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Unformatted text preview: BEHAVIOR OF GASES 1 Chapter 12 2 3 Hot Air Balloons —
How Do They Work? Importance
of Gases
• Airbags fill with N 2 gas in an accident.
• Gas is generated by the decomposition of
sodium azide, NaN3.
• 2 NaN3 > 2 Na + 3 N2 THREE
STATES
OF
MATTER 4 General Properties
of Gases
• There is a lot of “free”
space in a gas.
• Gases can be expanded
infinitely.
• Gases occupy containers
uniformly and completely.
• Gases diffuse and mix
rapidly. Page 1 5 6 Properties of Gases
Gas properties can be
modeled using math.
Model depends on—
• V = volume of the gas (L)
• T = temperature (K)
• n = amount (moles)
• P = pressure
(atmospheres) 7 Pressure 8 Pressure
Hg rises in tube until
force of Hg (down)
balances the force
of atmosphere
(pushing up). Pressure of air is
measured with a
BAROMETER
(developed by
Torricelli in 1643) Column height
measures P of
atmosphere
• 1 standard atm
= 760 mm Hg
= 29.9 inches
= about 34 feet of
water
SI unit is PASCAL,
Pa, where 1 atm =
101.325 kPa P of Hg pushing down
related to
• Hg density
• column height IDEAL GAS LAW PV=nRT
Brings together gas
properties.
Can be derived from
experiment and theory. 10 11 Boyle’s Law
If n and T are
constant, then
PV = (nRT) = k
This means, for
example, that P
goes up as V goes
down. Page 2 Pressure 9 Boyle’s Law
A bicycle pump is a good example of
Boyle’s law.
As the volume of the air trapped in the pump
is reduced, its pressure goes up, and air is
forced into the tire. Robert Boyle
(16271691).
Son of Early of
Cork, Ireland. 12 13 Charles’s
Law If n and P are
constant, then
V = (nR /P)T = kT
V and T are directly
related. 14 Jacques Charles (17461823). Isolated boron
and studied gases.
Balloonist. Modern longdistance balloon 16 Charles’s Law Avogadro’s Hypothesis
Equal volumes of gases at the same
T and P have the same number of
molecules.
V = n (RT/P) = kn
V and n are directly related. twice as many
molecules Page 3 15 Charles’s Law Charles’s original balloon Balloons immersed in liquid N 2 (at 196 ˚C) will
shrink as the air cools (and is liquefied). 17 18 Avogadro’s Hypothesis
Hypothesis The gases in this experiment are all
measured at the same T and P. Using PV = nRT 19 Using PV = nRT
R = 0.082057 L•atm/K•mol
0.082057 Solution
1. Get all data into proper units
V = 27,000 L
27,000
T = 25 oC + 273 = 298 K
25
P = 745 mm Hg (1 atm/760 mm Hg)
745
= 0.98 atm n= Strategy:
Calculate moles of H 2O2 and then
moles of O 2 and H2O.
Finally, calc. P from n, R, T, and V. Bombardier beetle
uses decomposition
of hydrogen peroxide
to defend itself. (0.98 atm)(2.7 x 104 L)
(0.0821 L• atm/K•mol)(298 K) n = 1.1 x 103 mol (or about 30 kg of gas) 22 Gases and Stoichiometry Gases and Stoichiometry Solution
2. Now calc. n = PV / RT R = 0.082057 L• atm/K•mol
0.082057 21 2 H2O2(liq) > 2 H2 O(g) + O2(g)
Decompose 1.1 g of H 2O2 in a flask with a
volume of 2.50 L. What is the pressure of O 2
at 25 oC? Of H2O? 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? 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 ? 2 H2O2(liq) > 2 H2 O(g) + O2(g)
Decompose 1.1 g of H 2O2 in a flask with a
volume of 2.50 L. What is the pressure of O 2
at 25 oC? Of H2O?
Solution 20 23 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 1.1 g H2 O2 • 0.032 mol H2O 2 • 1 mol
34.0 g 0.032 mol 1 mol O2
= 0.016 mol O2
2 mol H2 O 2 Page 4 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 = n RT/V
= (0.016 mol)(0.0821 L•atm/K•mol)(298 K)
2.50 L P of O2 = 0.16 atm 24 Gases and Stoichiometry 25 What is the total pressure in the flask? n at same T and P
at
n at same T and V
at Ptotal in gas mixture = P A + PB + ...
Therefore,
Ptotal = P(H2O) + P(O 2) = 0.48 atm There are 2 times as many moles of H 2O as
moles of O 2. P is proportional to n.
Therefore, P of H 2O is twice that of O 2.
P of H2O = 0.32 atm GAS DENSITY
Screen 12.5 Dalton’s Law: total P is sum of
PARTIAL pressures. 28 Low
Low
density
density GAS DENSITY
Screen 12.5 PV = nRT
nRT
P
n
=
V
RT
m
P
=
M•V
RT
where M = molar mass High
High
density
density Dalton’s Law
Dalton’s Law 2 H2O2(liq) > 2 H2 O(g) + O2(g)
0.32 atm
0.16 atm What is P of H 2O? Could calculate as above.
But recall Avogadro’s hypothesis.
P 27 Dalton’s Law of Partial Pressures 2 H2O2(liq) > 2 H2 O(g) + O2(g) V 26 d= m
PM
=
V
RT John Dalton
17661844 29 30 USING GAS DENSITY
The density of air at 15 oC and 1.00 atm is 1.23
g/L. What is the molar mass of air?
1. Calc. moles of air.
V = 1.00 L
1.00 P = 1.00 atm T = 288 K n = PV/RT = 0.0423 mol
2. Calc. molar mass d and M proportional
and Page 5 mass/mol = 1.23 g/0.0423 mol = 29.1 g/mol
mol ...
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This note was uploaded on 10/19/2011 for the course CHM 2210 taught by Professor Mcquade during the Fall '08 term at FSU.
 Fall '08
 McQuade
 Organic chemistry

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