1
The gaseous state. Gases and gas pressure. The gas laws. The ideal
gas law. Stoichiometric relationships with gases (McMurry pp. 308-325).
Kinetic-molecular theory of gases (McMurry pp. 326-328).
Liquid and solid states. Phase changes. Evaporation, vapor pressure, boiling
point (McMurry pp.357-367). The chemistry of water.
Week 7
2014
Anita Boratkó
Solid molecules are packed
closely together.
The
molecules are so rigidly
packed that they cannot
easily slide past each other.
Liquids molecules are
held closer together
than gas molecules,
but not so rigidly that
the molecules cannot
slide past each other.
Gas molecules are far
apart and do not
interact much with
each other.
2
» P: pressure; SI unit : pascal (Pa)
alternative pressure units:
millimeter of mercury (mm Hg) or also called torr
atmosphere (atm)
1 atm = 760 mmHg /torr = 101,325 Pa
» T = temperature given in Kelvin
273 K = 0 ◦C
» V= volume (liter)
3
» n = number of moles
Boyle’s Law PV=k
( T, n = 0)
Charles’ Law V/T=k
( P, n = 0)
Gay-Lussacs’ Law P/T=k
( V, n = 0)
Avogadro’s Law V/n=k
( P, T = 0)
4
A 600 mL sample of nitrogen is heated from 27 °C to 77 °C at
constant pressure. What is the final volume?
1. Change temperatures to Kelvin.
2. Use Charles' law to find the final volume.
T1 = initial temperature = 27 °C
T1 K = 273 + 27
T1 K = 300 K
Charles’ Law
V1/T1 = V2/T2
T2 = final temperature = 77 °C
T2 K = 273 + 77
T2 K = 350 K
where
V1 and T1 is the initial volume and temperature
V2 and T2 is the final volume and temperature
Solve the equation for V2:
V2 = V1T2/T1
Enter the known values and solve for V2.
V2 = (600 mL)(350 K)/(300 K)
V2 = 700 mL
5
At a pressure of 0.971atm a nitrogen gas occupies a volume of
500ml. What volume will the gas occupy at a pressure of 1.50
atm, assuming the temperature remains constant?
P1= 0.971 atm
V1= 500 ml
T= constant
P2= 1.50 atm
V2= ? ml
Boyle’s Law PV=k
P 1V 1= P 2V 2
0.971*500=1.5*V2
485.5=1.5*V2
323.67 ml=V2
6
A balloon with a volume of 2.0 L is filled with a gas at 3 atm. If the
pressure is reduced to 0.5 atm without a change in temperature,
what would be the volume of the balloon?
Since the temperature does not change, Boyle's law can be used.
P1V1=P2V2
To find the final volume, solve the equation for V 2 :
V 2 = P 1 V 1 /P 2
V 2 = (2.0 L)(3 atm)/(0.5 atm)
V 2 = 6 L/0.5
V 2 = 12 L
7
The ideal gas law…
8
The gas constant…
» The value of the gas constant 'R' depends on the units
used for pressure, volume and temperature.
R = 0.0821 liter·atm/mol·K
R = 8.3145 J/mol·K
R = 8.2057 m3·atm/mol·K
R = 62.3637 L·Torr/mol·K or L·mmHg/mol·K
9
If we had 1.0 mol of gas at 1.0 atm of pressure at 0°C
(273.15 K), what would be the volume? R=0.0821L atm/mol K
PV = nRT
V = nRT/P
V = (1.0 mol)(0.0821 L atm/mol K)(273 K)/(1.0 atm)
V = 22.41 L
0 °C and 1 atm pressure are referred to as the standard
temperature and pressure (STP).
The molar volume of an ideal gas (any ideal gas) is 22.4 liters
at STP
10
What size container do you need to hold 0.0459 mol N2
gas at STP?
1 mol = 22.4 L
0.0459 mol = X
X=22.4 L/mol x 0.0459 mol
X= 1.02816 L
A 30.6 g sample of gas occupies 22.414 L at STP. What is the
molecular weight of this gas?
Since one mole of gas occupies 22.414 L at STP, the
molecular weight of the gas is 30.6 g mol¯1
STP  V=22.414 L  1 mol gas
30.6 g / 1 mol = 30.6 g/mol
11
Determine the volume occupied by 2.34 grams of carbon
dioxide gas at STP. Molar mass of carbon dioxide is
44g/mol.
