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SOLUTIONSMANUAL TO ACCOMPANY ATKINS' PHYSICAL CHEMISTRY 65
E3A.5(b) �e e�ciency is de�ned in [3A.7–84], η = ∣w∣/∣qh∣, and for a Carnot cycle
e�ciency is given by [3A.9–84], η = 1 − (Tc/Th). �ese two are combined
and rearranged into an expression for the temperature of the hot source
∣w∣/∣qh∣ = 1 − (Tc/Th)
hence Th =
Tc
1 − ∣w∣/∣qh∣
= (273.15 K + 0 K)
1 − ∣3.00 kJ∣/∣ − 10.00 kJ∣
= 390 K .
E3A.6(b) �e e�ciency of a Carnot cycle is given by [3A.9–84], η = 1 − (Tc/Th).�is is
rearranged to give an expression for the temperature of the cold sink
Tc = (1 − η) × Th = (1 − 0.10) × (273.15 K + 40 K) = 282 K .
Note that the temperature must be in kelvins.
Solutions to problems
P3A.2 (a) �e �nal volume of the gas at 1.00 bar and 300 K temperature is found
using the perfect gas law [1A.4–8]:
Vf =
nRT
p
= (0.10 mol) × (8.3145 × 10−2 dm3 barK−1mol−1) × (300 K)
1.00 bar
= 2.49... dm3 = 2.5 dm3 .
(b) �e expansionwork against a constant external pressure is given by [2A.6–
40], w = −pex (Vf − Vi).
w = −pex (Vf − Vi)
= −(1.00 × 105 Pa) × (2.49... × 10−3 m3 − 1.25 × 10−3 m3)
= −1.24... × 102 J = −1.2 × 102 J .
�e work will be in joules if the pressure is expressed in pascals and the
volume in m3.
(c) For an isothermal process of a perfect gas ∆U = 0.�e First Law [2A.2–
38], ∆U = q +w, is then used to �nd the heat
q = ∆U −w = 0 − (−1.24... × 102 J) = +1.24... × 102 J = +1.2 × 102 J .
As explained in Section 3A.2(a) on page 80 the change in entropy of an isother-
mal expansion of an ideal gas can be calculated from ∆S = nR ln (Vf/Vi).