Part 2

1/29/2015
Chapter Outline







9.1
9.2
9.3
9.4
9.5
9.6
9.7
Energy as a Reactant or a Product
Transferring Heat and Doing Work
Enthalpy and Enthalpy Changes
Heating Curves and Heat Capacity
Heats of Reaction and Calorimetry
Hess’s Law and Standard Heats of Reaction
Heats of Reaction from Heats of Formation and
Bond Energies
 9.8 More Applications of Thermochemistry
Isolated, Closed and Open Systems
open systems
can exchange
mass and energy
closed systems
can only
exchange energy
isolated systems
can not exchange
mass or energy
1
1/29/2015
Exothermic and Endothermic Processes
• Exothermic process: Heat flows out
of the system to surroundings (q < 0).
• Endothermic process: Heat flows into the
system from surroundings (q > 0).
Pressure-Volume (P-V) Work
E = q + w
“internal
energy”
Heat flow
Work
(F x d)
E > 0 when heat flows in (q > 0)
or work is performed on the system
(w > 0)
E < 0 when heat flows out (q < 0)
or work is performed on the
surroundings (w < 0)
2
1/29/2015
(a) Intake – air and fuel enters the cylinder
(b) Compression – heats up reaction mixture and ignites (q > 0, w > 0, E > 0)
(c) Power Stroke (expansion) – combustion products perform work on the
surroundings (q < 0, w < 0, E < 0)
(d) Exhaust
Example of P-V Work, p. 368-369
Work = P x V
where V = Vf - Vi
and P =
F
A
Remember that energy (in Joules) is the capacity to do work, so P x V
has to have the same units as energy.
Units of a Joule = Newton x meters =
Units of P =
kg m2
s2
Newton kgm/s2 kg/s2
=
=
m
m2
m2
Units: P x V =
kg/s2
kg m2
x m3 =
= Joules
m
s2
So E = q + w = q - PV
There’s a minus sign because when
when V > 0, then system does work on
the surroundings (w < 0)
3
1/29/2015
Sample Exercise 9.2:
Calculating P-V Work
A tank of compressed helium is used to inflate 100 balloons for sale
at a carnival on a day when the atmospheric pressure is 1.01 atm. If
each balloon is inflated with 4.8 L, how much P-V work is done by the
compressed helium? Express your answer in Joules.
w = -PV
P = 1.01 atm
Conversion factor: L x atm = 101.325 J
V = Vf - Vi
= 4.8L – 0 = 4.8L
w = - 1.01 atm x 4.8L x 100 balloons x 101.325 J
balloon
Latm
= - 4.9 x 104 J
= - 49 kJ
Sample Exercise 9.3:
Relating E, q, and w
The racing cars in the figure are powered by V8 engines in which the motion of each piston in
its cylinder displaces a volume of 0.733 L. If combustion of the mixture of gasoline vapor and
air in one cylinder releases -1.68 kJ of energy, and if 33% of the energy does P-V work, how
much pressure, on average, does the combustion reaction mixture exert on each piston? How
much heat flows from the reaction mixture to its surroundings?
V =
33%
0.733 L
Plan: E  -PV  P
From
E = q - PV

q
4