Gases.ppt

Elements that exist as gases at 250C
and 1 atmosphere

• Gases assume the volume and shape of
their containers.
• Gases are the most compressible state of
matter.
• Gases will mix evenly and completely when
confined to the same container.
• Gases have much lower densities than
liquids and solids.
Physical Characteristics of Gases

Units of Pressure
1 pascal (Pa) = 1 N/m2
1 atm = 760 mmHg = 760 torr
1 atm = 101,325 Pa
1 atm = 101.3 kPa
Barometer
Pressure =
Force
Area
(force = mass x acceleration)

Units of Pressure
1 pascal (Pa) = 1 N/m2
1 atm = 760 mmHg = 760 torr
1 atm = 101,325 Pa/
1 atm = 1.01X105 Pa
1 kPa = 1000 Pa
Barometer
EXERCISES

Sea level 1 atm
4 miles 0.5 atm
10 miles 0.2 atm

Manometers Used to Measure Gas Pressures

As P (h) increases V decreases
Apparatus for Studying the Relationship Between
Pressure and Volume of a Gas

P a 1/V
P x V = constant
P1 x V1 = P2 x V2
Boyle’s Law
Constant temperature
Constant amount of gas

Question
Based on Boyle’s Law (p * V =
constant), when the temperature
(T) is held constant, pressure and
volume are:
A.Inversely proportional: if one goes up,
the other comes down.
B.Directly proportional: if one goes up,
the other goes up.
C. Not related

Boyle’s Law at Work…
Doubling the pressure reduces the volume by half.
Conversely, when the volume doubles, the pressure
decreases by half.

A sample of chlorine gas occupies a volume of 946 mL
at a pressure of 726 mmHg. What is the pressure of
the gas (in mmHg) if the volume is reduced at constant
temperature to 154 mL?
P1 x V1 = P2 x V2
P1 = 726 mmHg
V1 = 946 mL
P2 = ?
V2 = 154 mL
P2 =
P1 x V1
V2
726 mmHg x 946 mL
154 mL
=
= 4.46 X 103 mmHg
P x V = constant

1) A container holds 500 mL
of CO2 at 20.0° C and 742 torr.
What will be the volume of the
CO2 if the pressure is
increased to 795 torr?
2) A gas tank holds 2785 L of
propane, C3H8, at 830 mm Hg.
What is the volume of the
propane at standard pressure?

3) A sample of neon occupies a
volume of 461 mL at STP. What will
be the volume of the neon when the
pressure is reduced to 93.3 kPa?
4) 352 mL of chlorine under a pressure
of 680 mm Hg are placed in a
container under a pressure of 1210
mm Hg. The temperature remains
constant at 296K. What is the
volume of the container in liters?

If we place
a balloon
in liquid
nitrogen it
shrinks:
How Volume Varies With Temperature
So, gases shrink if cooled.
Conversely, if we heat a
gas it expands (as in a
hot air balloon).
Let’s take a closer look at
temperature before we
try to find the exact
relationship of V vs. T.

• If a volume vs. temperature graph is
plotted for gases, most lines can be
interpolated so that when volume is 0 the
temperature is -273 C.
• Naturally, gases don’t really reach a 0
volume, but the spaces between
molecules approach 0.
• At this point all molecular movement
stops.
• –273C is known as “absolute zero” (no
EK)
The Kelvin Temperature Scale

• Lord Kelvin suggested that a
reasonable temperature scale should
start at a true zero value.
• He kept the convenient units of C,
but started at absolute zero. Thus, K
= C + 273.
62C = ? K: K=C+273 = 62 + 273
= 335 K
• Notice that kelvin is represented as K
not K.
The Kelvin Temperature Scale

What is the approximate temperature for
absolute zero in degrees Celsius and kelvin?
Calculate the missing temperatures
0C = _______ K 100C = _______ K
100 K = _______ C –30C = _______ K
300 K = _______ C 403 K = _______ C
25C = _______ K 0 K = _______ C
Kelvin Practice
273 373
–173 243
27 130
298 –273
Absolute zero is –273C or 0 K

