Kinetic Energy

-         All matter is made of Particles

-         These Particles are in constant motion

-         Collisions in particles are understood to be completely elastic (no loss in E)

o       Remember, this is obviously not true, as any collision between two bodies will result in the releasing of heat energy produced by the friction between the two bodies.

-         We say that the “Mean free pass” or Average distance between the particles must travel in order to collide is High for gasses and Low for Solids

o       This is because the Intermolecular Forces (IMF) in Solids are the strongest – too strong to be broken by gravity alone

o       As one adds Energy to a solid, it will eventually reach its melting point, at which point, the solid will be unable to contain the Energy in the particles and the Energy is instead used to break Intermolecular Forces (IMF) within the substance

o       The same is true for the change from Liquid to Gas

-         Pressure

o       SI unit = Pascal (Pa)

o       STP = Standard Temperature and Pressure (“Normal” Conditions)

§         Average pressure of Air and sea levels under normal conditions

§         STP =

·        101.3 kPa = 760 mmHg = 1 atm = 14.7 lb/in²

o       (From this we derive that 1 kPa = 7.50 mmHg)

·        100°C = 373 K

o       Pressure can be measured by using a device called the Manometer

§         A Manometer consists of a Gas Chamber and an arm that is bent in a U-Shape.  Mercury is at the bottom of the U-Shaped arm, and, as the pressure in the gas chamber changes, the Mercury will move in regards to the difference in pressure from one side of the tube (in an open armed manometer – the air pressure) to the other (the gas chamber)

§         Close Arm – No Air

·        P_gas = P_Hg

·        Example:  The difference in Hg levels is 238mm Find P_g in kPa.

o       P_gas = P_Hg

o       P_gas = 238mm/1 * 1kPa/7.50 mmHg = 31.75 kPa

§         Open Arm – Arm is open to Air Pressure

·        P_gas = P_air ± P_Hg

o       If P_air > P_gas then

§         P_gas = P_air - P_Hg

o       Else if P_air < P_gas then

§         P_gas = P_air + P_Hg

o       Hint: “ALA” – Air less add

o       “AMS” – Air more subtract

·        Example: In an Open Arm Manometer, the difference in Hg levels is 6mm higher in the Arm connected to the gas cylinder.  Find the pressure of the gas in kPa if air pressure is 101.3kPa.

P_air > P_gas

(Draw picture to see this – if the Hg is higher on the side connected to the gas chamber then air pressure is greater than the gas pressure and vice versa if the Hg is higher on the side exposed to the Air Pressure.)

P_gas = P_air – P_Hg

P_gas = 101.3 kPa – (6mmHg/1 * 1 kPa/7.50 mmHg)

P_gas = 101.3 kPa – 0.80 kPa

P_gas = 100.5 kPa

Special Manometer – Barometer – measures Barometric pressure

-         Kinetic Energy (Cont.) = Energy of Motion

-         KE = ½mv²

o       m  = Mass (grams)

o       v = Velocity (Speed, Direction)

§         Velocity is a Vector Measurement meaning it includes Direction

-         Temperature – Measure of Average KE

o       Kelvin = SI unit of Temperature

§         K = °C + 273

o       (P₁V₁/T₁) = (P₂V₂/T₂)

§         P = Pressure

§         V = Volume

§         T = Temperature

o       Absolute Zero

§         The Temperature at which molecular movement stops completely

-         Arrangement of molecules in Compounds

o       Crystal Lattice – specific shape

§         NaClNaClNaCl
ClNaClNaClNa
NaClNaClNaCl
ClNaClNaClNa
NaClNaClNaCl

o       Amorphous – no specific arrangement of molecules

§         Glass

§         Asphalt

o       Allotrope – Different forms of the same ELEMENT

§         Carbon = Graphite and Diamond

-         Characteristics of the different Phases

               Shape        Volume     Particle Movement                          IMF

Solid       Definite       Definite       Vibrating                                       Strong

Liquid     Indefinite    Definite       Vibrating with Motion                Weaker

