GASES
9 Pressure & Boyle’s Law
9 Temperature & Charles’s Law
9 Avogadro’s Law
9 IDEAL GAS LAW
PV = nRT
N2
CH4
CO2
O2
N2 O
NO2
HCN
Earth’s atmosphere:
78% N2
21% O2
some Ar, CO2
Some Common Gasses
Formula
Name
Characteristics
N2
Nitrogen
Inert
O2
Oxygen
Explosion Hazard
Life-sustaining
HCN
Hydrogen
Cyanide
Very Toxic
H2S
Hydrogen
Sulfide
Very Toxic
Rotten Eggs
CO
Carbon
Monoxide
Toxic
Odorless
CO2
Carbon
Dioxide
Plant-Food
Odorless
CH4
Methane
Flamable
Odorless
N2O
Nitrous
Oxide
Laughing Gas
Sweet Odor
GAS MOLECULES’
CHARACTERISTICS
9 molecules are moving
9 they are far apart (10 times as
far apart as they are big)
9 move in straight lines
9 collisions with each other
9 collisions with walls: pressure
9 higher temp yields faster motion
9 lower temp yield slower motion
and eventually condensation
PRESSURE
Pressure = force per unit area = F / A
760 mm
at sea level
BAROMETER
UNDER CONSTANT PRESSURE
101,325 N/m2
14.7 lb/in2
your body has ~2 m2 surface area
~200,000 N or ~45,000 lb
pressing on you right now!
ORIGIN OF PRESSURE
Kinetic Theory of Gasses
N≡N
N≡N
N≡N
N≡N
O=O
UNDERSTANDING GASSES
3 foundational relationships
Boyle’s Law (P and V)
Charles’ Law (V and T)
Avogadro’s Law (V and moles)
PRESSURE / BOYLE’S LAW
PV = constant (fixed T,n)
1
V∝
P
1 atm
xL
2 atm
x
L
2
4 atm
x
L
4
Volume and pressure are inversely
related (for fixed T and n)
PRESSURE / BOYLE’S LAW
PV = constant (fixed T,n)
1
V∝
P
Volume and pressure are inversely
related (for fixed T and n)
TEMPERATURE
CHARLES’S LAW
V
T
= constant (fixed P, n)
V∝T
absolute temperature
K = oC + 273.15
TEMPERATURE
CHARLES’S LAW
V
T
= constant (fixed P, n)
V∝T
absolute temperature
K = oC + 273.15
AVOGADRO’S LAW
V
n
= constant (fixed P, T)
V∝n
2 mole
1 mole
Fixed P and T
AVOGADRO’S LAW
molar volumes nearly identical
for all gasses
If n = 1, then V = 22.41 L
for all gases at STP
STP: 1 atm
0 oC
IDEAL GAS LAW
PV = nRT
absolute temp K
gas constant
Units of R are important
R = 0.08206
L atm
mol K
R = 8.314
J
mol K
Simple problems: Given 3 quantities,
solve for the 4th
Problem:
What is the V of 2.35 moles of H2
at 22.0 oC and 700 torr ?
V=
=
nRT
P
(2.35 mol)(0.0820 L atm/mol K)(295 K)
= 61.7 L
0.921 atm
700 torr
760 torr
Most problems involve a change in
conditions (P, V, T) of a gas
CHANGES IN P, V, T
Given: Initial conditions P, V, T
Final conditions: any two
Find: Value for the third final value
Pi Vi
Ti
Pf Vf
Tf
= nR
Pi Vi
= nR
Ti
=
Pf Vf
Tf
The gas in a 750 mL vessel at 105 atm
and 27 oC is expanded into a vessel of
54.5 L and –10 oC. What is the final P ?
P1 V1
T1
P 2 = P1
=
P2 V2
T2
V1
V2
= (105 atm)
T2
T1
(0.750 L) (273 – 10 K)
(54.5 L) (273 + 27 K)
P2 = 1.3 atm
GAS DENSITY
AND MOLAR MASS
Start with
PV = nRT
m
but n =
M
m = mass (g)
M = molar mass (g/mol)
m
so
PV =
RT
M
m
density is d =
M
rearrange to get
m
PM = RT
V
P
d =M
RT
density ∝ molar mass
Weigh a 1.00 L bulb of air to find
air’s average molecular weight.
V = 1.00 L
T = 25 oC = 298 K
P = 760 torr = 1.00 atm
Weight of air = 1.20 g
M=d
RT
P
M = (1.20 g/L)
d=
1.20 g
1.00 L
= 1.20 g/L
(0.0821 L atm/mol K)(298 K)
1.00 atm
M = 29.3 g/mol
air is ~80% N2 + ~20% O2
(0.80)(28) + (0.20)(32) = 28.8 g/mol
PARTIAL PRESSURE
PV = nRT
P=
RT
V
n
what is this?
Number of moles of gas
What gas?
How about a mixture?
ntotal = n1 + n2 + •••
Dalton’s Law: total pressure is the sum
of partial pressures
PARTIAL PRESSURE
If 5 mol CO2, 2 mol N2, 1 mol Cl2 are mixed
in a 40 L vessel at 0 °C, what is P?