Variation of gas volume with temperature
at constant pressure.
V a T
V = constant x T
V1/T1 = V2 /T2
T (K) = t (0C) + 273.15
Charles’ &
Gay-Lussac’s
Law
Temperature must be
in Kelvin

Charles’ Law at Work…
As the temperature increases, the volume increases.
Conversely, when the temperature decreases, volume
decreases.
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MAIN

Question
Based on Charles’ Law (V / T = constant),
when the pressure (p) is held constant,
volume and temperature are:
a. Inversely proportional: if one goes up,
the other comes down.
b. Directly proportional: if one goes up,
the other goes up.
c. Not related

A sample of carbon monoxide gas occupies 3.20 L at
125 0C. At what temperature will the gas occupy a
volume of 1.54 L if the pressure remains constant?
V1 = 3.20 L
T1 = 398.15 K
V2 = 1.54 L
T2 = ?
T2 =
V2 x T1
V1
1.54 L x 398.15 K
3.20 L
= = 192 K
V1 /T1 = V2 /T2
T1 = 125 (0C) + 273.15 (K) = 398.15 K

2. A sample of gas occupies 3.5 L at 300 K.
What volume will it occupy at 200 K?
3. If a 1.00 L balloon is heated from 22.0°C to
100°C, what will its new volume be?
V1 = 3.5 L, T1 = 300K, V2 = ?, T2 = 200K
Using Charles’ law: V1/T1 = V2/T2
3.5 L / 300 K = V2 / 200 K
V2 = (3.5 L/300 K) x (200 K) = 2.3 L
V1 = 1.00 L, T1 = 22.0°C = 295 K
V2 = ?, T2 = 100 °C = 373 K
V1/T1 = V2/T2, 1 L / 295 K = V2 / 373 K
V2 = (1 L/295 K) x (373 K) = 1.26 L

Avogadro’s Law
V a number of moles (n)
V = constant x n
V1 / n1 = V2 / n2
Constant temperature
Constant pressure

• AVOGADRO’S LAW states
that equal volumes of different gases
at the same temperature and
pressure contain the same number
of molecules.
• If the amount of gas in a container is
increased, the volume is increased.
• If the amount of gas in a container is
decreased, the volume is decreased.

Ammonia burns in oxygen to form nitric oxide (NO)
and water vapor. How many volumes of NO are
obtained from one volume of ammonia at the same
temperature and pressure?
4NH3 + 5O2 4NO + 6H2O
1 mole NH3 1 mole NO
At constant T and P
1 volume NH3 1 volume NO

AVOGADRO’S LAW
• As you increase
the amount of gas
(i.e. through
inhalation) the
volume of the
balloon increases
likewise.

The mathematical form of Avogadro’s law
is:
V   ; V1 = V2
n n1 n2
SAMPLE PROBLEM 1
A sample of gas with a volume of 9.20 L is
known to contain 1.225 mol. If the amount of
gas is increased to 2.85 mol, what new
volume will result if the pressure and
temperature remain constant?

AVOGADRO’S LAW
SOLUTION
Given: V1 = 9.20 L V2 = ?
n1 = 1.225 mol n2 = 2.85 mol
Solution:
V2 = n2V1 = (2.85 mol) (9.20 L) = 21.4 L
n1 (1.225 mol)

AVOGADRO’S LAW
SAMPLE EXERCISES
1. If 0.25 mol of argon gas occupies
a volume of 7.62 mL at a
particular temperature and
pressure, what volume would
0.43 mol of argon have under
the same conditions?

AVOGADRO’S LAW
SAMPLE EXERCISES
2. At a certain temperature and
pressure, a balloon with 10.0 g
of oxygen has a volume of 7.00
L. What is the volume after 5.00
g of oxygen is added to the
balloon?

