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Gases ~ An Overview and Review of Concepts and Laws
J. Baumwirt, ChemistryGranada Hills Charter High School
From a compilation of different online and textbook resources for instructional purposes ONLY. Reproduction of this
PowerPoint is prohibited due to copyright laws.
States of Matter
• Gases are only one form of matter
• Note the relative distance in particles of a gas as compared to other states of matter
• This will be an important factor as we study the properties of gases
Solid Liquid
Gas Plasma
Properties of Gases• Gases are composed
of atoms or molecules• Gas particles are far
apart and therefore are the most compressible state of matter
• Gases have lower densities that solids or liquids
• Gases will mix evenly and completely when confined to the same container
Properties of Gases
• Gases expand to fill any container
• Because atoms and molecules are so small a container of gas is mostly empty space
• Gaseous particles have relatively few attractions or repulsions between particles under normal conditions
Variables that Affect theBehavior of a Gas
• Pressure
• Volume
• Amount of Gas
• Temperature
• Gases exert pressure on any surface they contact
• We can look at pressure simplistically as the number of times the particles hit the walls of the container
Pressure• Pressure =
Force per unit Area
P = Force Area
(Force = mass x acceleration)
( “Particle in a Box” model )
Units of Pressure• Pressure can be designated in a variety of
different units:– mmHg or torr– atmospheres, atm– Pascals, Pa– psi (pounds per square inch)– Bars
• Their equivalencies are as follows:
1 atm = 760 mmHg = 760 torr = 101,325 Pa = 14.7 psi
• In Chemistry these are the units we will use most commonly
mmHg or torr
atmospheres, atm
1 atm = 760 mmHg = 760 torr
How Pressure is Measured• The Barometer was
invented by EvangelistaTorricelli
• Origin of the pressure unit in mmHg is due to the use of the metric system to measure of the height of the mercury column. This unit is also known as a torr
• It is the atmosphericpressure that pushes the mercury up the inverted glass tube
• The height (h) above the level of the mercury in the dish is then read in mmHg Why was mercury used?
a) Pgas= Patm - h2 b) Pgas= Patm+ h2
Manometers (mă năh΄ mә tŭr)
• There are two types of manometers
• Open ended manometers with one end open tothe atmosphere
• Closed end manometers where one end is a vacuum
• Pressure readings require finding the difference between the “legs” of the manometer
Open ended devices require access to a barometerto find the pressure of the atmosphere.
(illustrated here as h1, h3)h2,
c) Pgas= h1
The two illustrations of an open ended manometer show: a) a gas with a pressure less than atmospheric pressureb) a gas with a pressure greater than atmospheric pressure
The Ideal Gas Law
• Most gases behave “ideally” under normal conditions
• The Ideal Gas Law equates the variables of pressure, volume, amount of gas and temperature that affect gas behavior together in one equation:
P V = n R T
Pressure is in the units of atmospheres, torr or mmHg
Units of the Ideal Gas Law
PV = nRT
Temperature in Kelvin
R = gas constant 0.0821 L•atm = 62.37 L•torr mol•K mol•K
Amount of gas is in the unit of moles = n
Volume in Liters (L)
Recall that C + 273.15 = K
The R constant is chosen to match the unit of Pressure used.
Recall that 1 torr = 1 mmHg
62.37 L• mmHg0.0821 L•atm
atmospheres
62.37 L•torr
torr mmHg
mmHg
760
atm15667.7
mmHg
A 12.25 L cylinder contains 75.5 g of neon at 24.5 oC. Determine the pressure of the cylinder of gas.
PV = nRT
P =V =
n =
R =
T =
?12.25 L
75.5 g mol = 374 mol20.18 g
62.4 L• mmHg mol • K
24.5 oC + 273 = 297.5 K
P = nRT V= (3.74 mol)(62.4 L•mmHg)(297.5K)
(12.25 L) mol•K
= 5667.7 mmHg
= 5670 mmHg
What is this in atm?
atm7.46=
Alterations of the Ideal Gas Law
• The Ideal Gas Law is used to find one aspect about a gas:– Pressure, Volume, number of moles or
Temperature
• Through mathematical substitutions, the variables can be extended to– Molecular mass and– Density
Using Subsitutions with PV=nRT
• n in the equation = moles
• But how do we find the number of moles of a substance?– moles =
– Substituting this back into the equation:
grams of substancemolar mass
PV = RTgrams Mwt
Molar mass (or molecular weight) = Mwt
The variables have now been extended to include mass and molecular weight.
More substitutions into PV=nRT
• Taking the previous substituted equation:
• The equation can be rearranged to solve for density. Density =
PV = RTgrams Mwt
MassVolume
PV = RTgrams Mwt
mass = grams
PMwtRT
grams V
=
The consideration here is to remember that density units are somewhat altered now as they are in grams per Liter.
What is the density of carbon dioxide gas at 25 oC and 725 mmHg pressure?
What do wedo now?
P =
V =
n =
R =
T =
725mmHg
62.4 L• mmHg
mol•K
25C + 273 = 298 K
grams ?Mwt
CO2 = 44.0 g/mol
grams V
=Density
725mmHg 44.0 g/mol 62.4 L• mmHg/mol K 298 K
grams V
=PMwt
RTgrams
V=
?1.72 g/Lgrams
V=
PV = RTgrams Mwt
Combined Gas Law• If there is a change in conditions such as a change in
pressure or volume of a gas, the ideal gas law can be converted to an equality called the Combined Gas Law:
Setting the equation equal to R:
• Then since R is constant for any gas, the following can be used to calculate a change for any of the variables.
• If a value is constant (does not change) it can be cancelled out and eliminated from the equation.
Condition 1 Condition 2
P1V1
n1T1 P2V2
n2T2
=
PVnT
= R
A balloon contains helium gas with a volume of 2.60 L at 25 oC and 768 mmHg. If the balloon ascends to an altitude where the helium pressure is 590 mmHg and the temperature is 15 oC, what is the volume of the balloon?
What type of problemis this?
There are 2 sets of conditions.
Yikes!
P1=V1=T1=
P2=V2=T2=
768 mmHg2.60 L 25 oC + 273 = 298 K
590 mmHg
15C + 273 = 288 K?
= (768 torr)(2.60 L)(288 K) (590 torr) (298 K)
= 3.27 L
A balloon contains helium gas with a volume of 2.60 L at 25 oC and 768 mmHg. If the balloon ascends to an altitude where the helium pressure is 590 mmHg and the temperature is 15 oC, what is the volume of the balloon?
Condition 1: Condition 2:
12
2112
TPTVP
V =
P1V1 = P2V2
n1T1 n1T2
P1V1 = P2V2
T1 T2
n isconstant
Avogadro’s Law andStandard Temperature and Pressure
Avogadro’s Law states that equal volumes of any two gases (Ideal) at the same temperature and pressure contain the same number of molecules.
STP:Pressure 1 atm (760 mm Hg)
Temperature 0oC (273 K)
Standard
At STP one mole of ideal gas occupies 22.4 L(Looks like another conversion factor to me!) 1 mol gas/22.4L
STANDARD TEMPERATURE & PRESSURE