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"You can dance anywhere, even if only in your heart." ~Unknown
"If dancing were any easier it would be called football."
~anonymous
KINETIC-MOLECULAR THEORY
DESCRIBES THE BEHAVIOR OF MATTER
IN TERMS OF PARTICLE MOTION
SOLIDS – PARTICLES ARE PACKED
CLOSE TOGETHER AND VIBRATE ABOUT
FIXED POINTS
LIQUIDS – PARTICLES ARE STILL
CLOSE TOGETHER, BUT CAN MOVE
PAST EACH OTHER
GASES ARE SEPARATED FROM EACH
OTHER BY EMPTY SPACE.
THE PARTICLE VOLUME IS SMALL
COMPARED TO THE VOLUME OF EMPTY
SPACE.
THE PARTICLES ARE FAR ENOUGH
APART SO THAT THEY HAVE LITTLE
EFFECT ON EACH OTHER.
GAS PARTICLES ARE IN CONSTANT,
RANDOM MOTION.
COLLISIONS WITH EACH OTHER
AND WITH THE WALLS OF THE
CONTAINER ARE PERFECTLY
ELASTIC.
IN AN ELASTIC COLLISION, NO
ENERGY IS LOST, BUT ENERGY CAN
BE EXCHANGED.
THE AVERAGE KINETIC ENERGY OF THE
PARTICLES IS A FUNCTION OF
TEMPERATURE.
KE = ½ mV2
where m = particle mass
V = velocity
THE KINETIC-MOLECULAR THEORY IS
USEFUL IN EXPLAINING THE
BEHAVIOR OF ALL STATES OF MATTER
AND THE TRANSITIONS FROM ONE
STATE TO ANOTHER, BUT IT IS
ESPECIALLY USEFUL WITH GASES.
WHY SOME SUBSTANCES ARE IN
DIFFERENT STATES (SOLID, LIQUID,
OR GAS) AT ROOM TEMPERATURE
WHEN THEY ALL HAVE THE SAME
AVERAGE KINETIC ENERGY HAS TO DO
WITH THE INTERMOLECULAR FORCES
OF ATTRACTION.
THE STRONGER THESE FORCES, THE
MORE LIKELY THE SUBSTANCE WILL
BE IN ONE OF THE CONDENSED
STATES (SOLID OR LIQUID).
EARLIER, WE USED THE EXAMPLE OF
WATER – THE MOLECULES HAVE SOME
ADDITIONAL “STICKIENESS” BECAUSE
OF THE POLAR BONDS.
COMPARED TO OTHER MOLECULES
WITH SIMILAR MOLECULAR MASS,
BUT WITH PURE POLAR BONDS,
WATER HAS AN UNUSUALLY HIGH
BOILING POINT AND MELTING POINT.
NOW, LET’S USE WATER AS AN
EXAMPLE AND CONSIDER WHAT
HAPPENS WHEN WE INCREASE THE
AVERAGE KINETIC ENERGY
(TEMPERATURE) ON A SUBSTANCE IN
ORDER TO CAUSE IT TO UNDERGO
CHANGES IN STATE.
THE HEATING CURVE FOR WATER IS
GIVEN ON THE NEXT SLIDE.
1) IN THE SOLID, THE PARTICLES OCCUPY
FIXED POSITIONS IN THE CRYSTAL
STRUCTURE. AS WE ADD ENERGY, THE
VIBRATIONS ABOUT THESE FIXED
POINTS INCREASE – THE TEMPERATURE
INCREASES.
2) WHEN THE MELTING POINT IS
REACHED, SOME PARTICLES HAVE
ENOUGH ENERGY TO BREAK AWAY FROM
THE SOLID STRUCTURE. AT THIS
POINT, ANY ENERGY ADDED WILL GO
TOWARDS MELTING. THE TEMPERATURE
WILL REMAIN CONSTANT.
3) AFTER THE SOLID HAS MELTED, ANY
ADDITIONAL ENERGY ADDED WILL GO
TOWARDS INCREASING THE
TEMPERATURE. THE PARTICLES CAN
MOVE RELATIVE TO EACH OTHER, AND
THEIR SPEEDS WILL INCREASE.
