Leis da Termodinâmica e Entropia

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70 3 THE SECOND AND THIRD LAWS
change for each block is found using this expression
∆S1 = mCV ,s ln(
Tf
T1
)
= (1.00 × 104 g) × (0.449 JK−1 g−1) × ln( 3.35... × 102 K
100 K + 273.15 K
)
= −4.75... × 102 JK−1 = −476 JK−1 .
∆S2 = mCV ,s ln(
Tf
T2
)
= (1.00 × 104 g) × (0.449 JK−1 g−1) × ln( 3.35... × 10
2 K
25 K + 273.15 K
)
= 5.31... × 102 JK−1 = +532 JK−1 .
�e total change in entropy is
∆Stot = ∆S1 + ∆S2 = (−4.75... × 102 JK−1) + (5.31... × 102 JK−1)
= 0.563... × 102 JK−1 = +56 JK−1 .
Because ∆Stot > 0 the process is spontaneous, in accord with experience.
E3B.6(b) Because entropy is a state function, ∆S between the initial and �nal states is
the same irrespective of the path taken.�us the overall process can be broken
down into steps that are easier to evaluate. First consider heating the initial
system at constant pressure to the �nal temperature. �e variation of entropy
with temperature at constant pressure is given by [3B.7–90], S(Tf) = S(Ti) +
Cp ln (Tf/Ti).�us the change in entropy, ∆S = S(Tf) − S(Ti), of this step is
∆S1 = Cp ln(
Tf
Ti
) = nCp ,m ln(
Tf
Ti
)
Next consider an isothermal change in pressure. As explained in Section 3A.2(a)
on page 80 the change in entropy of an isothermal expansion of an ideal gas
is given by ∆S = nR ln (Vf/Vi). Because for a �xed amount of gas at �xed
temperature p ∝ (1/V) an equivalent expression for this entropy change is
∆S2 = nR ln(
pi
pf
)
�erefore the overall entropy change for the system is
∆S = ∆S1 + ∆S2 = nCp ,m ln(
Tf
Ti
) + nR ln( pi
pf
)
= (2.00 mol) × (7
2
× 8.3145 JK−1mol−1) × ln(273.15 K + 135 K
273.15 K + 25 K
)
+ (2.00 mol) × (8.3145 JK−1mol−1) × ln(1.50 atm
7.00 atm
)
= (+18.2... J K−1) + (−25.6... J K−1) = −7.3 JK−1 .