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SOLUTIONSMANUAL TO ACCOMPANY ATKINS' PHYSICAL CHEMISTRY 543
0.0024 0.0026 0.0028 0.0030 0.0032
−2
−1
0
1
(1/T)/K−1
ln
(G
/S
)
Figure 15.9
P15E.6 It is assumed that the volume of the unit cell does not change on substitution
of Ca for Y. A tetragonal unit cell has volume V = a2c, where a and c are the
sides of the unit cell. �e mass density ρ is given by ρ = m/V , where m is the
total mass of the unit cell.�erefore ρ = NM/(NAV) where N is the number
of formula units per unit cell and M is the molar mass of a formula unit. It
follows thatM = ρNAV/N .
�ere are two formula units in each unit cell so N = 2 and
M = 2(200.59) + 2(137.33) + 88.91(1 − x) + 40.08x + 2(63.55)
+ 7.55(16.00)] gmol−1 = (1012.65 − 48.83x) gmol−1
From the data given
M = 1
2 ρNAa2c
= 1
2 (7.651 × 10
6 g m−3) × (6.0221 × 1023mol−1) × (0.38606 × 10−9 m)2
× (2.8915 × 10−9 m)
= 992.82 gmol−1
Equating the expression forMwith its numerical value gives 992.82 = (1012.65−
48.83x), hence x = (1012.65 − 992.82)/48.83 = 0.406 .
15F Themagnetic properties of solids
Answer to discussion question
Solutions to exercises
E15F.1(b) �emagneticmomentm is given by [15F.3–675],m = ge[S(S+1)]1/2µB, where
ge = 2.0023 and µB = eħ/(2me). For Mn2+, 5.3µB = ge[S(S + 1)]1/2µB, the
constant µB cancels leaving a quadratic which is solved for S
S2+S−(5.3/2.0023)2 = 0 S = 1
2 (−1±
√
1 + 4 × 1 × 7.006...) = −0.500±2.693...