Thermodynamics
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Thermodynamics: Understanding Temperature, Pressure, Density, and States of Matter, Slides of Civil Engineering
An introduction to thermodynamics, focusing on concepts such as temperature, pressure, density, and the states of matter. It includes definitions, explanations, examples, and problem-solving exercises. Students can use this document as study notes, summaries, or as a reference for understanding thermodynamics concepts.
Typology: Slides
2012/2013
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Thermodynamics
Temperature
Rapidly
moving
molecules
have
a
high
temperature
Slowly
moving
molecules
have
a
low
temperature
High T
Low T
Density
Density
mass
per
unit
volume
It
has
units
of
g/cm
3
M^ V
High density
Low density
Example
Problem
A
cube
of
osmium
measures
m
on
a
side.
It
sits
on
a
table.
At
the
contact
between
the
table
and
osmium,
calculate
the
pressure
(N/m
2
Note:
Densities
may
be
found
in
your
text
Solution
kg·m
N·s
1
m·s
kg
300 ,
44
m/s
kg
1
is
1N
that
recall
m·s
kg
300 ,
44
m s
- 9 m 2. 0 m
kg
610 ,
22
2
2
2
2
2
3
3 2
Lg
L
g
L
P
= 44,300 N/m
2
What
are
the
States
of
Matter?
Solid
Liquid
Gas
Plasma
Pressure,
Temperature,
and
State
Plasma
Gas
Vapor
Liquid
Solid
T
triple
T
critical
P
triple
P
critical
Pressure
Temperature
CriticalPoint
TriplePoint
Gas
Laws
Perfect
(ideal)
Gases
Boyle’s
Law
Charles’
Law
Gay
Lussac’s
Law
Mole
Proportionality
Law
Charles’
Law
(^21)
(^21)
T^ T
V V
T
1
V
1
T
2
V
2
P
= const
n
= const
Gay
Lussac’s
Law
(^21)
2 1
T^ T
P P
T
1
P
1
T
2
P
2
V
= const
n
= const
Perfect
Gas
Law
The
physical
observations
described
by
the
gas
laws
are
summarized
by
the
perfect
gas
law
(a.k.a.
ideal
gas
law)
PV
nRT
P
absolute
pressure
V
volume
n
number
of
moles
R
universal
gas
constant
T
absolute
temperature
Team
Exercise
min.)
A
balloon
is
filled
with
air
to
a
pressure
of
atm.
The
filled
balloon
has
a
diameter
of
m.
A
diver
takes
the
balloon
underwater
to
a
depth
where
the
pressure
in
the
balloon
is
atm.
If
the
temperature
of
the
balloon
does
not
change,
what
is
the
new
diameter
of
the
balloon?
3
3
1 2
1
2
1 2
3 1
(^32)
3 1
3
1
1
(^32)
3
2
2
1 2
1
2
1 2
2 1
atm
atm
- 1 m 3. 0 2
P P
D
D
P P
kD
kD
kD
D
V
kD
D
V
P P
V
V
P P
V V
= 0.235 m
Solution
P
1
= 1.1 atm
D
1
= 0.3 m
P
2
= 2.3 atm^ D
2