Chapter 13 States of Matter

Chapter 13States 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

2

2

1

1

A

F

A

For

A

FP

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 13States of Matter

Swimming under pressure

P F

A

mg

ABut d = m/v or m = dv

P F

A

mg

A

dvg

A

dAhg

Ahdg

P = hdg

Chapter 13States 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 13States 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. A body sinks if the weight of the fluid it displaces is less than the weight of the body.

2. A submerged body remains in equilibrium if the weight of the fluid it displaces exactly equals its own weight.

3. A body floats if it displaces a weight greater than that of its own weight

ABC

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 ?b. Will it float in gas gas = 700 kg/m3

d= m/v = 85g/100 cm3 = .85 g/cm3= 850 kg/m3

YESNO

Chapter 13States of Matter

What is the weight of a rock submerged in water if the rock weighs30 newtons and has a volume of .002 m3?

V = .002 m3

W = 30 Nwater = 1000 kg/m3

mg

vdgFnet = Weight - buoyant force

Fnet = mg - vdg

Fnet = 30 N - (.002 m3)(1000 kg/m3)(9.8 m/s2)

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 13States 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/m3Fnet = Weight - buoyant force

Fnet = mg - vdg

Fnet = (2m3)(.177kg/m3)(9.8m/s2)-(2m3)(1.2 kg/m3)(9.8 m/s2)

Fnet = 3.462 N - 23.52 Nmg

vdg

It can support 20 N

Chapter 13States of Matter

Bernouilli’s Principle For the horizontal flow of a fluid through atube, the sum of the kinetic energy per unit volume and the pressureis a constant.

Chapter 13States 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 13States of Matter Cohesion •Adhesion

Chapter 13 States of Matter

Surface Tension: The tendency of the surface of a liquid tocontract to the

smallestpossible area

Chapter 13States of Matter

Solid Liquid Gas

Melting

CondensationFreezing

Sublimation

Supercooled

Vaporization

Chapter 13States of Matter

Thermal Expansion: The increase in length or volume of asubstance when heated.

Volume expansion V2 = V1+βV1(T2-T1)

Linear expansion L2 = L1+αL1(T2-T1) Chart Pg 317

Chapter 13States of Matter

Linear expansion L2 = L1+αL1(T2-T1)

A iron bar is 3 m long at 21ºC. What is the length of the barat 100º C?

L2 = 3 m+(12 x 10 –6 (ºC-1)(3 m)(100ºC- 21ºC)

L2 = 3 m +.002844 m = 3.002844 m