Gas        Indefinite    Indefinite      Move freely in straight lines      Weakest

Plasma – e^- and ions⁺ in random motion at high speeds

Vapor – Gas state of substance normally a (lq) or (s) at room temperature

Melting – Solid to liquid (+E)

               Normal melting point for water = 0°C

Freezing – Liquid to solid (-E)

               Normal freezing point for water = 0°C

T_F = T_M

Boiling – Liquid to gas (+E)

               Normal boiling point for water = 100°C

Condensation – Gas to liquid

               Normal condensation point for water = 100°C

T_B = T_Cond

Sublimation – Solid to gas (without stopping at the liquid phase) (+E)

               (Also gas to solid (-E))

Specific Heat Capacity (Cp) – E needed to raise 1 g of substance 1 degree centigrade

Phase diagram – Shows the different phases of a substance in regards to Pressure and Temperature – it is divided into 3 sections- solid, liquid, and gas

Remember that STP = 0°C and 1atm or 101.3 kPa or 760 mmHg

Key Terms

1. Triple Point – point at which all 3 phases can exist in equilibrium
2. Critical Point – The point above which no amount of pressure will result in the liquefying of the gas
3. Melting/Freezing point = Line between Liquid and Solid
4. Sublimation point = Line between Gas and Solid
5. Boiling/Condensation point = Line between Liquid and Gas

a. Note – At any point on one of the dividing lines, the substance can exist in equilibrium with the phases that the line borders.

6. Normal Boiling Point – Temperature at which the substance boils under 101.3 kPa or 760 mmHg
7. Normal Melting Point – Temperature at which the substance melts under 101.3 kPa or 760 mmHg

This chart represents the change in state in a given substance.  As the substance gains Energy (Heat), the temperature of the substance will rise until it plateaus at certain points.

A – Solid

B – Solid à Liquid (melting)

C – Liquid

D – Liquid à Gas (Boiling)

E – Gas

In the chart above, the line plateaus at B.  During this point, the Energy being gained by the substance is used to break the intermolecular forces (IMF).  The same is true for section D.

By using the chart above, one can find the melting and boiling points of the given substance.

               T_M = 19.5°C

               T_B = 50.0°C

Change in Temperature problems

Enthalpy of fusion – Amount of Energy needed to melt 1 g of substance at normal melting point

Entalpy of vaporization – Amount of Energy needed to vaporize 1 g of substance at normal vaporization point

(Enthalpy = Heat)

ΔT formulas

q = mΔTCp

For melting/freezing

q = mH_fus

For Boiling/Cond

q = mH_vap

Eample:  How much Energy is needed to change the temperature of H₂O from –10.0°C to 150°C?

(You do not have to memorize these constants – the information you need will be given to you on the test.)

For H₂O: Cp(s) = 2.06 J/g°C      H_fus = 334 J/g                                                  

               Cp(l) = 4.18 J/g°C       H_vap = 2260 J/g

               Cp(g) = 2.02J/g°C

5 steps:

1)      Solid à Melting Point

a. q = mΔTCp

b.q = (10.0g)(10.0°C)(2.06J/g°C)

c. q = 206 J

2)      Melting Point à Liquid

a. q = mH_fus

b.q = (10.0g)(334J/g)

c. q = 3340 J

3)      Liquid à Boiling Point

a. q = mΔTCp

b.q = (10.0g)(100°C)(4.18 J/g°C)

c. q = 4180 J

4)      Boiling Point à Gas

a. q = mH_vap

b.q = (10.0g)(2260 J/g)

c. q = 22600 J

5)      Gas (100°C) à Gas (150°C)

a. q = mΔTCp

b.q = (10.0g)(50°C)(2.02 J/g°C)

c. q = 1010 J

6)      Total Energy Required

a. 206 J + 3340 J + 4180 J + 22600 J + 1010 J = 31336 J

b.Answer: 31300 J

c. Remember Signifigant digits for addition – round to least accurate digit