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
15.1 The First Law of Thermodynamics

A system’s internal energy can be changed by doing
work or by the addition/removal of heat:
ΔU = Q - W

W is negative if work is done on the system


What is the state of the system?

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Compression of the gas
Described by P, V, T, m, U
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15.2 Thermodynamic Processes and the First Law

Isothermal: T = constant → ΔU = 0 → W = Q

Adiabatic: Q = 0 → ΔU = -W
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15.2 Thermodynamic Processes and the First Law
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If pressure is constant then
W = Fd = PAd = P ΔV
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15.2 Thermodynamic Processes and the First Law
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The total work done during a process is equal to the
area under the PV diagram
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
15.4 The Second Law of Thermodynamics


Heat can flow spontaneously only from a hot object
to a cold object.
A reversible process is one that is always in
equilibrium and can return to its initial conditions
along the same path
Most natural processes are irreversible
 Sets an upper limit on efficiency of heat engines

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
15.5 Heat Engines

Heat engines convert U into other useful forms of
energy – mechanical, electrical, …
ΔUcycle = 0 → QH = W + QL
Automobile engines
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
15.5 Heat Engines

W
QL
 1
The efficiency of a heat engine is e 
QH
QH

Carnot (ideal) engine
Reversible processes
 Too slow for real engines

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15.6 Refrigerators, Air Conditioners and Heat Pumps

A heat engine in reverse.
QL
COP 
W
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
15.6 Refrigerators, Air Conditioners and Heat Pumps
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2. (a) The work done by a gas at constant pressure is found from
Eq. 15-3.
 1.01105 Pa 
3
3
5
W  PV  1 atm  
18.2
m

12.0
m

6.262

10
J



 1 atm 
(b) The change in internal energy is calculated from the first law
of thermodynamics
 4186 J 
5
6
U  Q  W  1400 kcal  

6.262

10
J

5.2

10
J

 1 kcal 
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26. Find the exhaust temperature from the original Carnot
efficiency, and then recalculate the intake temperature for the
new Carnot efficiency, using the same exhaust temperature.
e1  1 
TL
 TL  TH1 1  e    550  273 K 1  0.28   592.6 K
TH1
TL
e1 T1 
 TL T TH1 592.6
1  e  K 550  273 K 1o  0.28  o 592.6
L
e2  1  L T

 912 K  639 C  640 C
H1 TH2 
TH2
1  e2 1  0.35
TL
TL
592.6 K
e2  1 
 TH2 

 912 K  639o C  640o
TH2
1  e2 1  0.35
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