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