The Gas Laws
Chapter 14
The Kinetic Molecular Theory
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Gas particles do not attract or repel each
other
Gas particles are very small with large
amounts of space between them
Gas particles are in constant random motion
Gas particles have elastic collisions – no
energy lost
Gas particles have the same average kinetic
energy at the same temperature
Gas Pressure
Gas particles exert pressure when they
collide with the walls of their containers
 Temperature, volume, and the number of
moles affect the pressure that a gas exerts
Pressure Units
SI unit for pressure – Pascal (Pa)
101.3 kPa = 1 atm
760 mm Hg = 1 atm
760 torr = 1 atm
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Gas Laws
4 factors that affect gases – when one changes
it changes the other factors
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Volume
Temperature
Pressure
Moles (# of particles)
Boyle’s Law
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As the volume of a container of gas
decreases, then the pressure of that gas
increases
This is an inverse relationship (as one goes
up the other goes down)
Temperature remains constant
P1 V 1 = P2 V 2
Boyle’s Law – Sample Problem
P1V1 = P2V2
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P1 = 1.2 atm
check to make sure your
P2 = x
your units are the same
V1 = 3.5 L
V2 = 6.4 L
(1.2)(3.5) = P2(6.4)
(1.2)(3.5)
= P2
(6.4)
0.66 atm = P2
Boyle’s Law – Sample Problem
P1V1 = P2V2
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P1 = 7.5 atm
check to make sure your
P2 = 10.3 atm
your units are the same
V1 = x
V2 = 2.65 L
(7.5)V1 = (10.3)(2.65)
(10.3)( 2.65)
= V1
(7.5)
3.6 L = V1
Boyle’s Law
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Homework – pg. 422 # 1-5
Charles’ Law
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As the temperature of a gas increases so
does its volume
This is a direct relationship (the both change
in the same direction)
Pressure remains constant
V1
T1

=
V2
T2
Temperature must be in Kelvin
K = Celsius + 273
Charles’ Law – Sample Problem
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V1 = 3.4 L
V2 = 7.8 L
T1 = 45°C
T2 = x
check to make sure your
your units are the same
(45 + 273 = 318)
3 .4
7.8
=
318
T2
(7.8)(318)
= T2
(3.4)
cross multiply
729 K = T2
729 – 273 = 456°C
460°C
Gay Lussac’s Law
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As the pressure of a gas increases so does its
temperature
This is a direct relationship (the both change in the
same direction)
Volume remains constant
P1
T1

=
P2
T2
Temperature must be in Kelvin
Gay Lussac’s Law – Sample Problem
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P1 = 4.52 atm
check to make sure your
P2 = x
your units are the same
T1 = 22°C (22 + 273 = 295)
T2 = 315 K
4.52
295
(4.52)(315)
(295)
= P2
=
P2
315
cross multiply
4.8 atm = P2
Charles’ and Gay Lussac’s Law
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Homework pg. 425 # 6-8
pg. 427 # 9-13
Combined Gas Law
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Combines pressure, volume, and
temperature
Amount of gas (moles) is constant
Combined Gas Law - Example
A sample of nitrogen monoxide has a volume of
72.6 mL at a temperature of 16°C and a
pressure of 104.1 kPa. What volume will the
sample occupy at 24°C and 99.3 kPa?
P1 = 104.1 kPa
T1 = 16°C
P2 = 99.3 kPa
T2 = 24°C
V1 = 72.6 ml
V2 = X
Combined Gas Law - Example
First convert your
temperature to Kelvin
T1 = 16 + 273 = 289 K
T2 = 24 + 273 = 297 K
(104.1)( 72.6)
289
=
(99.3)( V 2)
297
(104.1)(72.6)(297) = (289)(99.3)V2
104.1 x 72.6 x 297
= V2
(289 x 99.3)
78.2ml = V2
Combined Gas Law - Example
P1 = 98.0 kPa
P2 = x
V1 = 1.5 L
V2 = 3.2 L
T1 = 25°C + 273 = 298
T2 = 60°C + 273 = 333
(98.0)(1.5)
298
98.0 x 1.5 x 333
(3.2 x 298)
= P2
=
(P 2)(3.2)
333
51.3 kPa = P2
Combined Gas Law - Homework
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Practice problems #19-23 pg. 430
Ideal Gas Law
Particles take up no space
 Particles have no intermolecular attractive
forces
 NO GAS IS TRULY IDEAL!
PV = nRT
P = pressure , V = volume, T = temperature
n = number of moles
R = ideal gas constant (depends on units of
pressure)
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Ideal Gas Law
The value of R depends on the pressure units:
Value of R Units of Pressure
Volume = Liter
0.0821
atm
Temp = Kelvin
8.314
kPa
n = moles
62.4
mm Hg ; torr
Ideal Gas Law – Example
P = 2.5 atm
V = 3.2 L
T = 47 °C + 273 = 320 K
PV = nRT
Which R value do you use? atm = 0.0821
(2.5)(3.2) = n (0.0821)(320)
(2.5)(3.2)
=n
(0.0821)(320)
0.30mol = n
Ideal Gas Law - Example
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n = 0.30 mol
If the molar mass of this gas is 20.2 g/mol.
How many grams of this gas do you have?
What is this gas?
20
.
2
g
0.30 mol x
= 6.06 g
1mol
If the molar mass is 20.2 g/mol then the gas is:
Neon
Ideal Gas Law – Example
P = 652 mm Hg
V = 17.5 L
T = 27 °C + 273 = 300 K
PV = nRT
Which R value do you use?
62.4
(652)(17.5) = n (62.4)(300)
(652)(17.5)
=n
(62.4)(300)
0.61 mol = n
Ideal Gas Law – Homework
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Practice problems # 41-45 pg. 437