50.00 g of a solid at o C is placed in g of its liquid in a g can at o C. The temp. equilibrates at o C. Calculate: H f of the solid C liquid = 2.00 J/gK C can = 1.00 J/gK MP solid = o C

Chapter 13 States of Matter

Fluid Any substance that flow and has no definite shape

Fluids Both liquids & gases are fluids

Pressure Applying a force to a surface results in pressure

Pressure (P) Force per unit area P = F/A

Common Units of P AtmTorcelli mm Hginches Hg Barmillibar Pascalkilopascal

Std Unit for Pressure Kilopascal (kPa) Pa = N/m 2

Standard Pressure kPa760 mmHg ~ 1 x 10 5 Pa 1.0 Atm 30 in Hg1013 mbar

If std. air pressure = kPa, calculate the force on a desk top that is 50.0 cm x cm.

Each car tire makes contact with the ground on an area that is 5.0 x 6.0 inches. The tire pressure is 40 psi. Calculate the weight of the car.

Fluids at Rest

Pascal’s Principle Pressure applied at any point in a confined fluid is transmitted undiminished throughout the fluid

Pressure Formula P = FAFA

In Constant Pressure P 1 = P 2

Thus F 1 F 2 A 1 A 2 =

Thus F 1 A 2 A 1 F 2 =

Thus A 2 A 1 F 2 =F 1 ( )

Using a car jack a person applies 50.0 lbs force on a 2.0 mm wide cylinder driving the fluid into a 2.0 cm wide piston. Calculate the force applied to the piston.

Density (  ) Mass/unit volume m V  =

In a fork lift, fluid is pumped through a 0.20 mm tube with a force of 15 N. The small tube opens up into a 20.0 cm cylinder driving a piston with ___ N

Gases

Boyle’s Law At constant temperature, pressure & volume are inversely proportioned

Boyle’s Law P 1 V 1 = P 2 V 2

Charles’ Law At constant pressure, volume & temperature are directly proportioned

Charles’ Law V 1 V 2 T 1 T 2 =

Guy L’ Law At constant volume, pressure & temperature are directly proportioned

GL’ Law P 1 P 2 T 1 T 2 =

C G Law P 1 V 1 P 2 V 2 T 1 T 2 =

Ideal Gas Law PV = nRT

Ideal Gas Constant R = 8.31 LkPa moleK

Ideal Gas Constant R = 8.31 J moleK

Ideal Gas Constant R= LAtm moleK

Calculate the new volume if the conditions of 20.0 L gas at 27 o C under 150 kPa is changed to 177 o C under 180 kPa:

Calculate the change in volume if the pressure on a gas is halved while its temperature is tripled:

Calculate the volume moles of gas at 27 o C under 150 kPa: R = 8.31 LkPa/moleK

Calculate the number of moles of gas occupying 831 mL at 77 o C under 175 kPa: R = 8.31 LkPa/moleK

Pressure in Water The pressure applied by the weight of water pressing down on an area

Pressure Formula P = WAWA

W = mg, thus: P = mg A

m =  V, thus: P =  Vg A

V = Ah, thus: P =  Ahg A

A’s cancel, thus: P =  hg

The Pressure of a Body of Water P =  hg

The Pressure of a Body of Water Because  & g are constant, P depends on only the height

Buoyant Force The upward force applied by water due to the pressure difference between the top & bottom of a submerged object

Buoyant Force F top = P top A =  hgA F btm = P btm A =  h + l)gA

Buoyant Force F buoy = F btm - F top F buoy =  h + l)gA -  hgA F buoy =  lgA =  Vg

Buoyant Force F buoy =  Vg F buoy = W water

Archimedes’ Principle A object immersed in fluid has an upward force equal to the weight of fluid displaced

1.0 m 3 of aluminum is immersed in water.  al = 2.70 g/cm 3. Calculate: W, F buoy, & F net acting on the aluminum block.

Calculate the number of moles of gas at 127 o C under 83.1 kPa pressure in a spherical hot air balloon with a 40.0 m diameter. (1 cm 3 = 1 mL)

Calculate the volume of 5.00 moles of gas at 127 o C under kPa pressure. R = 8.31 LkPa/moleK

Fluids in Motion

Bernoulli’s Principle As the velocity of a fluid increases, the pressure exerted by that fluid decreases

Cohesion Attractive forces between particles in the same substance or like particles

Adhesion Attractive forces between particles of different substances or different particles

Surface Tension Cohesive forces reduce surface area making a surface seem like it has a film holding it together

Capillary Action When a fluid rises up a thin tube due to adhesion being stronger than cohesion

Capillary Action When a fluid rises up a thin tube due to both adhesion & cohesion

Evaporation When particles gain enough kinetic energy to escape a surface

Volatile Evaporates easily

A boy pumps fluid with a force of 50.0 lbs through a 0.10 mm tube which opens up into a 40.0 cm cylinder driving a piston with ______ lbs of force.

Condensation Vapor particles clumping together enough to return to the liquid phase

Solids Rigid phase with a definite crystal structure

Solids All true solids have a definite crystal structure

Amorphous Solids Solids that do not have a definite crystal structure

Crystal Lattice Three dimensional arrangement of unit cells

Unit Cell The smallest repeating unit making up a solid

Elasticity The ability of a substance to return to its original form when bent or twisted

Thermal Expansion The increase in size of a substance due to an increase in temperature

Thermal Expansion Think of mercury in a thermometer, liquids just take up more space

Thermal Expansion Because solids are rigid, their length increases with increased temp.

Thermal Expansion  L =  L  T  L = change in length  = linear expansion coefficient

Thermal Expansion L f = L i +  L L f = L i +  L  T L f = L i +  L(T f - T i )

Thermal Expansion When dealing with volume: use   = volumetric expansion coefficient

Calculate the change in length of a 4.00 m strip of aluminum when the temperature rises from –25 o C to 75 o C.  Al = 25 x / o C

Calculate the length at its MP of 1221 o C of a 20.0 m strip of steel at 21 o C.  Fe = 12 x / o C

Calculate the length of a glass pole at 75 o C when it’s m at -25 o C.  glass = 3 x / o C

A 2.0 m by 2.0 m window is enclosed in an aluminum frame. Determine gap size.  glass = 3 x / o C  Al = 25 x / o C

Calculate the length of a steel pole at 175 o C when it’s m at -25 o C.  steel = 12 x / o C

A m rod expanded to m when heated from 25 o C to 275 o C. Calculate:  rod

Calculate the length of an aluminum pipe at 225 o C when it is 8.00 m at 25 o C.  Al = 25 x / o C

Calculate the increase in force when a fluid in a 2.0 mm line opens up into a 20.0 cm piston:

A 50.0 g rock at o C is placed into g of water in a 1.0 kg calorimeter both at o C. The temperature equilibrates at o C. Calculate the specific heat of the rock. C water = 4.18 J/gK C cal = 1.00 J/gK

50.0 g of steam at o C is pumped into 250 g of water in a g calorimeter both at 25.0 o C. Calculate the equilibrium temperature. C water = 4.18 J/gK, C cal = 1.00 J/gK, C steam = 2.02 J/gK, Hv = 2260 J/g