Ideal Gas Equation
Charles’ law: V a T (at constant n and P)
Avogadro’s law: V a n (at constant P and T)
Boyle’s law: V a (at constant n and T)
1
P
V a
nT
P
V = constant x = R
nT
P
nT
P
R is the gas constant
PV = nRT

The conditions 0 0C and 1 atm are called standard
temperature and pressure (STP).
PV = nRT
R =
PV
nT
=
(1 atm)(22.414L)
(1 mol)(273.15 K)
R = 0.082057 L • atm / (mol • K)
Experiments show that at STP, 1 mole of an ideal
gas occupies 22.414 L.
R = 0.0821 L • atm / (mol • K)

What is the volume
(in liters) occupied
by 49.8 g of HCl at
STP?

What is the volume (in liters) occupied by 49.8 g of HCl
at STP?
PV = nRT
V =
nRT
P
T = 0 0C = 273.15 K
P = 1 atm
n = 49.8 g x
1 mol HCl
36.45 g HCl
= 1.37 mol
V =
1 atm
1.37 mol x 0.0821 x 273 K
L•atm
mol•K
V = 30.7 L

Argon is an inert gas used in
lightbulbs to retard the
vaporization of the filament.
A certain lightbulb containing
argon at 1.20 atm and 18 0C
is heated to 85 0C at
constant volume. What is the
final pressure of argon in the
lightbulb (in atm)?

Argon is an inert gas used in lightbulbs to retard the
vaporization of the filament. A certain lightbulb
containing argon at 1.20 atm and 18 0C is heated to
85 0C at constant volume. What is the final pressure of
argon in the lightbulb (in atm)?
PV = nRT n, V and R are constant
nR
V
=
P
T
= constant
P1
T1
P2
T2
=
P1 = 1.20 atm
T1 = 291 K
P2 = ?
T2 = 358 K
P2 = P1 x
T2
T1
= 1.20 atm x 358 K
291 K
= 1.48 atm

EXERCISE
What pressure will be
exerted by 0.400 mol of
gas in a 5.00 L container
at 17.0°C?

What pressure will be exerted by 0.400
mol of gas in a 5.00 L container at
17.0°C?
Given:
n = 0.400 mol V= 5.00 L
T= 17.0 °C + 273 = 290 K
Solution: L  atm
P = nRT = (0.400 mol) (0.0821 mol  K) (290 K)
V (5.00 L)
P = 1.9atm

Density (d) Calculations
n
V
=
P
RT
m is the mass of the gas in g
M is the molar mass of the gas
m
M
n =
m
M V
=
P
RT

Density (d) Calculations
d =
m
V
=
PM
RT
m is the mass of the gas in g
M is the molar mass of the gas
Molar Mass (M ) of a Gaseous Substance
dRT
P
M = d is the density of the gas in g/L

Calculate the density
of carbon dioxide
(CO2) in grams per
liter (g/L) at 0.990
atm and 550C.

Calculate the density of carbon
dioxide (CO2) in grams per liter
(g/L) at 0.990 atm and 550C.
d =
PM
RT
d =
(0.990 atm) (44.01 g/mol)
(0.0821 L.atm/K.mol) (328K)
d = 1.62 g/L

A 2.10-L vessel
contains 4.65 g of a
gas at 1.00 atm and
27.0 0C. What is the
molar mass of the
gas?

A 2.10-L vessel contains 4.65 g of a gas at 1.00 atm
and 27.0 0C. What is the molar mass of the gas?
dRT
P
M = d = m
V
4.65 g
2.10 L
= = 2.21
g
L
M =
2.21
g
L
1 atm
x 0.0821 x 300.15 K
L•atm
mol•K
M = 54.6 g/mol

Gas Stoichiometry
What is the volume of CO2 produced
at 37 0C and 1.00 atm when 5.60 g
of glucose are used up in the
reaction:
C6H12O6 (s) + 6O2 (g) 6CO2 (g)
+ 6H2O (l)
HINT: 1 mol C6H12O6 = 180 g C6H12O6