P=
nRT
V
PT = (nco2 + nN2 + nCl2) RT
V
nCO2 RT = PCO
2
V
Partial Pressure
of CO2
Dalton’s Law: Total pressure is the sum
of partial pressures
PT = (5 + 2 + 1)
(0.0821)(273)
40
= 4.5 atm
EXAMPLE
Collect N2 over water
Barometric pressure = 742 torr
Volume = 55.7 mL
Temp = 23 °C
How much N2 is collected ?
n=
PV
RT
R = 0.0821 L atm/mol K
T = (273 + 23) = 296 K
V = 55.7 mL = 0.0557 L
P two gases
PH2O at 23 °C is 21 torr (from Tables)
PN2 = 742 – 21 = 721 torr or 0.95 atm
n=
PV
RT
=
(0.95)(0.0557)
(0.0821)(296)
= 0.0022 mol N2
0.0022 mol x 28 g/mol = 0.0616 g or 62 mg N2
KINETIC MOLECULAR
THEORY
• PV = nRT
explains how gases behave
• Need more to explain why
• Look at gases on the molecular level
Five key postulates of KMT:
• straight-line motion, random direction
• molecules are small
• no intermolecular forces
• elastic collisions
• mean kinetic energy ∝ T (in K)
Ek = ½ mv2
KINETIC MOLECULAR THEORY
So…. Kinetic molecular theory provides
understanding of why gases behave
as they do
Increase T at const. V ⇒ P increase
T increase, E increase, v increase,
more collisions per unit time and
harder collisions, so P increases
Increase V at const. T ⇒ P decrease
Const. T means const. E and const. v,
longer distances between collisions,
fewer collisions per unit time with walls,
so P decrease
Temperature and
Molecular Speeds
N2 at 0 °C
N2 at 100 °C
v
v
Some molecules move slowly
Some molecules move fast
Mean speed in middle
Entire curve (and mean) shift upwards
for higher temperature
Temperature and
Kinetic Energy
Average kinetic energy of a molecule
E = ½ m v2
root mean square speed
mass
E=½m
v2 =
3 RT
2 N
N Avogadro’s number
At a given T, all gases have the
same average kinetic energy, E
v=
3RT
M
½
M = molar mass
EFFUSION &
DIFFUSION
From KMT:
v=
rms speed
3RT
M
molar mass
lighter gases have higher rms molecular speed
Graham’s Law
r1
of effusion
=
r2
M2
M1
r is rate
M is molar mass
effusion = escape of gas through pinhole
diffusion = spread of one gas through another
Even though diffusion is much more complicated
(due to gas molecular collisions) it still obeys
Graham’s Law well
EFFUSION RATES
time
N2: M = 28 Æ v is smaller
He: M = 4 Æ v is larger
1
r∝v∝
M
rN2 smaller
rHe larger
IDEAL GASES
PV = nRT
Ideal gas law
Works best at low P & high T
Assumptions:
• no intermolecular interactions
• no molecular volume
REAL GASES
Deviations from ideal gas law
Reasons:
(1) Molecules have finite size,
they occupy space
(2) Molecules have attractive forces
that become stronger when they
are close together
INTERMOLECULAR
POTENTIAL
repulsive
ENERGY
distance
d
attractive
High pressure pushes gas molecules
close together, so the attract each other
Very high pressure pushes gas molecules
even closer so they repel each other
High Pressure
o
o o o oo
o
o
o o oo oo oo
at high P
attractive
forces lead
to the
appearance
of a smaller n
Example: CO2 at 200 atm
Very High Pressure
o
o o o oo
o
o
o o o o oo o o
At very high P,
finite molecular
volume leads to
repulsion and the
appearance of a
larger n
Example: CO2 at 800 atm
NON-IDEAL BEHAVIOR
Real Gas
For a gas, measure P, V, T
You find, at higher P that P is too small
(due to attraction between molecules)
The amt of P missing is proportional to
(1) size of attractive interactions (a)
(2) freq of collisions (n/V)2
To compensate use:
n2
a
P+
2
V
a = constant
At high P, V is too large due to
excluded volume
The actual V = Vmeasured– Vexcluded
So, use:
V – nb
b = constant
Real Gas
van der Waals Equation
adjustments relative to measurement
n2
(P+
a ) ( V – nb) = nRT
V2
Correction
for attractive
forces between
gas molecules
Pressure correction
Adjusts P ↑
Correction
for excluded
volume of gas
molecules
Volume correction
Adjusts V ↓
van der Waals Constants
molecular shape and interactions
Substance
a(L2-atm/mol2)
b(L/mol)
He
Xe
0.0341
4.19
0.02370
0.0510
H2
Cl2
0.244
6.49
0.0266
0.0562
CH4
CCl4
2.25
20.4
0.0428
0.1383
dispersion
interactions
increase with
molar mass
CCl4 is largest
molecule, has
largest b value
Real Gas Example
What is the P of 1.0 mol Cl2 in 2.0 L at 273 K?
Ideal
nRT
P=
Gas Law
V
van der
Waals
(P+
n2
V2
=
(1)(0.0821)(273)
a ) ( V – nb) = nRT
nRT
2 a
n
P=
–
V2
V – nb
P=
2
(1)(0.0821)(273)
2 – (1)(0.0562)
–
a = 6.49
L2 atm
mol2
b = 0.0562 L/mol
(1)(6.49)
22
P = 11.5 – 1.6 = 9.9 atm
The term containing a is more important
% difference:
= 11.2 atm
11.2 – 9.9
x 100% = 12%
11.2