Solutions to Problem Set 7
6.4
(a) The work done is given by :
(4 cylinders).(60 cycles/min).(1400 bar. 10
5
Pa/bar). (2.5 L. 0.001m
3
/L) = 8.4 x
10
7
J
(b) The expanding gas in the cylinder does work on the pistons. Therefore, the
work done by the gas is negative.
6.8.
To calculate this, we can use the relationship
Ä
U = q + w, which arises from the
first law of thermodynamics. Because the system releases heat, q will be a
negative number as will
Ä
U, because it is a decrease in internal energy:
125 kJ = 346 kJ + w
w = 221 kJ.
Because work is positive, surrounding will do work on the system.
6.16
(a) The heat change will be made of two terms: one term to raise the temperature
of the stainless steel and the other to raise the temperature of the water:
q = 450.0 x 4.18 x (100 – 25) + 500.0 x 0.51 x (100 25)
= 141 kJ + 19 kJ = 160 kJ
(b) The percentage of heat attributable to the raising of the temperature of the
water will be (141/160)x100 = 88.1%
(c) The use of copper kettle is more efficient as a larger percentage of the heat
goes into heating the water and not the container holding it.
6.22
(a) Because the process is isothermal,
Ä
U = 0 and q = w. For a reversible process,
w = nRT ln(V2/V1)
n is obtained from the ideal gas equation:
n = PV/RT = 0.359 mol
w = (0.359)(8.314)(298)ln(7.39/3.42) = 685 J
q = +685 J
(b) For step 1, because volume is constant, w = 0 ,
Ä
U = q
In step 2, there is an irreversible expansion against constant pressure,
W = P(V
final
 V
initial
)=  1.19 atm (7.39 L – 3.42 L) = 4.72 L.atm = (4.72
L.atm)(101.325 J/1 L.atm) = 479 J.
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 Fall '06
 CRANE, B
 Chemistry, Thermodynamics, Heat

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