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SOLUTIONSMANUAL TO ACCOMPANY ATKINS' PHYSICAL CHEMISTRY 77
∆S3 =
m
M
Cp ,m(Cu(s)) ln(
Tf
Ti
)
= 2.00 × 103 g
63.55 gmol−1
× (24.44 JK−1mol−1) × ln(3.30... × 10
2 K
273.15 K
)
= +1.46... × 102 JK−1
�e overall entropy change for water is
∆S1 + ∆S2 = (−1.09... × 102 JK−1) + (−9.19... J K−1) = −118 JK−1 .
�e entropy change change for copper block is ∆S3 = +146 JK−1 .
�e total entropy change is therefore
∆Stot = ∆S1 + ∆S2 + ∆S3
= (−1.09... × 102 JK−1) + (−9.19... J K−1) + (+1.46... × 102 JK−1)
= +27.8... J K−1 = +28 JK−1 .
P3B.10 (a) Because the working substance is assumed to be a perfect gas, its internal
energy depends on temperature only. DU as a function of temperature
is given by [2A.15b–45], ∆U = CV∆T . Because the steps 1 and 3 are
adiabatic, q1 = 0 and q3 = 0 .�us, from the First Law, ∆U = q +w, the
work done in these steps is equal to the change in internal energy
w1 = ∆U1 = CV(TB − TA) = nCV ,m(TB − TA) .
w3 = ∆U3 = CV(TD − TC) = nCV ,m(TD − TC) .
Because the steps 2 and 4 are at constant volume, nowork is done: w2 = 0
andw4 = 0 .�us, from the First Law, ∆U = q+w, the heat in these steps
is equal to the change in internal energy
q2 = ∆U2 = CV(TC − TB) = nCV ,m(TC − TB) .
q4 = ∆U4 = CV(TA − TD) = nCV ,m(TA − TD) .
(b) �e e�ciency is de�ned as η = ∣wcycle∣/∣q2∣, thus
η =
∣wcycle∣
∣q2∣
= ∣w1 +w3∣
∣q2∣
= nCV ,m × ∣(TB − TA) + (TD − TC)∣
nCV ,m × ∣TC − TB∣
= ∣(TD − TA) − (TC − TB)∣
∣TC − TB∣
= ∣TD − TA
TC − TB
− 1∣ = ∣1 − TD − TA
TC − TB
∣ .