Gas Stoichiometry
What is the volume of CO2 produced at 37 0C and 1.00
atm when 5.60 g of glucose are used up in the reaction:
C6H12O6 (s) + 6O2 (g) 6CO2 (g) + 6H2O (l)
g C6H12O6 mol C6H12O6 mol CO2 V CO2
5.60 g C6H12O6
1 mol C6H12O6
180 g C6H12O6
x
6 mol CO2
1 mol C6H12O6
x = 0.187 mol CO2
V =
nRT
P
0.187 mol x 0.0821 x 310.15 K
L•atm
mol•K
1.00 atm
= = 4.76 L

1. Consider the formation of
nitrogen dioxide from nitric
oxide and oxygen:
2NO(g) + O2(g) 2NO2(g)
If 9.0L of NO are reacted with
excess O2 at STP, what is the
volume in liters of the NO2
produced?

2. Methane, the principal
component of natural gas, is
used for heating and cooking.
The combustion process is,
CH4(g) + 2O2(g) CO2(g) + 2H2O(l)
If 15.0 moles of CH4 are
reacted, what is the volume of
CO2 (in liters) produced at
23.0°C and 0.985 atm?

Dalton’s Law of Partial Pressures
V and T
are
constant
P1 P2 Ptotal = P1 + P2

Consider a case in which two gases, A and B, are in a
container of volume V.
PA =
nART
V
PB =
nBRT
V
nA is the number of moles of A
nB is the number of moles of B
PT = PA + PB XA =
nA
nA + nB
XB =
nB
nA + nB
PA = XA PT PB = XB PT
Pi = Xi PT mole fraction (Xi) =
ni
nT

A sample of natural gas
contains 8.24 moles of CH4,
0.421 moles of C2H6, and
0.116 moles of C3H8. If the
total pressure of the gases is
1.37 atm, what is the partial
pressure of propane (C3H8)?

A sample of natural gas contains 8.24 moles of CH4,
0.421 moles of C2H6, and 0.116 moles of C3H8. If the
total pressure of the gases is 1.37 atm, what is the
partial pressure of propane (C3H8)?
Pi = Xi PT
Xpropane =
0.116
8.24 + 0.421 + 0.116
PT = 1.37 atm
= 0.0132
Ppropane = 0.0132 x 1.37 atm = 0.0181 atm
1. What is the partial pressure of methane (CH4)?
2. what is the partial pressure of ethane (C2H6)?

Kinetic Molecular Theory of Gases
1. A gas is composed of molecules that are
separated from each other by distances far
greater than their own dimensions. The
molecules can be considered to be points;
that is, they possess mass but have
negligible volume.
2. Gas molecules are in constant motion in
random directions, and they frequently
collide with one another. Collisions among
molecules are perfectly elastic.

Kinetic Molecular Theory of Gases
3. Gas molecules exert neither attractive
nor repulsive forces on one another.
4. The average kinetic energy of the
molecules is proportional to the
temperature of the gas in kelvins. Any
two gases at the same temperature will
have the same average kinetic energy
KE = ½ mu2

Kinetic theory of gases and …
• Compressibility of Gases
• Boyle’s Law
P a collision rate with wall
Collision rate a number density
Number density a 1/V
P a 1/V
• Charles’ Law
Collision rate a average kinetic energy of gas
molecules
Average kinetic energy a T
P a T

Kinetic theory of gases and …
• Avogadro’s Law
Collision rate a number density
Number density a n
P a n
• Dalton’s Law of Partial Pressures
Molecules do not attract or repel one another
P exerted by one type of molecule is unaffected
by the presence of another gas
Ptotal = SPi

Gas diffusion is the gradual mixing of
molecules of one gas with molecules of
another by virtue of their kinetic
properties.
NH3
17 g/mol
HCl
36 g/mol
NH4Cl

Gas effusion is the is the process by
which gas under pressure escapes from
one compartment of a container to
another by passing through a small
opening.

Gases.ppt

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