1) Rearrange PV = nRT
V = nRT / P
2) Substitute:
n=m/M = (2.34 g / 44.0 g mol¯1) = 0.0532 mol
V = (0.0532 mol)(0.08206 L atm mol¯1 K¯1) (273.0 K) ] / 1.00 atm
V = 1.192 L
12
A sample of argon gas at STP occupies 56.2 liters.
Determine the number of moles of argon and the mass in
the sample. Atomic weight of Ar 39.948 g/mol.
1) Rearrange PV = nRT
n = PV / RT
2) Substitute:
[ (1.00 atm) (56.2 L)
n=
(0.08206 L atm mol¯1 K¯1) (273.0 K)
n = 2.50866 mol
3) Multiply the moles by the atomic weight of Ar to get the grams:
2.50866 mol times 39.948 g/mol = 100. g
13
What is the molar mass of a gas if 0.460 g of the
gas occupy 315 mL at 0.52 atm and 50.0°C?
n = PV = (0.52 atm) (0.315 L)
= 0.00618 mol
RT (0.0821 L atm/mol K)(323K)
Molar mass =
g =
0.460 g = 74.4 g/mol
mol 0.00618 mol
14
Dalton’s Law…
• Dalton’s Law: in a gas mixture the total pressure is
given by the sum of partial pressures of each
component:
Pt = P1 + P2 + P3 + …
15
•Each gas obeys the ideal gas equation:
•Combining equations:
Pt
RT
n1 n2 n3 
V
16
A mixture of oxygen, hydrogen and nitrogen gases exerts a
total pressure of 278 kPa. If the partial pressures of the
oxygen and the hydrogen are 112 kPa and 101 kPa
respectively, what would be the partial pressure exerted by
the nitrogen.
PO2=112 kPa
PH2=101 kPa
PN2= ?
Ptotal=278 kPa
Ptotal = PO2 + PN2 + PH2
278 kPa = 112 kPa + 101 kPa + Pnitrogen
Pnitrogen = 278 kPa - (112 kPa + 101kPa)
Pnitrogen = 65 kPa
17
» A 2.00L flask contains 0.00125 mol O2 and 0.00325 mol He at 15°C.
What are the partial pressures of these gases and what is the total
pressure?
(n O 2 )R T
(n He )R T
P
=
P
=
O2
He
n O 2 0.00125 mol
VT
VT
n He 0.00125 mol
L atm
(0.00125mol) 0.08206
288 K
V 1 2.00 L
mol K
P O2 =
=1.48 •10 -2 atm
T1 15 C
2.00 L
P O2 ?
L atm
(0.00325mol ) 0.08206
288 K
P He ?
mol
K
P He =
= 3.84 • 10-2 atm
2.00 L
1.48 •10 -2 atm + 3.84 •10-2 atm = 5.32 • 10-2 atm
PT
P O2 + P He
PT =
(n O 2 )R T
(n He )R T (n O2 + n He )R T
+
=
VT
VT
VT
Or
PT =
(0.00125mol + 0.00325mol)R T
-2
= 5.32 •10 atm
VT
18
For Ideal Gases
19
The distribution of speeds for a gas sample at different
temperatures
20
How the Kinetic Molecular Theory Explains the Gas Laws
Gas pressure is a measure
of
the
number
and
forcefulness of collisions
between gas particles and
the walls of their container.
The smaller the volume at
constant n and T, the more
crowded
together
the
particles are and the
greater the frequency of
collisions. Thus, pressure
increases
as
volume
decreases.
Temperature
is
a
measure of the average
kinetic energy of the gas
particles. The higher the
temperature at constant n
and P, the faster the gas
particles move and the
more room they need to
move around in to avoid
increasing their collisions
with the walls of the
container. Thus, volume
increases as temperature
increases.
The more particles
there are in a gas
sample, the more
volume the particles
need at constant P
and
T
to
avoid
increasing
their
collisions with the
walls of the container.
Thus,
volume
increases as amount
increases.
The chemical identity of
the particles in a gas is
irrelevant. Total pressure of
a fixed volume of gas
depends only on the
temperature T and the total
number of moles of gas n.
The pressure exerted by a
specific kind of particle
thus depends on the mole
fraction of that kind of21
particle in the mixture.
