b. THIS PART OF THE PROBLEM IS TO SERVE AS AN ILLUSTRATION. YOU
ARE NOT RESPONSIBLE FOR THIS MATERIAL IN THIS COURSE.
It is necessary to rotate the properties of the crystal to coincide with the orientation
of the crystal axes with respect to the laboratory reference frame. This is done using
the matrix of cosines, A~. Let ~L = A~~ be the dielectric constant referred to the lab
reference frame.
0
1
cos sin 0
A~ = B@ ; sin cos 0 CA
0
And so,
0
1
0
10
1
cos sin 0
0 0
11
~L = B
@ ; sin cos 0 CA B@ 0 22 0 CA
0
0 1
0 0 33
0
1
11 cos 22 sin 0
~L = B@ ;11 sin 22 cos 0 CA
0
0
1
And the change in internal energy is given by the following.
U = E12 11 cos + E22 22 cos + E3233 + E1E2 (22 ; 11 ) sin c. When the crystal has tetragonal symmetry the following holds.
Ua = 11 (E12 + E22 ) + 33 E32
Ub = 11 (E12 + E22 ) cos + 33 E32
On inspection these are equal when cos = 1 or = =2. We've shown that if a
tetragonal crystal is rotated 90 degrees about its c-axis the crystal properties are
indistinguishable from those before the rotation. This is an illustration of Neumanns
Principle.
Example Problem 4.2
The equation of state for an ideal gas is PV = nRT . The equation of state for one
mole of a van der Waals gas is below. Calculate the change in internal energy for one mole
of an adiabatically contained van der Waals gas during an isothermal transformation from
(P0; V0) to (P1; V1).
a
P + V 2 (V ; b) = RT
Solution 4.2 From the rst law: dU = dQ + dW where dU is the change in internal
energy, dQ is heat added to the system and dW is the work done on the system. For an
adiabatic system dQ = 0.
dU =
Z V;1 PdV
U = ; PdV
V
0
We also need P (V ) for one mole of gas.
"
a
;
P = (VRT
; b) V 2
#
#
Z V0 " RT
a
U =
; dV
(V! ; b) V 2
V1
b +a 1 ; 1 U = RT ln VV0 ;
;b
V V
1
0
1
Contrast this with the expression for an ideal gas, U = RT ln
V0 V1
.
Example Problem 4.3
Find the equation of the adiabats in the P-V plane for a gas with an equation of state
given by U = AP 2 V where A is a positive constant.
Solution 4.3
dU = dQ + dW
dU = ;PdV
AP 2dV + 2APV dP = ;PdV
2APV dP = ;(AP 2 + P )dV
dP = ; dV
AV
Z P1 dPAP + 1 Z V1 2dV
=;
P0 AP + 1
V0 2AV
1 ln V0 = 1 ln P1
2A V1 A P0
V0 12 P1 V1 = P0
1
1
V02 P0 = V12 P1 = constant
The equation of the adiabats for this gas is V 12 P = constant. Note that the equation for
the adiabats of the ideal gas is similar, PV = constant.