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64 3 THE SECOND AND THIRD LAWS
(i)
∆S = qrev
T
= 250 kJ
(273.15 K + 20 K)
= 0.853 kJ = +853 J
(ii)
∆S = qrev
T
= 250 kJ
(273.15 K + 100 K)
= 0.670 kJ = +670 J
E3A.3(b) As explained in Section 3A.2(a) on page 80 the change in entropy for an isother-
mal expansion of a gas is calculated using
∆S = nR ln(Vf
Vi
) = m
M
R ln(Vf
Vi
)
= ( 4.00 g
28.02 gmol−1
) × (8.3145 JK−1mol−1) × ln(750 cm
3
500 cm3
)
= +0.481 JK−1 .
E3A.4(b) �e change in entropy for an isothermal expansion of a gas is ∆S = nR ln (Vf/Vi)
as explained in Section 3A.2(a) on page 80.
(i) Isothermal reversible expansion
∆S = ( 14 g
28.02 gmol−1
) × (8.3145 JK−1mol−1) × ln(4.60 dm
3
1.20 dm3
)
= +0.81 JK−1 .
Because the process is reversible ∆Stot = 0 .
Because ∆Stot = ∆S + ∆Ssur
∆Ssur = ∆Stot − ∆S = −0.81 JK−1 .
(ii) Isothermal irreversible expansion against pex = 0 Because entropy is a
state function and the initial and �nal states of the system are the same as
in (a), ∆S is the same.
∆S = +0.81 JK−1 .
Expansion against an external pressure of 0 does no work, and for an
isothermal process of an ideal gas ∆U = 0. From the First Law if follows
that q = 0 and therefore ∆Ssur = 0 .
∆Stot = ∆S + ∆Ssur = +0.81 JK−1 .
(iii) Adiabatic reversible expansion For an adiabatic expansion there is no heat
�owing to or from the surroundings, thus ∆Ssur = 0 . For a reversible
process ∆Stot = 0 , therefore it follows that ∆S = 0 as well.