The Ideal Gas Law in Chemistry
Chapter 7-4
III. Ideal Gas Law
()
A. Avogadro’s Principle
•
Equal volumes of gases contain equal
numbers of moles
–
at constant temp & pressure
–
true for any gas
P
V
T
V
n
P
V
n
T
A. Ideal Gas Law
UNIVERSAL GAS CONSTANT
R=0.0821 L
atm/mol
K
R=8.315 dm
3
kPa/mol
K
=
R
You don’t need to memorize these values!
Merge the Combined Gas Law with Avogadro’s Principle:
A. Ideal Gas Law
UNIVERSAL GAS CONSTANT
R=0.0821 L
atm/mol
K
R=8.315 dm
3
kPa/mol
K
P
V
=
n
R
T
You don’t need to memorize these values!
GIVEN:
P = ? atm
n = 0.412 mol
T = 16°C = 289 K
V = 3.25 L
R =
0.0821
L
WORK:
PV = nRT
P(3.25)=(0.412)(0.0821)(289)
L mol L
K
P = 3.01 atm
C. Ideal Gas Law Problems
•
Calculate the pressure in atmospheres of 0.412
mol of He at 16°C & occupying 3.25 L.
GIVEN:
V
=
?
n
=
85 g
T
=
25°C = 298 K
P
=
104.5 kPa
R
=
8.315
dm
3
C. Ideal Gas Law Problems
•
Find the volume of 85 g of O
2
at 25°C and
104.5 kPa.
= 2.7 mol
PV = nRT
(104.5)V=(2.7) (8.315) (298)
kPa mol
dm
3
K
V = 64 dm
3
IV. Gas Stoichiometry at Non-
STP Conditions
A. Gas Stoichiometry
•
Moles
Moles
Liters of a Gas
Liters of a Gas
:
:
–
STP - use 22.4 L/mol
–
Non-STP - use ideal gas law
•
Non-
Non-
STP
STP
–
Given liters of gas?
•
start with ideal gas law
–
Looking for liters of gas?
•
start with stoichiometry conv.
1 mol
CaCO
3
100.09g
CaCO
3
B. Gas Stoichiometry Problem
•
What volume of CO
2
forms from 5.25 g of
CaCO
3
at 103 kPa & 25ºC?
5.25 g
CaCO
3
=
.
0
5
2
5
m
o
l
C
O
2
CaCO
3
CaO + CO
2
1 mol
CO
2
1 mol
CaCO
3
5.25 g
? L
non-STP
Looking for liters: Start with stoich
and calculate moles of CO
2
.
Plug this into the Ideal
Gas Law to find liters.
WORK:
PV = nRT
(103 kPa)V
=(.0525mol)(8.315
dm
3
)
(298K)
V
=
1
.
2
6
d
m
3
C
O
2
B. Gas Stoichiometry Problem
•
What volume of CO
2
forms from 5.25 g of
CaCO
3
at 103 kPa & 25ºC?
GIVEN:
P
=
103 kPa
V = ?
n
=
.0525 mol
T
=
25°C = 298 K
R
=
8.315
dm
3
WORK:
PV = nRT
(97.3 kPa) (15.0 L)
= n (8.315
dm
3
) (294K)
n
=
0
.
5
9
7
m
o
l
O
2
B. Gas Stoichiometry Problem
•
How many grams of Al
2
O
3
are formed from 15.0 L
of O
2
at 97.3 kPa & 21
°
C?
GIVEN:
P
=
97.3 kPa
V = 15.0 L
n
=
?
T
=
21°C = 294 K
R
=
8.315
dm
3
4 Al + 3 O
2
2 Al
2
O
3
15.0 L
non-STP
? g
Given liters: Start with
Ideal Gas Law and
calculate moles of O
2
.
NEXT
2 mol
Al
2
O
3
3 mol O
2
B. Gas Stoichiometry Problem
•
How many grams of Al
2
O
3
are formed from
15.0 L of O
2
at 97.3 kPa & 21
°
C?
0.597
mol O
2
=
4
0
.
