Chapter 13
States of Matter
Fluid - A material flows and have no definite shape of their own.
Pascal’s Principle: The change in pressure applied at any point in a
confined fluid is transmitted undiminished throughout the fluid.
Examples
•Toothpaste
•Hydraulics
F
F1
F2
P  or

A A1 A 2
How much can be lifted?
F1/A1 =F2/A2
F2 = F1A2/A1
F2 = (20 N)(.1 m2)/(.05 m2) = 40 N
If the little piston moves 1 meter, how
far does the big one move?
V1=V2
A1H1=A2H2
H2 = A1H1/A2 H2= (.05 m2)(1m)/(.1 m2)
H2= .5 m
Chapter 13
States of Matter
Swimming under pressure
F
mg
P

A
A
But d = m/v or m = dv
F
mg dvg dAhg
P



 hdg
A
A
A
A
P = hdg
Chapter 13
States of Matter
Taking P = hdg and multiplying both sides by A gives
PA = Ahdg or F = vdg
Where F = vdg is the buoyant force
Chapter 13
States of Matter

Archimedes’ Principle - An object immersed
in a fluid has an upward force on it equal to
the weight of the fluid displaced by the
object.
1.
2.
3.
A body sinks if the weight of the fluid it displaces
is less than the weight of the body.
A submerged body remains in equilibrium if the
weight of the fluid it displaces exactly equals its
own weight.
A body floats if it displaces a weight greater than
that of its own weight
A
B
C
Chapter 13
States of Matter
A block of wood has a volume of 100 cm3 and a mass of 85 grams.
a. Will it float in water water = 1000 kg/m3 ? YES
b. Will it float in gas
gas = 700 kg/m3 NO
d= m/v = 85g/100 cm3 = .85 g/cm3= 850 kg/m3
Chapter 13
States of Matter
What is the weight of a rock submerged in water if the rock weighs
30 newtons and has a volume of .002 m3?
V = .002 m3
W = 30 N
Fnet = Weight - buoyant force 
3
water = 1000 kg/m
vdg
Fnet = mg - vdg
Fnet = 30 N - (.002 m3)(1000 kg/m3)(9.8 m/s2)
mg
Fnet = 30 N - 19.6 N = 10.4 N
The acceleration of the rock will be a = F/m
A = 10.4 N/3.06 kg = 3.2 m/s2
Chapter 13
States of Matter
What the maximum weight a helium balloon of volume 2 m3 can
support in air?
V
= 2 m3
air = 1.2 kg/m3
helium = .177 kg/m3
vdg Fnet = Weight - buoyant force
Fnet = mg - vdg
Fnet = (2m3)(.177kg/m3)(9.8m/s2)-(2m3)(1.2 kg/m3)(9.8 m/s2)
mg Fnet = 3.462 N - 23.52 N
It can support 20 N
Chapter 13
States of Matter
Bernouilli’s Principle For the horizontal flow of a fluid through a
tube, the sum of the kinetic energy per unit volume and the pressure
is a constant.
Chapter 13
States of Matter




Cohesion: The force of attraction between
like particles.
Adhesion: The force of attraction between
unlike particles.
Capillary action: The rise of a liquid in a
narrow tube because the adhesive force is
stronger than the cohesive force.
Volatile Liquid: A liquid that evaporates
quickly.
Chapter 13
States of Matter

Cohesion
•Adhesion
Chapter 13
States of Matter
Surface Tension:
The tendency of the
surface of a liquid to
contract to the smallest
possible area
Chapter 13
States of Matter
Sublimation
Melting
Solid
Vaporization
Liquid
Freezing
Condensation
Supercooled
Gas
Chapter 13
States of Matter
Thermal Expansion: The increase in length or volume of a
substance when heated.
Linear expansion L2 = L1+αL1(T2-T1)
Volume expansion V2 = V1+βV1(T2-T1)
Chart Pg 317
Chapter 13
States of Matter
A iron bar is 3 m long at 21ºC. What is the length of the bar
at 100º C?
Linear expansion L2 = L1+αL1(T2-T1)
L2 = 3 m+(12 x 10 –6 (ºC-1)(3 m)(100ºC- 21ºC)
L2 = 3 m +.002844 m = 3.002844 m