Transcript Chemistry: McMurry and Fay, 5th Edition

Homework Answers
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9.29 – Would look same (c) only pressure higher
9.33 – Levels would be even
9.41 – .649 atm
9.43 – 423 mm of CHCl3
9.45 - 4.56 amu
Copyright © 2008 Pearson Prentice Hall, Inc.
Chapter 1/1
Ideal Gas Law
and
Stoichiometry
Copyright © 2008 Pearson Prentice Hall, Inc.
The Ideal Gas Law
Summary
Boyle’s Law:
Charles’ Law:
Avogadro’s Law:
PinitialVinitial = PfinalVfinal
Vinitial
Tinitial
Vinitial
ninitial
=
=
Vfinal
Tfinal
Vfinal
nfinal
Copyright © 2008 Pearson Prentice Hall, Inc.
Chapter 9/3
The Ideal Gas Law
Is there a mathematical relationship between
P, V, n, and T for an ideal gas?
Copyright © 2008 Pearson Prentice Hall, Inc.
Chapter 9/4
The Ideal Gas Law
Ideal Gas Law:
PV = nRT
R is the gas constant and is the same for all gases.
L atm
R = 0.082058
K mol
Standard Temperature and
Pressure (STP) for Gases
T = 0 °C (273.15 K)
P = 1 atm
Copyright © 2008 Pearson Prentice Hall, Inc.
Chapter 9/5
The Ideal Gas Law
What is the volume of 1 mol of gas at STP?
(1 mol) 0.082058
V=
nRT
P
=
L atm
K mol
(273.15 K)
(1 atm)
Copyright © 2008 Pearson Prentice Hall, Inc.
= 22.414 L
Chapter 9/6
Stoichiometric Relationships
with Gases
The reaction used in the deployment of automobile
airbags is the high-temperature decomposition of sodium
azide, NaN3, to produce N2 gas. How many liters of N2 at
1.15 atm and 30.0 °C are produced by decomposition of
45.0 g NaN3?
2NaN3(s)
2Na(s) + 3N2(g)
Copyright © 2008 Pearson Prentice Hall, Inc.
Chapter 9/17
Stoichiometric Relationships
with Gases
2NaN3(s)
2Na(s) + 3N2(g)
Moles of N2 produced:
45.0 g NaN3
x
1 mol NaN3
65.0 g NaN3
x
3 mol N2
2 mol NaN3
= 1.04 mol N2
Volume of N2 produced:
V=
nRT
P
L atm
(1.04 mol) 0.082058
(303.15 K)
K mol
=
(1.15 atm)
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= 22.5 L
Chapter 9/18
To find density of gases…
To find density of gases…
Another
Way?
H40 – C9
• 9.6 – 9.13, 9.49, 9.51, 9.57, 9.63*, 9.67,
9.71, 9.73*