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504 14MOLECULAR INTERACTIONS
attraction of the form −C6/R6 and performing the integration gives
U = 1
2 N
2∫
∞
R=d
∫
π
θ=0
∫
2π
ϕ=0
−C6
R6
R2 sin θ dR dθ dϕ
= 1
2 N
2∫
∞
d
−C6
R4
dR × ∫
π
0
sin θ dθ × ∫
2π
0
dϕ
= 1
2 N
2 × [ C6
3R3
]
∞
d
× [− cos θ]π
0 × [ϕ]2π
0 = 1
2 N
2 × (− C6
3d3
) × 2 × 2π
= − 23πN 2 C6
d3
�e number densityN is related to themass density ρ by noting that in volume
V there areNV molecules and henceNV/NAmoles.�emass of this amount
is M ×NV/NA, where M is the molar mass. �erefore the mass density ρ is
ρ = M ×NV/NAV = MN /NA; rearranging this gives N = NAρ/M. Using
this expression forN in the expression for U gives
U = − 23π (NAρ
M
)
2 C6
d3
= − 23π
N2A
d3M2
ρ2C6
14C Liquids
Answers to discussion questions
D14C.2 �is is discussed in Section 14C.4 on page 611.
Solutions to exercises
E14C.1(b) �e vapour pressure of a liquid when it is dispersed as spherical droplets of
radius r is given by the Kelvin equation [14C.15–611], p = p∗e2γVm(l)/rRT , where
p∗ is the vapour pressure of bulk liquid to which no additional pressure has
been applied. Because the mass density of a substance with molar volume
Vm and molar mass M is given by ρ = M/Vm, it follows that Vm = M/ρ.
Substituting this into the Kelvin equation gives p = p∗e2γ(M/ρ)/rRT .
�e surface tension γ of water at 35 ○C is not given in Table 14C.1 on page 605,
so the value at this temperature is estimated from the graph in Fig. 14C.7 on
page 607 as being approximately 71 mNm−1, which is equal to 71×10−3 Jm−2.
Hence, taking p∗ as 5.623 kPa andM as 18.0158 gmol−1(18.0158×10−3 kgmol−1),
p = p∗ exp(2γM/ρ
rRT
) = (5.623 kPa)
× exp(2×(71 × 10
−3 Jm−2)×(18.0158 × 10−3 kgmol−1)/(994.0 kgm−3)
(20 × 10−9 m)×(8.3145 JK−1mol−1)×([35 + 273.15] K)
)
= 5.9 kPa