becomes attractive, thereby bringing the mo- lecular configuration to its original state. If the solid temperature is raised, the molecular vi-

brational energy increases. The consequent rise in the vibration amplitude tends to stretch the molecules over greater distances although l<l min . Since intermolecular attractive forces increase weakly as compared to repulsive forces, molecules can be spaced farther apart at higher tem- peratures, leading to their thermal expansion.

m. Example 13 Water is contained inside a piston-cylinder assembly. Assuming the water to be gaseous, determine both the rms and average velocities and internal energies of the molecules at 293 K and 3000 K. If 1 kmole of water is contained in a piston–cylinder–weight assembly, the volume of

which is either 1041.5 m 3 or 0.0805 m 3 , determine the average volume around each molecule. Assuming these volumes to be spheres of radius r´, determine the sphere radii and the

intermolecular spacing for the two cases. If the collision diameter ( ≈l 0 ) of water molecules is ≈2.56 Å (1 Å = 10 –10 m), deter- mine l/l max for each of the two cases. Express the answers in terms of l/l max . If it is as- sumed that 1 kmole of H 2 O behaves as an ideal gas at 1041.5 m 3 , determine the mean

free path at 293 K. Comment on the results.

Solution The Boltzmann constant,

B = R /N Avog = 8314 J K –1 kmole –1 /(6.023

×10 26 molecule kmole )

×10 –23 J molecule –1 K –1 .

The molecular mass m = M/N Avog = 18.02 kg kmole –1 /(6.023

×10 26 molecule kmole –1 )

×10 –26 kg molecule –1 .

V = (3k T/m) 1/2 = (3

V –1/2 avg = (2 π )V rms = 718.5 m s –1 . The energy per molecule,

2 ×10 –21 ×637 = 6.066 ×10 J molecule –1 , and u = 6.066

u´ = (1/2)m(V rms ) 2 = (1/2) 2.99

26 molecule kmole ×6.023×10 –1 =3654 kJ kmole ×10 –1 . At 3000 K,

–21 J molecule –1

rms = (3 ×1.38×10 × 3000/2.99 ×10 ) = 2037.4 m s , and V avg = 29 ms –1 .

u’=7.901x10 -21 J/molecule , u= 4760 kJ/kmole

For a volume of 1041.5 m 3 ,

v´ = v /N Avog = 1041.5 m 3 kmole –1 /(6.023 26 ×10 –1 molecule kmole ) = 1.73

×10 –24 m 3 molecule –1 .

r´ = (3v´/4 1/3 π) –10 = 74.46 ×10 m or 74.46 Å. l = 2r´ = 2 ×74.46 Å = 148.9 Å.

3 o ecu es For a volume of 0.0805 m ed ,

v´ = 13.4 ×10 –29 m 3 molecule –1 .

r´ = 3.172 ×10 –10 m or 3.172 Å. l = 6.34 Å.

Furthermore, since l min /l 0 =1.1225, l max /l 0 =1.2445, and

l 0 ≈σ = 2.56 Å, l max = 3.19 Å, and l/l max = 148.91/3.19 = 46.68 at v = 1041.5 m 3

kmole –1 , and l/l max = 3.85/3.19 = 1.21 at v = 0.018304 m 3 kmole –1 .

Finally l mean =1/(n’ 2 πσ 26 }, n’ =6.023x10 /1041.5

= 5.783 x10 23 ,l mean =1/(5.783x10 23 Figure 23. Illustration of ice * π *(2.53x10 -10

molecules that exist in a fixed

configuration with respect to =8.6x10 -06 m or 86000 Å!

each other.

10N / cm

Remarks Attractive forces are negligible for the larger

volume (l/l dx max = 46.68) and, hence, the water

molecules behave as in an ideal gas. However, 10N / cm upon compression to the smaller volume l/l 2

1.34, and attractive forces become strong

(a) (b)

enough for the water to exist as either liquid or

a solid (ice). Figure 24. Schematic illustration If velocity of a hypothetical ideal gas tends to

of work being done. zero, so does u´, and, consequently, u=0. In re-

ality, as the molecular momentum becomes negligibly small, matter is drawn together due to the intermolecular attractive forces, thereby condensing it into a liquid or a solid. The molecules are located farther apart in the larger volume at a specified temperature or average molecular velocity. Consequently, the number of molecules per unit vol- ume is lower than in the smaller volume, resulting in a lower pressure. Using the ideal gas law, the pressure exerted by the larger volume at T = 293 K is 0.023 bar. How- ever, this law cannot be applied once the molecules are relatively closely spaced as in

a liquid or solid, and cannot be used to predict the pressure under these conditions, since attractive forces are not considered in its development. In Chapter 6 we will dis- cuss real gas equations of state which consider the effect of attractive forces on pres- sure.