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268 7 QUANTUM THEORY
(b) It is evident from eqn 7.2 that dU , and hence the work, goes as n2.
(c) �ework of expansion against an external pressure pex is given by [2A.5a–
39], dw = −pexdV . In eqn 7.2 the term A which multiplies dV refers to
the sample itself, and somust presumably in someway re�ect the pressure
of the sample, not the external pressure. However, if the expansion is
reversible, the external pressure is equal to the internal pressure and the
term A can then be identi�ed as the pressure. �erefore, if it is assumed
that the expansion is both adiabatic and reversible
p = NAh
2n2
12mV 5/3
�e expression can be rewritten in terms of the average energy of each
particle which, because they all occupy the same level, is simply Eav =
n2h2/8mL2 = n2h2/8mV 2/3, hence
p = NAh
2n2
12mV 5/3
= 8NA
12V
Eav³¹¹¹¹¹¹¹¹¹¹¹¹·¹¹¹¹¹¹¹¹¹¹¹µ
n2h2
8mV 2/3
= 2NAEav
3V
�is expression is reminiscent of the form of the pressure derived using
the kinetic theory of gases (Topic 1B): pV = 1
3nMυ2rms, where n is the
amount in moles,M is the molar mass, and υrms is the root-mean-square
speed. Because M = mNA, where m is the mass of a molecule, the ex-
pression can be rewritten
pV = 1
3nmNAυ2rms = n 23NA
Ek³¹¹¹¹¹¹¹¹¹¹·¹¹¹¹¹¹¹¹¹¹µ
1
2mυ2rms
�e term 1
2mυ2rms is identi�ed as the average kinetic energy of onemolecule
and, because in the kinetic theory the only energy a molecule possesses is
kinetic, Ek can further be identi�ed as the average energy, Eav.�us, for
one mole (n = 1)
pV = 2
3NAEav hence p = 2NAEav
3V
�e two expressions for the pressure are therefore directly comparable
within the restrictions imposed.
(d) For an isothermal expansion heat would have to enter the system in order
to maintain its temperature, and this would involve promoting particles
to higher energy levels. As the volume increases the energy levels move
closer together, so the promotion of particles to higher levels needs to
o�set this e�ect as well.
I7.6 (a) In Problem P7D.6 and Problem P7D.7 it is shown that for a particle in a
box in a state with quantum number n
∆x = L(1/12 − 1/2n2π2)1/2 and ∆px = nh/2L