chapter 5 solid and liquid (bb101)

Chapter 5

Solids & Fluid

Solids, Liquids and Gases

Pressure

Characteristics of Solid, Liquid and Gas

Density and relative density

Contents

Pascal Principle

Archimedes Principle

Three states of matter

solid liquid gas

At room temperature most substances exist in one of three physical states.

The particle model

The difference between solids, liquids and gases can be explained by the…

All substances are made up of particles.

The particles are attracted to each other. Some particles are attracted strongly to each other and others weakly.

The particles move around. They are described as having kinetic energy.

The kinetic energy of the particles increases with temperature.

Particles in a solid

Particles in a liquid

Particles in a gas

Properties of solids, liquids and gases

How do smells spread out?

Where is the smell coming from and how does it spread out?

Diffusion occurs in liquids and gases but hardly at all in solids.

What is diffusion?

Diffusion is the movement of particles that allows them to spread out and mix with other particles.

For example, the smell of aftershave or perfume diffuses and is detected by people on the other side of the room.

Use the particle model to explain these facts about diffusion:

Diffusion happens more quickly for gases than for liquids.

Diffusion happens more quickly at warm temperatures than at cooler temperatures.

Across:

2. Change from liquid to solid

5. Force caused by collisions of particles

6. Arranged in a 3-dimensional pattern

9. Spacing of particles in solids and liquids

10. All substances are this state at very low temperatures

11. Only liquids and gases do this

1. Closely packed particles 6. Particles are close but disorganised

3. Change from gas to liquid 7. Particles widely spread out

4. Change from solid to liquid 8. Change from liquid to gas

Crossword

DEFINE DENSITY, ρ

Density is defined as ratio of the mass of substance to its volume. It is a measure of how tightly packed and how heavy the

molecules are in an object. Density is the amount of matter within a certain volume.

Proof that water and ice have different densities

To find the density

• Find the mass of the object

• Find the volume of the object

DEFINE DENSITY

Units for density usually express in kg/m3

ρ = m (kg) V (m3)ρ = m (kg) V (m3)

Example 1:

A big box has mass of 20N and size 30cmx30cmx30cm, Using all the information, calculate the density of the box.

Solution:

W = mg

20 N = m (9.81)

m = 20 / 9.81

= 2.04 kg

Volume = 30cm x 30cm x 30cm

= 0.3m x 0.3m x 0.3m

= 0.027m3

ρ = m = 2.04 kg = 75.57 kg/m3

V 0.027m3

• Also known as Specific Gravity

• Specific gravity is ratio of the density of a sunstance to the density t of water.

DEFINE RELATIVE DENSITY, ρb

ρb = ρsubstance (kg/m3) ρwater (kg/m3)ρb = ρsubstance (kg/m3) ρwater (kg/m3)

No unit

1000 kg/m31000 kg/m3

Example 2:

If the density of an object is 4000 kg/m3 ,calculate the specific gravity of the object. ( Density of water = 1000 kg/m3 )

ρb = ρsubstance (kg/m3) ρwater (kg/m3)

ρb = 4000(kg/m3) 1000 (kg/m3)

PRESSURE, P

Pressure is defined as Force per unit Area acting on a surface.

Factors that affect the pressure acting on a surface

• Contact area ( Smaller contact area greater pressure)

•Force acting on the surface ( Large force greater pressure)

P = F (N) A (m2)P = F (N)

A (m2)

Unit in N/m2 or Pascal (Pa)

APPLICATION OF PRESSURE

• High Pressure • Low Pressure

Example 3:

How many Pascal’s are exerted by an elephant of weight 50 000 N standing on his feet of total area 0.8m2 ?

Solution:

F = 50000 N

Area = 0.8m2

P = F / A

= 50000 / 0.8

= 62 500 Pa

Example 4:

What Pressure is exerted by an apple of weight 1 N sitting

on area 20mm2 ?

Solution:

F = 1 N

Area =20 mm2 = 0.00002m2

P = F / A

= 1 / 0.00002

= 50 000 Pa

PRESSURE IN LIQUID

A liquid in a container exerts pressure because of it weight (Force).

