# 10. Gases (1)

```Chapter 10
Gases
1
Gases
What gases are important for each of the
following: O2, CO2 and/or He?
A.
B.
C.
D.
Gases
What gases are important for each of the
following: O2, CO2 and/or He?
A. CO2
B. O2/CO2
C. O2
D. He
Section 13.1
Pressure
Pressure
•
•
•
force
Pr essure =
area
SI units = Newton/meter2 = 1 Pascal (Pa)
1 standard atmosphere = 101,325 Pa
1 standard atmosphere = 1 atm =
760 mm Hg = 760 torr
1 atm = 760 torr
1 atm = 101, 325 Pa
Section 13.1
Pressure
6
Section 13.1
Pressure
Pressure Conversions: An Example
The pressure of a gas is measured as 2.5
atm. Represent this pressure in both torr
and pascals.
How do we get there?
 760 torr 
3
=
1.9

10
torr
 2.5 atm  

 1 atm 
 101,325 Pa 
5
2.5
atm

=
2.5

10
Pa

 

 1 atm 
Section 13.1
Pressure
Exercise
The vapor pressure over a beaker of hot water is
measured as 656 torr. What is this pressure in
atmospheres?
a)
b)
c)
d)
1.16 atm
0.863 atm
0.756 atm
0.500 atm
656 torr 
1 atm
 0.863 atm
760 atm
Four Physical Quantities for Gases
Phys. Qty. Symbol
SI unit
Other common units
pressure
P
Pascal
(Pa)
atm, mm Hg, torr,
psi
volume
V
m3
dm3, L, mL, cm3
temp.
T
K
°C, °F
moles
n
mol
Section 13.2
Pressure and Volume: Boyle’s Law
Boyle’s Law (1st )
•
•
In constant temperature (T), and number of moles
of gas (n).
PV = k (k is a constant for a given sample of air at
a specific temperature)
P1  V1 = P2  V2
Section 13.2
Pressure and Volume: Boyle’s Law
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?
P x V = constant
P1 x V1 = P2 x V2
P2 =
P1 = 726 mmHg
P2 = ?
V1 = 946 mL
V2 = 154 mL
P1 x V1
V2
726 mmHg x 946 mL
=
= 4460 mmHg
154 mL
Section 13.2
Pressure and Volume: Boyle’s Law
Exercise
A sample of helium gas occupies 12.4 L at
23°C and 0.956 atm. What volume will it
occupy at 1.20 atm assuming that the
temperature stays constant?
9.88 L
P1V1 = P2V2
(0.956 atm) (12.4 L) = (1.20 atm) (V2)
V2 = 9.88 L
Section 13.3
Volume and Temperature: Charles’s Law
Charles’s Law (2nd)
•
•
•
•
(constant Pressure and number of moles).
V=bT (b is a proportionality constant)
K = °C + 273
0 K is called absolute zero.
V1 V2
=
T1 T2
Variation in Gas Volume with Temperature at Constant Pressure
As T increases
V increases
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 /T1 = V2 /T2
V1 = 3.20 L
V2 = 1.54 L
T1 = 398.15 K
T2 = ?
T1 = 125 (0C) + 273.15 (K) = 398.15 K
T2 =
V2 x T1
V1
=
1.54 L x 398.15 K
3.20 L
= 192 K
Section 13.3
Volume and Temperature: Charles’s Law
Exercise
Suppose a balloon containing 1.30 L of air at
24.7°C is placed into a beaker containing liquid
nitrogen at –78.5°C. What will the volume of the
sample of air become (at constant pressure)?
0.849 L
V1
V2

T1
T2
(V2 )
(1.30 L)

(24.7 + 273 K)
(  78.5 + 273 K)
V2 = 0.849 L
Section 13.4
•
•
(constant T and P).