6
g
A
l
2
O
3
4 Al + 3 O
2
2 Al
2
O
3
101.96 g
Al
2
O
3
1 mol
Al
2
O
3
15.0L
non-STP
? g
Use stoich to convert moles
of O
2
to grams Al
2
O
3
.
Explore the concepts of the ideal gas law, including Avogadro's principle and gas stoichiometry at non-STP conditions. Learn how to apply the law to calculate pressure, volume, and moles of gases under various conditions with practical examples provided.
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Presentation Transcript
A. Avogadros Principle Equal volumes of gases contain equal numbers of moles at constant temp & pressure true for any gas V= k V n n
A. Ideal Gas Law Merge the Combined Gas Law with Avogadro s Principle: PV T n nT PV V = k = R UNIVERSAL GAS CONSTANT R=0.0821 L atm/mol K R=8.315 dm3 kPa/mol K You don t need to memorize these values!
A. Ideal Gas Law PV=nRT UNIVERSAL GAS CONSTANT R=0.0821 L atm/mol K R=8.315 dm3 kPa/mol K You don t need to memorize these values!
C. Ideal Gas Law Problems Calculate the pressure in atmospheres of 0.412 mol of He at 16 C & occupying 3.25 L. GIVEN: P = ? atm n = 0.412 mol T = 16 C = 289 K V = 3.25 L R = 0.0821L atm/mol K WORK: PV = nRT P(3.25)=(0.412)(0.0821)(289) L mol L atm/mol K K P = 3.01 atm
C. Ideal Gas Law Problems Find the volume of 85 g of O2 at 25 C and 104.5 kPa. GIVEN: V=? n=85 g T=25 C = 298 K P=104.5 kPa R=8.315 dm3 kPa/mol K WORK: 85 g 1 mol = 2.7 mol 32.00 g PV = nRT (104.5)V=(2.7) (8.315) (298) kPa mol dm3 kPa/mol K K V = 64 dm3 = 2.7 mol
IV. Gas Stoichiometry at Non- STP Conditions
A. Gas Stoichiometry Liters of a Gas: STP - use 22.4 L/mol Non-STP - use ideal gas law Moles Non-STP Given liters of gas? start with ideal gas law Looking for liters of gas? start with stoichiometry conv.
B. Gas Stoichiometry Problem What volume of CO2 forms from 5.25 g of CaCO3 at 103 kPa & 25 C? CaCO3 CaO + CO2 5.25 g Looking for liters: Start with stoich and calculate moles of CO2. ? L non-STP 5.25 g CaCO3 1 mol CaCO3 100.09g CaCO3 1 mol CO2 1 mol CaCO3 = .0525 mol CO2 Plug this into the Ideal Gas Law to find liters.
B. Gas Stoichiometry Problem What volume of CO2 forms from 5.25 g of CaCO3 at 103 kPa & 25 C? GIVEN: P=103 kPa V = ? n=.0525 mol T=25 C = 298 K R=8.315 dm3 kPa/mol K WORK: PV = nRT (103 kPa)V =(.0525mol)(8.315dm3 kPa/mol K) (298K) V = 1.26 dm3 CO2
B. Gas Stoichiometry Problem How many grams of Al2O3 are formed from 15.0 L of O2 at 97.3 kPa & 21 C? 4 Al + 3 O2 2 Al2O3 15.0 L non-STP ? g GIVEN: P=97.3 kPa V = 15.0 L n=? T=21 C = 294 K R=8.315 dm3 kPa/mol K WORK: PV = nRT (97.3 kPa) (15.0 L) = n (8.315dm3 kPa/mol K) (294K) Given liters: Start with Ideal Gas Law and calculate moles of O2. NEXT n = 0.597 mol O2
B. Gas Stoichiometry Problem How many grams of Al2O3 are formed from 15.0 L of O2 at 97.3 kPa & 21 C? 4 Al + 3 O2 2 Al2O3 15.0L non-STP ? g Use stoich to convert moles of O2 to grams Al2O3. 0.597 mol O2 2 mol Al2O3 3 mol O2 101.96 g Al2O3 1 mol Al2O3 = 40.6 g Al2O3
