Transcript Intermolecular Forces

Chapter 5 - Gases
1) fluidity
– ability to flow
– mainly empty space
– random arrangement
2) low density
– part. very spread out
•
1000x further apart than solid or liquid
–
–
–
–
example – oxygen density
D–O2(gas) = 1.43 g/L or 0.00143 g/mL
D–O2(liquid) = 1140 g/L or 1.14 g/mL
D-O2(solid) = 1429 g/L or 1.429 g/mL
3) highly compressible
– no definite volume
– mainly empty space
4) fill any container(no def. shape)
– no attractive forces, independent part.
– expand to occupy all spaces of container
5) exert pressure
– gas pressure – force created when gas
particles collide with a surface
•
press = force/area
–
•
force – measured in newtons
» 1 N = 1 kg▪m/s2
pascal(Pa) = newton/m2
– standard measures of pressure
•
•
•
•
•
101,325 Pa or 101.325 kPa
760 mm Hg = 29.92 in. Hg
760 torr
14.7 psi
1 atm
– measured w/ barometer
•
•
discovered by Torricelli
air pressure, on an avg. day at
sea level, will support
760 mm(29.92 in.) of Hg
• aneroid barometer
• kinetic molecular theory(KMT)
– molecular model that describes behavior of
gases
– postulates of KMT
1) gas part. in constant random motion
–
no intermolecular forces
2) volume of a gas particle is negligible
3) collisions are elastic, no attractive or repulsive
forces
4) avg. KE is directly related to Kelvin temp.
• measurable properties of gases
1) pressure(P)
•
kPa, mm Hg, atm
2) temp.(T)
•
K, oC
3) volume(V)
•
mL, L, cm3
4) moles(n)
• Boyle’s law
• VP or VP
– volume of a gas is inversely related to the
pressure
• k = PV
or
V1P1 = V2P2
sample problem:
A balloon has a volume of 5.75 L and a pressure
of 0.975 atm. If Aunt Edith sits on the balloon
and decreases the volume to 1.75 L what is the
pressure of the gas inside the balloon?
k = PV
P1 V1 = P2 V2
P1 V1 = P2 V2
k= 0.975 atm x 5.75 L
V2
V2
k = 5.61 L x atm
P2 = P1 V1
P=k
V
2
V
P2 = 0.975 atm x 5.75 L
P = 5.61 L x atm
1.75 L
1.75 L
P2 = 3.20 atm
P = 3.21 atm
• Gay-Lussac’s Law
TP or TP 
– the pressure of a gas is directly related to the
temperature of the gas
• k=P
T
or
P1
P2
=
T1
T2
Sample problem –
A balloon has a pressure of 135.5 kPa at 25o
C, what is the pressure if the balloons
temperature was at -15o C?
k=P
T
k = 135.5 kPa
298
25o K
C
o C
k = 0.455
5.4 kPa/
kPa/K
kxT=PxT
T
P=kxT
o C
P = 5.4
kPa
x -15
0.455
kPa
x 258
K
o C
K
P = -81
117 kPa
kPa
Sample problem – A spray paint can has a
bursting pressure of 5.65 atm. If the can
has a pressure of 1.30 atm at 25o C, what
temperature would the can burst?
k=P
T
k = 1.30 atm
298 K
k = 0.00436 atm/K
Txk=P xT
T
Txk=P
k
k
T=P
k
T = 5.65 atm
0.00436 atm/K
T = 1.30 x 103 K = 1020o C
When a gas is heated, the volume of the gas
INCREASES
??
__________
• Charles’ Law
TV or TV 
– the volume of a gas is directly related to
the temperature of the gas
• k=V
T
or
V1
V2
=
T1
T2
• Charles also proved the change in volume
of a gas was 1/273 the original volume for
every o C change
– what would happen to 1.0 L of gas if the
temperature decreased by 273o C????
Sample problem:
If a weather balloon contains 45 L of
hydrogen gas at 15o C, what is the volume
when it gets up in the atmosphere and the
temperature is -45o C?
k=V
kxT=V
T
V = 0.16 L x 228 K
k = 45 L
K
288 K
V = 36 L
k = 0.16 L
K
Boyle’s Law k = PV
Gay-Lussac’s Law - k = P/TT
Charles’ Law k = V/T
T
Combined Gas Law - k = _____
or
P1 V1 = P2 V2
T1
T2
Sample problem –
An aerosol can has propane gas as a
propellant. What would the volume of
propane be at STP if the volume of
propane in the can is 135 mL at a
temperature of 22o C and pressure of 1.73
atm?
Sample problem –
When I was 6, I accidentally let go of my
helium balloon from the circus. The
volume of the balloon when I let go was
2.49 liters at 18o C and 1.22 atm of
pressure. When the balloon was struck by
a jet passing over at higher altitude, what
was its volume if the temperature was 24o
C less and the pressure was 674 torr?
Avogadro’s law
– equal volumes of gases, at the same T and P,
contain the same # of particles(Avogadro’s
principle)
– V of a gas is directly related to the particles of
gas(moles)
• nV or nV 
k=V
or
V1
V2
=
n
n1
n2
– basis for Avogadro’s #(6.023 x 1023 )
– molar volume – the volume of 1 mol of any
gas @ STP
• 22.41 L of any gas @ STP = 1 mol any gas
Ideal gas law
- describes the behavior of a gas that always
and perfectly follows the gas laws
- ideal gas properties
- no IM forces
- particles have no volume
- no condensation
- real gases follow gas laws most of the time
- except at high P
- except at low T
- except any time intermolecular forces affect gas part.
Ideal Gas Law  PV = nRT
P = pressure
V = volume
R = PV
n = moles
nT
T = temperature
R = 0.08206 L• atm/mol • K
R = 8.314 L • kPa/mol • K
R = 62.36 L • torr/mol • K
Sample problem –
How many moles of carbon dioxide gas are
there if 35.6 liters has T = 35o C and 96.3
kPa?
Sample problem –
How many liters of helium gas are there if
15.5 grams has T = 15o C and 105.3 kPa?
Dalton’s law of partial pressure –
• the total P of a mixture of gases is the sum
of the partial P of each gas
Ptotal = P1 + P2 + P3 + Pn
Sample problem
diffusion
– the even spreading out of a substance from
an area of high concentration to areas of low
concentration
effusion – passage of a gas thru a small
opening or hole
Graham’s law of diffusion
– the rate of diffusion of a gas is inversely
related to the square root of the gases molar
mass
– compare rates of diffusion of different gases
‫ט‬A
MB
=
‫ט‬B
MA
Sample problem –
If helium gas and carbon dioxide gas effuse
out of the same hole, how do their rates
compare?
rate He
44.01 g/mol
3.32
=
=
rate CO2 4.00 g/mol
1.00
• He effuses 3.32 times faster than CO2
Gas Stoichiometry
• gas volumes correspond to mole ratios
3 H2(g) + N2(g)  2 NH3(g)
• coefficients represent
– 3 moles hydrogen react with 1 mole nitrogen
to form 2 moles ammonia
– 3 liters hydrogen react with 1 liter nitrogen to
form 2 liters ammonia(for gases only)
– 67.23 liters hydrogen @ STP react with 22.41
liters nitrogen @ STP to form 44.82 liters
ammonia @ STP
sample problem:
How many liters of hydrogen gas will be
produced at 280.0 K and 96.0 kPa if 1.74
moles of sodium react with water?
answer: 21.1 L H2
sample problem:
What volume of oxygen, collected at 25o C
and 101 kPa, can be prepared by
decomposing 37.9 grams of potassium
chlorate?
answer: 11.4 L