V = an (a is a proportionality constant)
n1
n2
=
V1
V2
Section 13.4
The Relationship Between Volume and Moles
Section 13.4
Exercise
If 2.45 mol of argon gas occupies a volume of 89.0 L,
what volume will 2.10 mol of argon occupy under the
same conditions of temperature and pressure?
76.3 L
V1
V
= 2
n1
n2
V2
89.0 L
=
2.45 mol
2.10 mol
V2 = 76.3 L
If n and V are
constant,
then P α T
P and T are directly
proportional.
P1 P2

T1 T2

If one temperature
goes up, the
pressure goes up!
Joseph Louis GayLussac (1778-1850)
Gay Lussac’s Law Summary
The pressure and temperature of a gas are
directly related, provided that the volume
remains constant.
P1 P2

T1 T2
P1
P2
T1
T2
3.00 atm
298 K
V2
325 K
3.27 L
P1
P2
T1
T2
1.8 atm
1.9 atm
273 K
T2
288 K or
15.2 oC
Gay-Lussac’s Law: Pressure
and Temperature
We can simplify this relationship by the
formula:
P1 P2
=
T1 T2
Where,
P1, P2 = pressure in any unit (atm, kPa, or
mmHg), BUT they must match!
T1, T2 = temperature is always in Kelvin!
(Recall, just add 273 + °C)
Gay-Lussac’s Law: Example
A gas has a pressure of 103kPa at 25°C. What will the
pressure be when the temperature reaches 928°C?
P1= 103kPa
P1 P2
=
T1= 25°C +273= 298K
T
T
1
2
P2= ?
T2= 928°C+273= 1201K
(P2)
(103kPa)
=
(1201K)
(298K)
P2 = 415kPa
Section 13.5
The Ideal Gas Law
•
We can bring all of these laws together
into one comprehensive law:
V = bT (constant P and n)
V = an (constant T and P)
V = k (constant T and n)
P
PV = nRT
(where R = 0.08206 L·atm/mol·K, the
universal gas constant)
The Combined Gas Law
A gas occupies 3.78L at 529mmHg and 17.2°C. At what
pressure would the volume of the gas be 4.54L if the
temperature is increased to 34.8°C?
P1 V1 P2 V2
=
P1= 529mmHg
T
T
1
2
V1= 3.78L
T1= 17.2°C + 273= 290.2K
(P2)(4.54L)
(529mmHg)(3.78L)
P2= ?
=
(307.8K)
(290.2K)
V2= 4.54L
P2 = 467mmHg
T2= 34.8°C + 273= 307.8K
Using PV = nRT
How much N2 is req’d to fill a small room with a volume of 960
cubic feet (27,000 L) to P = 745 mm Hg at 25 oC?
R = 0.082057 L•atm/K•mol
Solution
2. Now calc. n = PV / RT
4
(0.98 atm)(2.7 x 10 L)
n =
(0.0821 L• atm/K • mol)(298 K)
n = 1.1 x 103 mol (or about 30 kg of gas)
29
30
Gases and Stoichiometry
2 H2O2(liq) ---> 2 H2O(g) + O2(g)
Decompose 1.1 g of H2O2 in a flask with a
volume of 2.50 L. What is the pressure of O2
at 25 oC? Of H2O?
Solution
Strategy:
Calculate moles of H2O2 and then
moles of O2 and H2O.
Finally, calc. P from n, R, T, and V.
31
Gases and Stoichiometry
2 H2O2(liq) ---> 2 H2O(g) + O2(g)
Decompose 1.1 g of H2O2 in a flask with a volume of 2.50 L.
What is the pressure of O2 at 25 oC? Of H2O?
Solution
1 mol
1.1 g H2 O2 •
 0.032 mol
34.0 g
1 mol O2
0.032 mol H2O2 •
= 0.016 mol O2
2 mol H2 O 2
Gases and Stoichiometry
2 H2O2(liq) ---> 2 H2O(g) + O2(g)
Decompose 1.1 g of H2O2 in a flask with a volume of 2.50 L.
