Ch15 Thermodynamics
Zeroth Law of Thermodynamics
If two systems are in thermal equilibrium with a third system, then
they are in thermal equilibrium with each other.
First Law of Thermodynamics
The Internal Energy of a closed system will be equal to the energy
added to the system by heating minus the work done by the system
its surrounding
Second Law of Thermodynamics
Heat flows out from hot objects to cold; heat does NOT
flow from cold to hot
Internal Energy
• The sum of all the energy of all the molecules
in an object (thermal energy)
• Internal Energy of an Ideal Gas
U  32 nRT
nR  Nk
Ch15 Thermodynamics
• Heat – Transfer of energy due to ΔT
• Work – Transfer of energy NOT due to ΔT
• Q – Heat
• W – Work
– W done on the system is negative (Giancoli)
– W done by the system is positive (Giancoli)
• ΔU – Change in energy
Ch15 Thermodynamics
First Law of Thermodynamics
U  Q  W
Law of Conservation of Energy
U  Q  W
Heat added is
Heat lost is
Work on system is
Work by system is
+
–
+
AP equation sheet
Ch15 Thermodynamics
The distinction between work done on the gas and
work done by the gas is one that is often made on the
AP Exam
The area under the P-V curve will always be the work done
by the gas during the process
First Law of Thermodynamics
2500J of heat is added to a system, and 1800J of work is
done on the system. What is the changed in internal
energy of the system?
(Q) 2500J of heat will increase the Internal Energy
(W) 1800J of work done ON the system will …
U  Q  W
Is the work positive or negative? Why?
U  2500 J  (1800 J )
U  4300J
Did the temperature increase or decrease?
Ch15 Thermodynamics
• Isothermal process: Constant temperature
– The system is in contact with a heat reservoir
Isothermal
– Change of phase
4
Pressure
U  32 nRT  0
U  Q  W
Q W
5
3
2
1
0
0
1
2
3
Volume
• The work done by the gas in an
isothermal process equals the heat
added to the gas
4
5
Isothermal process:
Constant temperature, i.e. PV is constant
Isothermal
Which Isothermal process is
at a higher Temperature?
5
Which Isothermal process
does more work?
Pressure
4
3
2
1
0
0
1
2
3
Volume
4
5
Adiabatic
• Adiabatic Process: No heat in or out of the system
– Well insulated (like a thermos)
– The process happens very quickly (firing of a car
cylinder
Adiabatic
Isothermal
9
8
7
Pressure
U  Q  W
U  W
10
6
5
4
3
2
1
0
0
1
2
3
4
5
Volume
6
7
8
9
10
Work
Given the following two processes: Isothermal and Adiabatic.
Both processes start at 10Pa and end with a volume of 10m3
During which process is more work done?
Estimate the work done in each process.
Adiabatic
Isothermal
PV  5.5Pa(10m  3m )
3
PV  38.5J
3
9
8
7
Pressure
PV  6.5Pa(10m3  3m3 )
PV  45.5J
10
6
5
4
3
2
1
0
0
1
2
3
4
5
Volume
6
7
8
9
10
Ch15 Thermodynamics
• Isovolumetric: (Isochoric) No change in volume
– Inside a ridged container
Isovolumetric
10
W 0
8
7
Pressure
W  PV
W  P(0)
9
6
5
4
3
2
1
0
0
1
2
3
4
5
Volume
6
7
8
9
10
Ch15 Thermodynamics
• Isobaric: No change in pressure
– Movable piston
Isobaric
10
W  PV
9
8
Pressure
7
6
5
4
3
2
1
0
0
1
2
3
4
5
Volume
6
7
8
9
10
Internal Energy ΔU
U  32 nRT
PV  nRT
U  32 PV
U  32 400 Pa (60m3 )
Pressure (Pa)
• 1 mole of an ideal gas is brought from point a to point c
by 3 different process paths. Which path has the
highest change in internal energy?
• 1) U abc
• 2) U adc
400
b
• 3) U ac
300
• 4) All the same
• 5) Unknown
c
200
100
d
a
10
20
30
Volume (m3)
40
50
60
Work (W)
W  PV
Wabc  400(50)
Pressure (Pa)
• 1 mole of an ideal gas is brought from point a to point c
by 3 different process paths. During which path did
the gas do the most work?
• 1) Wabc
• 2) Wadc
400
b
• 3) Wac
300
• 4) All the same
• 5) Unknown
c
200
100
d
a
Wadc  100(50)
Wac  250(50)
10
20
30
Volume (m3)
40
50
60
Heat (Q)
U  Q  W
Q  U  W
Qabc  U  400(50)
Qadc  U  100(50)
Qac  U  250(50)
Pressure (Pa)
• 1 mole of an ideal gas is brought from point a to point c
by 3 different process paths. During which path was
the most heat added?
• 1) Qabc
• 2) Qadc
400
b
• 3) Qac
300
• 4) All the same
• 5) Unknown
c
200
100
d
a
10
20
30
Volume (m3)
40
50
60
•
One mole of monatomic ideal gas is enclosed under a
frictionless piston. A series of processes occur, and
eventually the state of the gas returns to its initial state with
a P-V diagram as shown below. Answer the following in
terms of P0, V0, and R.
• Find the temperature at each vertex.
PV
o o  nRTA
TA 
PV
o o
nR
TB 
4 PV
o o
nR
• Find the change in internal energy for each
process.
U  nRT
3
2
U A B  32 nR (
4 PoVo PoVo

)
nR
nR
• Find the work by the gas done for each
process.
WAB  PVAB  Po (4Vo  V0 )
•
a)
b)
Calculate the total work done by the gas
Calculate the total heat flow into the gas
Pressure Pa
•
An ideal gas is slowly compress at constant pressure (2.0 ATM) from 10.0L
to 2.0L
Heat is then added to the gas holding the volume constant and the
pressure and temperate are allowed to rise until the temperature reaches
its original value.
2
Volume m3
10
In an engine 0.25 moles of an ideal gas in the cylinder expands rapidly
and adiabatically against the piston. In this process, the temperature of
the gas drops from 1150K to 400K.
a)What type of process is this?
b) How much work does the gas do?
U  Q  W
U  W
3
U  nRT
2
U  32 (0.25moles)(8.314 molJ K (400 K  1150 K )
 2300J
Is the work done by the gas positive or
negative?
W  2300J
Efficiency
• Efficiency (e): the ratio of work W done by the
system to the input heat QH
W
e
QH
W
e
QH

QH  W  QL
QH  QL
QH
QL
 1
QH
TL
 1
TH
An automobile engine has an efficiency of 20% and produces an average
of 23,000J of mechanical work per second.
a) How much input heat is required?
b) How much heat is discharged as wasted per second?
b)
a)
W
e
QH
QL
 1 e
QH
W
QH 
e
QL  (1  e)QH  .8(115kJ )
23, 000 J
QH 
0.20
QH  1.15 x105 J
QL  92kJ
• Is the car’s efficiency higher or lower as
the car warms up?