INSTRUCTOR’S MANUAL
62
E4.16(b)
(a) (b)
q = 0 [adiabatic]
w = −pex V = −(1.5 atm) ×
1.01 × 105 Pa atm
2
× (100.0 cm ) × (15 cm) ×
1 m3 106 cm3
= −227.2 J = −230 J (c)
U = q + w = 0 − 230 J = −230 J
(d)
U = nCV ,m T T =
(e)
U −227.2 J = nCV ,m (1.5 mol) × (28.8 J K −1 mol−1 )
= −5.3 K Vf Tf + nR ln S = nCV ,m ln Ti Vi Tf = 288.15 K − 5.26 K = 282.9 K nRT (1.5 mol) × (8.206 × 10−2 L atm K−1 mol−1 ) × (288.2̄ K) Vi = = pi 9.0 atm = 3.942 L
Vf = 3.942 L + (100 cm2 ) × (15 cm) ×
1L 1000 cm3
= 3.942 L + 1.5 L = 5.44 L S = (1.5 mol) × (28.8 J K
−1
mol
−1
) × ln
282.9
288.2 5.44̄ −1 −1 + (8.314 J K mol ) × ln 3.942
= 1.5 mol(−0.5346 J K−1 mol−1 + 2.678 J K−1 mol−1 ) = 3.2 J K−1 E4.17(b)
− vap H − 35.27 × 103 J mol−1 = = + 104.58 J K−1 = 104.6 J K−1 Tb (64.1 + 273.15) K (b) If vaporization occurs reversibly, as is generally assumed
(a)
− vap S − =
Ssys + Ssur = 0 E4.18(b)
(a)
so Ssur = −104.6 J K−1
− − − − − − − − − r S − = Sm (Zn2+ , aq) + Sm (Cu, s) − Sm (Zn, s) − Sm (Cu2+ , aq)
= [−112.1 + 33.15 − 41.63 + 99.6] J K −1 mol−1 = −21.0 J K−1 mol−1 (b)
− − − − − − − − − = 12Sm (CO2 , g) + 11Sm (H2 O, l) − Sm (C12 H22 O11 , s) − 12Sm (O2 , g) r S −
= [(12 × 213.74) + (11 × 69.91) − 360.2 − (12 × 205.14)] J K −1 mol−1 = + 512.0 J K−1 mol−1 E4.19(b)
(a)
− − − r H − = f H − (Zn2+ , aq) − f H − (Cu2+ , aq)
= −153.89 − 64.77 kJ mol−1 = −218.66 kJ mol−1 − r G − = −218.66 kJ mol−1 − (298.15 K) × (−21.0 J K −1 mol−1 ) = −212.40 kJ mol−1
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