What is the pressure of O2 at 25 oC? Of H2O?
Solution
P of O2 = nRT/V
(0.016 mol)(0.0821 L• atm/K • mol)(298 K)
=
2.50 L
P of O2 = 0.16 atm
32
What is the volume (in liters) occupied by 49.8 g of HCl at STP?
T = 0 0C = 273.15 K
P = 1 atm
PV = nRT
nRT
V=
P
1 mol HCl
n = 49.8 g x
= 1.37 mol
36.45 g HCl
1.37 mol x 0.0821
V=
V = 30.7 L
L•atm
mol•K
1 atm
x 273.15 K
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
= P = constant
T
V
P1
P2
=
T1
T2
P1 = 1.20 atm
T1 = 291 K
P2 = ?
T2 = 358 K
T2
= 1.20 atm x 358 K = 1.48 atm
P2 = P1 x
291 K
T1
DALTON’S LAW
(of Partial Pressures)
Partial Pressures:
The gas in each tank
on the left exerts a pressure
Ptotal = Pgas1 + Pgas2 + Pgas3 …
Dalton’s Law states: The total pressure of a
mixture of gases is equal to the sum of the
partial pressures of the gases in the mixture.
Example Problem 1
An automobile tire contains a mixture of nitrogen, oxygen,
and carbon dioxide with partial pressures of 25 psi, 7 psi,
and 3 psi respectively. What is the total pressure inside the
tire?
Example Problem 1
An automobile tire contains a mixture of nitrogen, oxygen,
and carbon dioxide with partial pressures of 25 psi, 7 psi,
and 3 psi respectively. What is the total pressure inside the
tire?
Ptotal = 25 psi + 7 psi + 3 psi = 35 psi
Example Problem 2
A football has a mixture of nitrogen and oxygen gases. The
pressure inside the football is 760. mmHg. The partial
pressure of nitrogen (PN2) is 600. mmHg. What is the
partial pressure of oxygen (PO2)?
Example Problem 2
A football has a mixture of nitrogen and oxygen gases. The
pressure inside the football is 760. mmHg. The partial
pressure of nitrogen (PN2) is 600. mmHg. What is the
partial pressure of oxygen (PO2)?
Ptot = PN2 + PO2
760mmHg - 600mmHg = PO2
160mmHg = PO2
Example 3
A gas mixture containing oxygen, nitrogen, and carbon
dioxide has PO2 = 20.1 kPa, PN2 = 18.3 kPa, and PCO2 =
34.4 kPa. What is Ptotal?
Ptotal = P1 + P2 + P3
Ptotal = PO2 + PN2 + PCO2
Ptotal = 20.1 kPa + 18.3 kPa + 34.4 kPa
Ptotal = 72.8 kPa
Example 4
A gas mixture containing oxygen, nitrogen, and argon
has a total pressure of 50.2 kPa. If PO2 = 20.1 kPa and
PN2 = 18.3 kPa what is PAr?
Ptotal = P1 + P2 + P3
Ptotal = PO2 + PN2 + PAr
PAr = Ptotal - PO2 - PN2
PAr = 50.2 kPa - 20.1 kPa - 18.3 kPa
Ptotal = 11.8 kPa
Mole fraction

n1
nTotal
Mole fraction = Pressure fraction
n1

n1  n2  n3 ....
E.g. A 1.00L sample of dry air at 25°C and 786mmHg contains
0.925g of N2, plus other gases. (a) What is the partial pressure of N2
in the air sample? (b) What is the mole fraction and mole percent of
N2 in the air?
