Chapter 10
States of Matter
• Solid – definite shape, definite volume
– def. shape  part. fixed position
• part. locked in place by IMF
• vibrate in place
– def. vol.  part. close together
• held together by IMF
• Liquid – no definite shape, definite volume
– no def. shape  part. not locked in one
position
• randomly arranged/lack crystal structure
• part. have enough NRG to move past each other
– fluidity – ability to flow
– viscosity – resistance to flow
– def volume  part. close together
• part. held by IMF
• density similar to solid
• cohesion – prop. caused by particles
attracting to one another
– beads of water
– raindrops
– liquid Hg - video
• surface tension – “skin” on surface of
liquid due to cohesion - video
• adhesion – prop. caused by a liquid
attracting to solid surfaces
– water meniscus
– capillary action in plants(xylem) - video
Cohesion and surface tension
Chapter 11
5
• Gas/vapor – no definite shape, no definite
volume
– no def. shape  part. randomly arranged
• fluidity
– no def volume  part. very spread out
• part. 1000 times further apart than liquid/solid
• IMF virtually nonexistent
– independent part.
– part. NRG high enuf to break all IMF
» part. “zip” around, collide, bounce off, repeat
• phase/state changes
– difference in three phases – NRG
• kinetic NRG - NRG of motion
– temp – measure of avg. KE of a systems particles
• potential NRG – NRG of position
– stored NRG
– adding NRG to a sub. causes its part. to move
more(i.e. increase in temp)
– at some point, NRG causes part. to move
enuf to cause a phase change
Solid + NRG  liquid
melting
- solid part. gain enuf NRG to break free
from fixed position
- move too much to stay in same spot
- still attracted to other particles(IMF)
- melting point – temp. and press. conditions @
which a solid changes to a liquid
- simulation
liquid  solid + NRG
freezing
- liquid loses NRG causing part. to move so
little they become stuck on one spot
- part. still moving – vibrate in fixed position
- freezing point – temp. and press. conditions @
which a liquid changes to a solid
- quantity of NRG added to melt solid =
quantity of NRG released to
freeze liquid
liquid + NRG  gas
evaporation
- evaporation occurs when part. on surface
gain enuf NRG to overcome IMFs and
break free from the liquid
gas  liquid + NRG
condensation
- gas part. collide with surface(liquid or
solid), lose NRG and slow down
- slow moving part. stick together w/ IMFs
- independent gas part. form groups of
particles(droplet)
solid + NRG  gas/vapor
sublimation
• direct change from solid to gas/vapor
– no liquid phase
• mothballs
• dry ice
• ice on very cold day
gas/vapor  solid + NRG
deposition
• direct change from gaseous phase to solid
– no liquid phase
•
•
•
•
frost
synthetic diamonds
coatings on jet wings leading edge(CVD)
electrostatic paint
gas/vapor
Δ KE
liquid
Δ KE
solidΔ KE
Δ PE
melting/
freezing
point
boiling/
condensing
point
Δ PE
(potential NRG
change)
• supercooling
– phenomena in which the temp of a liquid is
below its freezing pt. without the liquid
freezing
ice chandelier
• superheating
– aka – flash boiling, boiling delay
– phenomena in which a liquid is heated to a
temperature higher than the liquids boiling
point without boiling
• requires a homogeneous substance and no
nucleation sites
• specific heat
– NRG needed to change 1 g of a substance by
1o C or 1 K
– 1 calorie = NRG needed to change 1 g of H2O
by 1oC
– 1 cal = 4.18 joules(J)
– 4.18 J = NRG needed to change 1 g of H2O
by 1oC
– 1 food Cal = 1 kcal = 1000 cal = 4180 J
• molar heat
– NRG need to change 1 mol of a substance by
1oC or 1 K
– molar heat H2O = 75.3 J/mol x K
• enthalpy change in phase changes
– enthalpy of fusion(melting) – NRG needed to
change a quantity of a solid at its melting
point to liquid at same temp
• enthalpy change in phase changes
• molar heat of fusion(ΔHfus-H2O = 6.01 kJ/mol)
• or heat of fusion = 334 J/g
– sample problem
How many joules need to be added to 350.0 grams
of ice @ 0oC to create water at 0oC?
