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SOLUTIONSMANUAL TO ACCOMPANY ATKINS' PHYSICAL CHEMISTRY 267
Answers to integrated activities
I7.2 Macroscopic synthesis and material development always contains elements of
randomness at the molecular level. Crystal structures are never perfect. A
product of organic synthesis is never absolutely free of impurities, although
impurities may be at a level that is lower than measurement techniques make
possible. Alloys are grainy and slightly non-homogeneous within any particu-
lar grain. Furthermore, the random distribution of atomic/molecular positions
and orientations within, and between, macroscopic objects causes the conver-
sion of energy to non-useful heat during manufacturing processes. Production
e�ciencies are di�cult to improve. Nanometre technology on the 1 nm to
100 nm scale may resolve many of these problems. Self-organization and pro-
duction processes by nanoparticles and nanomachines may be able to exclude
impurities and greatly improve homogeneity by e�ective examination and se-
lection of each atom/molecule during nanosynthesis and nanoproduction pro-
cesses. Higher e�ciencies of energy usage may be achievable as nanomachines
produce idealizedmaterials at the smaller sizes and pass their products to larger
nanomachines for production of larger scale materials.
�e directed, non-random, use of atoms and molecules by nanotechniques
holds the promise for the production of smaller transistors and wires for the
electronics and computer industries. Unusual material strengths, optical prop-
erties, magnetic properties, and catalytic properties may be achievable. Higher
e�ciencies of photo-electronic conversion would have a great impact.
I7.4 (a) �e �rst step is to compute the total energy of the system of NA particles,
which is identi�ed as the internal energyU .�e energy levels for a parti-
cle in a cubic box of side L are given by [7D.13b–267], En = h2n2/8mL2,
where n2 = n21 + n22 + n23 . If there are NA particles, all occupying the
level corresponding to a particular value of n, the internal energy of the
system is U = NAEn = NAh2n2/8mL2. Using V = L3 the length is
written in terms of the volume as L = V 1/3, hence L2 = V 2/3 and therefore
U = NAh2n2/8mV 2/3.
If the expansion is adiabatic (that is, not heat �ows into or out of the
system) then from the First Law, dU = dq + dw, it follows that dU = dw.
�e work done on expansion is therefore computed by �nding how U
changes with volume, speci�cally by �nding ∂U/∂V .
∂U
∂V
= ∂
∂V
(NAh
2n2
8mV 2/3
)
adia
= −2
3
× NAh
2n2
8mV 5/3
= − NAh
2n2
12mV 5/3
�e change in internal energy on expansion through dV will therefore be
dU = (∂U
∂V
)
adia
dV = −
A
³¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹·¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹µ
NAh2n2
12mV 5/3
dV (7.2)
�e work is equal to this change in internal energy. For a �nite change
the expression is integrated with respect to V between limits V1 and V2,
with ∆V = V2 − V1.