0.925gN2 x 1mol N2 = 0.330 molN2
28.0g N2
PN2 = nN2RT/V
= 0.330molx0.0821 Latm/(K mol) x 298K
1.00L
= 0.807 atm (=613mmHg)
(b) Mole fraction of N2 = PN2 = 613mmHg = 0.780
P
786mmHg
Air contains 78.0 mole percent of N2
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
PT = 1.37 atm
0.116
Xpropane =
8.24 + 0.421 + 0.116
= 0.0132
Ppropane = 0.0132 x 1.37 atm = 0.0181 atm
GAS DENSITY
Screen 12.5
PV = nRT
n
P
=
V
RT
m
P
=
M• V
RT
where M = molar mass
m
PM
d =
=
V
RT
d and M proportional
47
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
M=
P
M=
g
2.21
L
M = 54.5 g/mol
4.65 g
m
=
= 2.21
d=
2.10
L
V
x 0.0821
L•atm
mol•K
1 atm
x 300.15 K
g
L
Kinetic Energy of Gas Particles
At the same conditions of temperature, all
gases have the same average kinetic energy.
1 2 m = mass
KE  mv
v = velocity
2
 At the same temperature, small
molecules move FASTER than large
molecules
The Meaning of Temperature
3
( KE ) avg  RT
2
Kelvin temperature is an index of the
random motions of gas particles (higher T
means greater motion.)
Kinetic Molecular Theory
Maxwell’s equation
u2
3RT
M
root mean square speed
u
where
is the speed and M is
the molar mass.
• speed INCREASES with T
• speed DECREASES with M
51
Diffusion
 Diffusion describes the mixing
of gases. The rate of diffusion
is the rate of gas mixing.
 Diffusion is the result of
random movement of gas
molecules
 The rate of diffusion
increases with temperature
 Small molecules diffuse faster
than large molecules
Effusion
Effusion: describes the passage of gas
into an evacuated chamber.
GAS DIFFUSION AND
EFFUSION
•Diffusion - The rate at
which two gases mix.
•Effusion - The rate at
which a gas escapes
through a pinhole into
a vacuum.
54
GAS DIFFUSION AND
EFFUSION
Graham’s law
governs effusion and
diffusion of gas
molecules.
Rate for A
Rate for B
M of B
M of A
Rate of effusion is
inversely proportional
to its sq. root molar
mass.
Thomas Graham, 1805-1869.
Professor in Glasgow and London.
55
Graham’s Law of Effusion
Distance traveled by gas 1
M2

Distance traveled by gas 2
M1
M1 = Molar Mass of gas 1
M2 = Molar Mass of gas 2
Graham’s Law of Effusion
The rate of effusion of a gas is inversely proportional to the square root of its
molar mass.”
Rate A

Rate B
MB
MA
Rate = Rate of effusion
M=Molar Mass of gas
Example: What is the relative rate of effusion of H2 vs. O2?
Rate H 2
Rate O2
M
32
32



 16  4
M
2
2
O2
H2

At what temperature does 16.3 g of nitrogen gas have a
pressure of 1.25atm in a 25.0 L tank??

Calculate the molecular weight of a gas if 35.44 g of the
gas stored in a 7.50 L tank that exerts a pressure of 60.0
atm at a constant temperature of 35.5°C.

An unknown gas has a density of 3.167 g/L at STP. What is
the identity of the gas? (Ar, O2, Cl2, HF, H2O)?
Problem example: O2 generated in the
decomposition of KClO3 is collected over water.
The volume of the gas collected at 24 oC and at an
atmospheric pressure of 762 torr is 128 ml.
Calculate the number of moles of O2 obtained. The
vapor pressure of H2O at 24 oC is 22.4 torr.
61
First step: Calculate the partial pressure of O2.
PTotal  Pgas  PH O (Dalton’s law)
2
62
Pgas  PTotal  PH O
2
= 762 torr – 22.4 torr
= 739.6 torr (extra sig. fig)
 739.6 torr 1 atm
760 torr
= 0.973 atm
63
From the ideal gas equation PV = nRT,
PV
n=
RT
(0.973 atm) (0.128 l)
=
(0.08206 l atm mol-1K-1)(297 K)
= 0.00511 mols
64
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