X J = 1000 J x 6.01 kJ x 1 mol x 350.0 g
1 kJ
1 mol
18.02 g
X J = 117,000 J
– enthalpy of vaporization – NRG needed to
change a quantity of liquid @ its boiling point
to gas/vapor @ the same temperature
• molar heat of vaporization(ΔHvap – H2O = 40.7
kJ/mol)
• or heat of vaporization = 2260 J/g
– enthalpy of vaporization – NRG needed to
change a quantity of liquid @ its boiling point
to gas/vapor @ the same temperature
• molar heat of vaporization(ΔHvap – H2O = 40.7
kJ/mol or 2260 J/g)
sample problem
What is the temperature of 450.0 grams of
water that started at 125oC steam if it
loses 1.06 x 106 joules?
– need 3 constants
1. specific heat of steam = 2.06 J/gCo
2. molar heat of vaporization = 40.7 kJ/mol
3. specific heat of water = 4.18 J/gCo
step #1 – calc NRG given off when steam
changes temp from 125oC to 100oC.
x J = 2.06 J x 450.0 g x 25oC
g Co
x J = 2.32 x 104 J
step #2 – calculate the NRG released when
450.0 grams of steam @ 100oC changes
to 450.0 grams of water @ 100oC
x J = 1000 J x 40.7 kJ x 1 mol x 450.0 g
1 kJ
1 mol
18.02 g
x J = 1.02 x 106 J
step #3 – with remaining joules released,
calculate ΔT for the 100oC water
total NRG released = 1.06 x 106 J
step #1 =
2.32 x 104 J
step #2 =
1.02 x 106 J
NRG released when water cools = 1.68 x 104 J
from 100oC to ???
x Co = g Co x 1.68 x 104 J
4.18 J 450.0 g
x Co = 8.93oC = ΔT
water temp = 100oC – 8.93oC
water temp = 91.07oC
• for tomorrow(10 pts)
– what phase of matter is H2O in and what is
the temperature if 75.0 grams of ice @
-35.0oC gains 75,500 J?
10.4 Phase equillibrium
• 2 phase system
• consist of two phases of matter
• ice water
• clouds(air and water)
• equillibrium – a process in which two
opposing changes occur simultaneously
• melting and freezing
solid + NRG ↔ liquid
• evaporation and condensation
liquid + NRG ↔ gas
• in a closed container, a liquid will establish
equilibrium with its vapor
• creates vapor pressure
– force caused by collisions of gas particles on
the surface of its liquid
• boiling point - temperature @ which the
vapor pressure of the liquid = external
pressure on the surface of liquid
– normal boiling point - temperature when vapor
pressure = air pressure
• LeChatlier’s principle – if a stress is
applied to a system @ equilibrium, the
system will absorb the stress and reach
equilibrium at new conditions
• phase diagrams
– diagram of phases of matter and equilibrium
conditions
– graphed of temp as indep. variable and press
as dep. variable
• three main points of phase diagram
1) triple point - temperature and pressure at
which all three phases are in equilibrium.
•
•
•
solid ↔ liquid ↔ gas
video
video
2) critical point - the temperature and pressure
conditions at which the distance between the
liquid particles and gas particles are the
same.
• usually very high pressure(squeezing the gas
particles close together)
3) normal melting and boiling points
– temperature at which the solid melts and
liquid boils @ normal pressure(1 atmosphere)
– each substance has its own unique phase
diagram
•
caused by IMFs and molar masses
•
link
• Why is ice slippery?
– At temperatures as low as 200K, the surface of ice is
highly disordered and water-like. As the temperature
approaches the freezing point, this region of disorder
extends farther down from the surface and acts as a
lubricant.
– The illustration is taken from from an article in the April 7, 2008
issue of C&EN honoring the physical chemist Gabor Somorjai
who pioneered modern methods of studying surfaces.
Water Pseudoscience
• Water pseudoscience and quackery
1 atmosphere