"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 mV 2 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 1 o C. USUALLY MEASURED IN CALORIES OR JOULES. 1 CALORIE = THE ENERGY REQUIRED TO RAISE THE TEMPERATURE OF 1 GRAM OF LIQUID WATER 1 o C. 1 CALORIE = 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  H f = heat of fusion = 6.01 kJ/mol  H vap = heat of vaporization = 40.7 kJ/mol  H = heat capacity of water = 4.18 J/g deg  H for ice = J/g deg  H for steam = J/g deg

PROBLEM: HOW MUCH ENERGY WOULD BE REQUIRED TO RAISE THE TEMPERATURE OF 20 g OF WATER FROM 25 o TO 35 o C?  H = 4.18 J/g deg

PROBLEM: HOW MUCH ENERGY WOULD BE REQUIRED TO MELT 80 g OF ICE?  H F = 6.01 kJ/mol

PROBLEM: HOW MUCH ENERGY WOULD BE REQUIRED TO BOIL 1000 g OF WATER?  h vap = 40.7 kJ/mol

PROBLEM: HOW MUCH ENERGY WOULD BE REQUIRED TO CONVERT 50 g OF ICE AT 0 o C TO STEAM AT 100 o C?  H f = 6.01 kJ/mol  h vap = 40.7 kJ/mol  H = 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.