P = ρghP = ρgh Unit in N/m2 or Pascal (Pa)

•Volume, V = Ah

•Density, ρ = m V

•Mass, m = ρV

•Weight,w = Force,F = mg = ρVg

= ρAhg

•Pressure = F = ρAhg = ρhg A A

CHARACTERISTICS OF PRESSURE IN A LIQUID

• Depth to pressure in liquid

• Liquid pressure increase with depth

• The pressure of water is the lowest at the highest point of the cylinder and the pressure of water is highest at the lowest point of the cylinder.

• Density to pressure in liquid– Pressure of liquid is increases with density.

• Water • Oil

x1 > x2 Density of cooking oil is less than water

Fluid exerts forces in many directions. Try to submerse a rubber ball in water to see that an upward force acts on the

float.

• Fluids exert pressure in all directions.

• Water seeks its own level, indicating that fluid pressure is independent of area and shape of its container.

• At any depth h below the surface of the water in any column, the pressure P is the same. The shape and area are not factors.

Water level

Independence of Shape and Area

Example 5:

The figure shows a cross section of a dam. Calculate;

a.The pressure exerted by the water at X, if the density of water is 1000kg/m 2

b.Explain why the bottom of the dam is built with thick wall?

c.If there is air exerted and given that Patm is 101.3Kpa. What is absolute pressure at point x?

6.5mx0.5m

Solution:

a) Px = ρwatergh

= (1000)(9.81)(6.5-0.5)

= 580860 Pa

b) This is because the water pressure increase as the depth of water increases. So, a grater pressure is exerted at the bottom of the dam.

c) Pabsolute = Patm + Px

= 101.3 kPa + 580860 Pa

= 682.16 kPa

APPLICATION OF PRESSURE IN LIQUIDS

• Public water supply systems ( Water Tank)

• The wall of a dam

PASCAL PRINCIPLE

• Pascal’s principle states that pressure exerted on an enclosed fluid is transmitted equally to every part of the fluid.

P1 = P2

F1 = F2

• Pressure in Pascal can be expressed:

A1d1 = A2d2

Transmitting force

An applied force F1 can be “amplified”:

Hydraulic press

P F1A1

F2 F1A2A1

A common application of this is a hydraulic lift used to raise a car off the ground so it can be repaired at a garage.

Example

The cylindrical piston of a hydraulic jack has a cross-sectional area of 0.06 m2 and the plunger has a cross-sectional area of 0.002m2.a.The upward force for lifting a load placed on top of the large

piston is 9 000 N. calculate the downward force on the plunger required

b.If the distance moved by the plunger is 75cm, what is the distance moved by the large piston?

Solution:

a) F1 = F2

F1 = F2 A1

= 0.002 x 9 000

= 300 N

b) A1d1 = A2d2

d2 = A1 d1

= 0.002 x 75

= 2.5cm

APPLICATION OF PASCAL PRINCIPLE

• Hydraulic Brake• Hydraulic Jack

ARCHIMEDES PRINCIPLE

• Archimedes' principle is the law of buoyancy.

• It states that "Any object partially or completely submerged in a fluid is buoyed up by a force equal to the weight of the fluid displaced by the body."

Buoyant force = Weight of fluid displaced

Buoyant force = Weight object in air – weight in water

Buoyant force = ρVg

Volume of the submerged = Volume of the liquid object displaced

• Related buoyant force with the actual weight and apparent weight

Buoyant force makes things seem to be lighter.The weight of an object is its actual weight.The weight measured when the object is immersed in fluid is its apparent weight.The apparent weight loss of the object is due to buoyant force

Buoyant force = Apparent weight Loss

Law of Floatation

• A floating object displaces its own weight of fluid in which it floats.

• Buoyancy explains why some objects sink and others float.

• Objects that are less density than water will float.• Objects that are more density than water will sink.

• Objects that are the same density as water will neither sink nor float.

Law of Floatation

Which ball will sink in water?Which ball will float in water?

APPLICATION OF ARCHIMEDES PRINCIPLE

chapter 5 solid and liquid (bb101)

liquid liquid pressure

p pressure

pressure of water

characteristics of pressure

density of water

gas density

n area

ah density

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