Hydrogen sulfide reacts with sulfur dioxide to give H2O and S,
H2S + SO2 = H2O + S(solid), unbalanced.
If 6.0 L of H2S gas at 750 torr produced 3.2 g of sulfur, calculate the
temperature in C.
» Balanced reaction:
2 H2S(g) + SO2 (g)= 2 H2O + 3 S(s)
2 mol
3*32 = 96 g
X mol
3.2g
3.2g/96g * 2mol = 0.067 mol H2S
P = 750/760 = 0.987 atm
T=
PV
nR
=
(0.987 atm * 6 L)
= 1085 K
( 0.067 mol * 0.08205 atm L /(mol K)
1085 K = 812°C
22
Properties of liquid
•particles in a liquid are subject to intermolecular
attraction
•the attraction between the particles in a liquid
keeps the volume of the liquid constant
•Cohesion is the tendency for the same kind of
particles to be attracted to one another. This
cohesive "stickiness" accounts for the surface
tension of a liquid.
•Adhesion is when forces of attraction exist
between different types of particles. Particles of
the liquid are drawn up above the surface level
of the liquid at the edges where they are in
contact with the sides of the container.
23
•Because the particles of a liquid are in
constant motion, they will collide with
one another, and with the sides of the
container. Such collisions transfer
energy from one particle to another.
When enough energy is transferred to
a particle at the surface of the liquid, it
will eventually overcome the surface
tension holding it to the rest of the
liquid.
•The pressure exerted by the
vapor/liquid equilibrium in the closed
container is called the vapor
pressure.
24
Volatility, Vapor Pressure, and Temperature
consider the weight of a substance and intermolecular forces of attraction
the boiling point of a liquid at 1 atm
pressure is called its normal boiling
point.
the boiling point decreases as the
vapour pressure increases
25
Which one would you expect to have the larger vapor
pressure at a given temperature.
smaller mass
weaker London forces
F2 or Br2
CH3CH2OH or CH3OCH3
smaller mass
weaker London forces
no H-bonding
no H-bonding
He or Kr
H2Se or H2O
26
Consider the kinds of intermolecular forces present in the
following compounds, and rank the substances in likely
order of increasing boiling point:
H2S (34 amu), CH3OH (32 amu), C2H6 (30 amu), Ar (40 amu)
H2S: dipole-dipole, London
CH3OH: H-bonding, dipole-dipole, London
C2H6: London
Ar: London
C2H6 < Ar < H2S < CH3OH
27
Phase changes
All of these may be explained by an appreciation of the energy overcoming the
forces holding the particles together or becoming insufficient to keep them
apart.
28
endothermic
Objects gains
energy
Added energy
results is less
restrictive
connections
between atoms,
molecules
exothermic
Objects loses
energy
Results is more
restrictive
connections
between atoms,
molecules
29
» During a phase change energy is added, but the temperature does
not increase.
» The energy goes toward breaking up weak intermolecular forces
30
between the particles.
A phase diagram is “a relationship between physical
states that deals with temperature and pressure.”
 Indicates critical
temp and pressure.
 Pressure and temperature where a
substance exists as a solid, liquid, and gas.
When two phases exist at the same time it is called equilibrium.
Equilibrium is a dynamic condition in which two opposing changes
occur in equal rates in a closed system
31
Why does carbon dioxide
sublime?
Water
CO2
32
Answer the following questions based on this phase
diagram.
• A phase change from Phase C to Phase B
is known as:
Melting
• A phase change from Phase B to Phase C
is known as:
Freezing
• A phase change from Phase A to Phase B
is known as:
Condensation
• A phase change from Phase C to Phase A
is known as:
Sublimation
• The area of the graph that represents the
liquid phase is:
B
gas phase is :
A
solid phase is:
C
33
Answer the following questions based on this phase
diagram.
• If the temperature of the substance is held
constant at -15 ºC, the phase change that
would occur with a pressure increase
from 1 atmosphere to 30 atmospheres is:
deposition
• At 30 atmospheres pressure, the melting
point of this substance is: 0 C.
• The triple point of this substance occurs
at:
- 15 ºC and 6 atmospheres
• At STP, this substance can exist as:
a gas
34
Chemistry of water
hydrogen bond in water
35
Why does ice float on top of liquid water?
36
The molecular arrangement taken by ice leads to an increase
in volume and a decrease in density.