T2 = V2 • T1

78
81.
Avogadro's Law of Combining Volumes
Standard Temperature & Pressure:
22.4 L at 0
o
C and 1 atmosphere pressure
Standard Ambient Temperature & Pressure: 24.8 L at 20
o
C and 1 atmosphere pressure
1.
A sample of carbon dioxide gas has a volume of 55.0 mL at 45
o
C and 85.0 kPa. Determine the volume at STP and SATP.
2.
What pressure will 37.18 grams of CO
2
gas exert on a container at standard temperature?
3.
Find the mass of 543 mL of acetylene gas, C
2
H
2
, collected at a pressure of 85.0 kPa and standard temperature.
4.
What is the density of CO
2
gas measured at 5
o
C and 200 kPa?
5.
A sample of cooking gas, taken from a cylinder, was collected and its density measured at 27
o
C and 100 kPa. The density at those conditions was
1.768 g/L. What was the molar mass of the cooking gas?
6.
At STP, how many molecules of hydrogen are in 22.4 L?
7. Given 4.80 g of O2gas and 2.80 g of N2gas. Calculate for each of these samples:
82.
Answers - Avogadro's Law of Combining Volumes
1.
@ STP V
1
= 55.0 mL V
2
= ?
P
1
= 85.0 kPa P
2
= 101.325 kPa
T
1
= 45
o
C = 318.15 K
T
2
= 0
o
C = 273.15 K
V
2
= V
1
• P
1
• T
2
= 55.0 mL
• 85.0 kPa • 273.15 K
= 39.61 mL at STP
P
2
• T
1
101.325 kPa • 318.15 K
@ SATP V
1
= 55.0 mL V
2
= ?
P
1
= 85.0 kPa P
2
= 100 kPa
T
1
= 45
o
C = 318.15 K
T
2
= 25
o
C = 298.15 K
V
2
= V
1
• P
1
• T
2
=
55.0 mL • 85.0 kPa • 298.15 K
=
42.51 mL at SATP
P
2
• T
1
100 kPa • 318.15 K
2.
Pressure of 37.18 of CO
2
at standard temperatures
Solution steps
Step #1 Find the number of moles of carbon dioxide present
Step #2 Using the STP definition find the value of pressure
Step #3 Convert to all other standard pressures values
Step #1 Moles of CO
2
present
n = m
= 37.18 g
= 0.845 moles of CO
2
M 44.01 g/mol
Step #2 Pressure at STP
@ STP 1 mole
= 1 atm
therefore 0.845 mol x x = 0.845 atm
Step #3 Convert to all other pressure units
1 atm
= 101.325 kPa
= 760 Torr
0.845 atm x y x = 85.60 kPa y = 642.05 Torr
3.
@ STP V
1
= 543 mL
V
2
= ?
P
1
= 85.0 kPa P
2
= 101.325 kPa
T
1
= 0
o
C = 273.15 K
T
2
= 0
o
C = 273.15 K
V
2
= V
1
• P
1
=
543 mL • 85.0 kPa
= 0.456 L at STP
P
2
101.325 kPa
@ STP 1 mol
= 22.4 L
therefore x 0.456 L x = 0.02 mol of gas present
m = n • M = 0.02 mol • 26.04 g/mol = 0.52 grams
C
2
H
2
= 2 C = 2 • 12.01 =
24.02 g/mol

79
2 H = 2 •
1.01 = 2.02
g/mol
26.04 g/mol
4.
1 mole of any gas at STP = 22.4 L
@ STP V
1
= 22.4 L
V
2
= ?
P
1
= 101.325 kPa
P
2
= 200 kPa
T
1
= 0
o
C = 273.15 K
T
2
= 5
o
C = 278.15 K
V
2
= V
1
• P
1
=
22 4 L • 101.325 kPa • 278.15 K
= 11.56 L at STP
P
2
101.325 kPa • 273.15 K
D = m
= 44.01 g
= 3.81 g/L
V 11.56 L
5.
The gas has a density of 1.768 g/L
Therefore two pieces of information are given: m = 1.768 g and V
1
= 1 L
@ STP V
1
= 1 L
V
2
= ?
P
1
= 100 kPa
P
2
= 101.325 kPa
T
1
= 27
o
C = 300.15 K
T
2
= 0
o
C = 273.15 K
V
2
= V
1
• P
1
• T
2
=
1 L • 100 kPa • 273.15 K
=
0.90 L
P
2
• T
1
101.325 kPa • 300.15 K
Therefore @ STP 1 mole
= 22.4 L
x 0.90 L x = 0.04 mol
Therefore M = g/mol = 1.76 g/0.04 mol = 44.01 g/mol
6.
At STP 22.4 L = 1 mole and 1 moles has 6.02 X 10
23
molecules

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