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The Gas Laws
Chapter 14
The Kinetic Molecular Theory
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 UnitsSI unit for pressure – Pascal (Pa)101.3 kPa = 1 atm760 mm Hg = 1 atm760 torr = 1 atm
Gas Laws
4 factors that affect gases – when one changes it changes the other factors
Volume Temperature Pressure Moles (# of particles)
Boyle’s Law
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
P1V1 = P2V2
Boyle’s Law – Sample Problem
P1V1 = P2V2
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)
= P2 0.66 atm = P2
)4.6(
)5.3)(2.1(
Boyle’s Law – Sample Problem
P1V1 = P2V2
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)
= V1 3.6 L = V1
)5.7(
)65.2)(3.10(
Boyle’s Law
Homework – pg. 422 # 1-5
Charles’ Law 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
=
Temperature must be in Kelvin K = Celsius + 273
1
1
T
V
2
2
T
V
Charles’ Law – Sample Problem
V1 = 3.4 L check to make sure your
V2 = 7.8 L your units are the same
T1 = 45°C (45 + 273 = 318) T2 = x = cross multiply
= T2 729 K = T2
729 – 273 = 456°C
460°C
318
4.3
2T
7.8
(3.4)
)318)(8.7(
Gay Lussac’s Law
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
=
Temperature must be in Kelvin
1
1
T
P
2
2
T
P
Gay Lussac’s Law – Sample Problem
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
= cross multiply
= P2 4.8 atm = P2
295
52.4
315
P2
(295)
)315)(52.4(
Charles’ and Gay Lussac’s Law
Homework pg. 425 # 6-8
pg. 427 # 9-13
Combined Gas Law
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
297
)V)(3.99( 2
First convert your temperature to Kelvin
T1 = 16 + 273 = 289 K
T2 = 24 + 273 = 297 K
289
)6.72)(1.104(=
(104.1)(72.6)(297) = (289)(99.3)V2
)3.99 289(
297 6.72 1.104
x
xx= V2
78.2ml = V2
Combined Gas Law - Example
P1 = 98.0 kPa T1 = 25°C + 273 = 298
P2 = x T2 = 60°C + 273 = 333
V1 = 1.5 L
V2 = 3.2 L 298
)5.1)(0.98(333
)2.3)(P( 2=
)298 2.3(
333 5.1 0.98
x
xx= P2 51.3 kPa = P2
Combined Gas Law - Homework
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 = nRTP = pressure , V = volume, T = temperaturen = number of molesR = ideal gas constant (depends on units of pressure)
Ideal Gas Law
The value of R depends on the pressure units:
Value of R Units of Pressure
0.0821 atm
8.314 kPa
62.4 mm Hg ; torr
Volume = Liter
Temp = Kelvin
n = moles
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)
)320)(0821.0(
)2.3)(5.2(= n 0.30mol = n
Ideal Gas Law - Example
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?
0.30 mol x mol1
g2.20= 6.06 g
If the molar mass is 20.2 g/mol then the gas is:
Neon
Ideal Gas Law – ExampleP = 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)
)300)(4.62(
)5.17)(652(= n 0.61 mol = n
Ideal Gas Law – Homework
Practice problems # 41-45 pg. 437