4) AS THE TEMPERATURE OF THE
LIQUID INCREASES, THE VAPOR
PRESSURE OF THE LIQUID WILL
INCREASE.
5) AT THE BOILING POINT, THE VAPOR
PRESSURE OF THE LIQUID EQUALS THE
EXTERNAL PRESSURE.
6) AS ENERGY IS PUT INTO THE
SYSTEM AT THE BOILING POINT, IT
WILL GO TO CONVERTING THE LIQUID
TO VAPOR. THE TEMPERATURE WILL
STAY CONSTANT UNTIL ALL OF THE
LIQUID HAS BEEN VAPORIZED.
7) AFTER THE SUBSTANCE HAS BOILED,
ANY ADDITION OF ENERGY WILL GO TO
INCREASING THE TEMPERATURE OF THE
VAPOR.
SOME DEFINITIONS
HEAT CAPACITY – THE AMOUNT OF
HEAT REQUIRED TO RAISE THE
TEMPERATURE OF 1 GRAM OF A
SUBSTANCE 1o C.
USUALLY MEASURED IN CALORIES OR
JOULES.
1 CALORIE = THE ENERGY REQUIRED TO
RAISE THE TEMPERATURE OF 1 GRAM OF
LIQUID WATER 1o C.
1 CALORIE = 4.184 JOULES
THE FOOD CALORIE IS ACTUALLY 1
KILOCALORIE OR 1000 CALORIES.
HEAT OF FUSION – THE HEAT
REQUIRED TO MELT 1 GRAM OF A
SUBSTANCE AT ITS MELTING POINT.
HEAT OF VAPORIZATION – THE HEAT
REQUIRED TO CONVERT 1 GRAM OF A
LIQUID TO A VAPOR AT ITS BOILING
POINT
CONSTANTS FOR WATER
DHf = heat of fusion = 6.01 kJ/mol
DHvap = heat of vaporization = 40.7 kJ/mol
DH = heat capacity of water = 4.18 J/g deg
DH for ice = 2.108 J/g deg
DH for steam = 1.996 J/g deg
PROBLEM: HOW MUCH ENERGY WOULD
BE REQUIRED TO RAISE THE
TEMPERATURE OF 20 g OF WATER
FROM 25o TO 35o C?
DH = 4.18 J/g deg
PROBLEM: HOW MUCH ENERGY
WOULD BE REQUIRED TO MELT 80 g
OF ICE?
DHF = 6.01 kJ/mol
PROBLEM: HOW MUCH ENERGY
WOULD BE REQUIRED TO BOIL 1000 g
OF WATER?
Dhvap = 40.7 kJ/mol
PROBLEM: HOW MUCH ENERGY
WOULD BE REQUIRED TO CONVERT
50 g OF ICE AT 0o C TO STEAM AT
100o C?
DHf = 6.01 kJ/mol
Dhvap = 40.7 kJ/mol
DH = 4.18 J/g
A PHASE DIAGRAM REPRESENTS THE
EFFECTS OF TEMPERATURE AND
PRESSURE ON THE STATE OF A
SUBSTANCE IN A CLOSED CONTAINER.
THE LINES SEPARATING THE REGIONS
REPRESENT THE TEMPERATURES AND
PRESSURES AT WHICH THE PHASES
WOULD BE IN EQUILIBRIUM.
NOTE THAT THE LINE BETWEEN ICE
AND WATER HAS A SLIGHT NEGATIVE
SLOPE – MOST SUBSTANCES HAVE A
SLIGHT POSITIVE SLOPE.
THE TRIPLE POINT REPRESENTS THE
ONLY CONDITION OF TEMPERATURE
AND PRESSURE THAT A SUBSTANCE
CAN EXIST IN ALL THREE STATES AT
THE SAME TIME.
THE CRITICAL POINT REPRESENTS THE
LAST POINT AT WHICH A LIQUID AND
A GAS CAN COEXIST IN EQUILIBRIUM.
BEYOND THE CRITICAL POINT, THE
LIQUID AND GAS ARE NO LONGER
DISTINGUISHABLE.