Boost PC performance: How more available memory can improve productivity
Imfa
1. INTERMOLECULAR FORCES OF ATTRACTION
Outside and
weak
attraction
Physical
properties
Physical
states
Boiling
point
compounds
Partial
charges
Farther
distance
solubility
are
between
explains
2. KIND OF IMFA COMPOUNDS THAT FORMED THEM
electrostatic Ionic & ionic
Ion dipole Ionic & polar
H bond
Polar H terminal & lone pair of small
Electronegative atom
Dipole dipole Polar & polar
Ion induced dipole Ionic & non polar
Dipole induced diipole Polar & non polar
dispersion Non polar & non polar
Decreasing
strenght
11. IMFA and Solubility
1. Dissolved substance
(solute) must separate
(IMFA breaking-
endothermic heat)
2. Dissolving substance
(solvent) must separate
(IMFA breaking-
endothermic heat)
3. Solute and solvent must
mix (IMFA forming-
exothermic heat)
For dissolving to
happen, 3
processes must
occur
12. IMFA between solute and solvent > IMFA among solute and or IMFA among solvent
_
• Summarized as : LIKE DISSOLVES LIKE ;
• polar solvent dissolves polar solute
• Non polar solvent dissolves non polar solute
18. MELTING FREEZINGEVAPORATION CONDENSATION
S to L L to SG to LL to G
Temp > IMFA Temp >>> IMFA
Temp < IMFA Temp <<<IMFA
ENDO EXO
Sorrounding
cold
Sorrounding
Warm
IS IS IS IS
WHENWHEN WHEN WHEN
makes makes
is is
20. TE MP
E NE R GY
-4
0
s
l
l - g
s - l
g
80 cal/g
540 cal/g
100
25
Heat of Fusion- amount of
energy needed to melt 1 gram
of a substance at its melting
point
H fusion water = 80 cal/g
Q= mass X H fusion
Heat of Vaporization – amount
of energy needed to evaporate
1 gram of a substance at its
boiling point
H vap water = 540 cal/g
Q = mass X H vap
Heat of Freezing = energy
released to change 1 gram of
liquid to solid
Heat of Condensation= energy
released to convert 1 gram of gas to
liquid
SPECIFIC HEATS
Energy involved to change the
temperature of 1 gram of a
substance, 10Celsius
21. LATENT HEATS
• Heat of Fusion- amount of energy needed to
melt 1 gram of a substance at its melting
point
• H fusion water = 80 cal/g
• Q= mass X H fusion
22. LATENT HEATS
• Heat of Freezing = energy released to change
1 gram of liquid to solid
•
• Heat of Fusion (endo) = Heat of Freezing (exo)
23. LATENT HEATS
• Heat of Vaporization – amount of energy
needed to evaporate 1 gram of a substance at
its boiling point
• H vap water = 540 cal/g
• Q = mass X H vap
24. LATENT HEATS
• Heat of Condensation= energy released to
convert 1 gram of gas to liquid
•
• Heat of Vaporization (endo) = Heat of Condensation(exo)
25. SPECIFIC HEATS
• Energy involved to change the temperature of
1 gram of a substance, 10Celsius
• For water: Sp. Heat = 1 cal/g-0C
27. IMFA & Boiling Point
Because of the KE energy of
evaporating liquids, they break their
IMFA and result in some of them going
into the gaseous state
and exert a pressure called vapor
pressure.
Under a given temperature each liquid
has its own vapor pressure.
When the vapor pressure of escaping
liquid molecules become equal to the
pressure of the air above it. The
temperature at that point is called the
boiling point of the liquid.
The stronger the IMFA, the harder to
break, the longer time the liquid to
evaporate the higher the boiling point
Vapor pressure
Air pressure
• Vapor pressure = air pressure : Boiling point
28. IMFA & Boiling Point
• IMFA increases, boiling point increases
• MWt increases, IMFA increases, boiling point
increases
• Branching increases, IMFA decreases, boiling
point decreases