KKinetic inetic MMolecular olecular TTheoryheoryKMTKMT

THEORY THAT DISCRIBES THE PROPERTIES OF AN

IDEAL GAS

Ideal GasesIdeal Gases•Particles move rapidly in constant random motion

•Far apart

•All collisions perfectly elastic

IDEAL GASESIDEAL GASES• NEVER Stick togetherOr interact.

Therefore can not be Solids or liquids

X

X

KKinetic inetic MMolecular olecular TTheoryheoryKMTKMT

M : All matter is made of very tiny molecules. Space between the piece much large than the size of molecules.

K: Always in motion or kinetic. The higher the temperature, the faster

the particles move.

T: Elastic collisions: Do NOT interact. No attractions or repulsions between.

4 Variables to describe4 Variables to describe• P

• T

• V

• n

•Pressure

•Temperature

•Volume

•Number / amount

•Atm, torr or mmHg, Kpa•Celsius or Kelvin•mL, L, cc, cm3

•Molecules or moles

P: PressureP: Pressure– Collisions or Hits

– The more collisions, the greater the pressure

UNIT FOR PRESSUREUNIT FOR PRESSURE• Atm, torr, mmHg, KPa

Standard Pressure

101.3 kPa = 760 torr, = 760 mm Hg = 1 atm

Pressure ConversionsPressure Conversions• How many kPa’s are in 1.50 atm? 1 atm = 101.3 kPa

1.50 atm x 101.3kPa = 1 atm• How many kPa’s are in 6.0 mm Hg?

6.0 mm Hg x 101.3 kPa = 760 mm Hg

152 kPa

0.80 kPa

Converting from CelsiusConverting from Celsius00 Celsius = 273 K

K = 0C + 273 or 0C = K – 273

Convert 191 K to Celsius0C = 191 – 273 = -82 0C

Convert 191 C to KelvinK = 191 + 273 = 464 K

SSTTPPStandard Temperature and Pressure

•ST 00C or 273 K

•SP 101.3 kPa, 760 torr,

760 mm Hg, or 1 atm

IDEAL GASESIDEAL GASES

How do we use A, B, How do we use A, B, C, & D laws to relate C, & D laws to relate

PPressure, ressure, TTemperature, emperature, VVolume olume

and Amount (and Amount (nn)?)?

Gas Relationships:Gas Relationships:A, B, C, & DA, B, C, & D

• Relationships between P, V, T, and n

•Avogadro’s•Boyle’s

•Charles’s•Dalton’s

A: Avogadro's LawA: Avogadro's Law• Amount (n) is directly proportional to

the space occupied (V).

•n1 = n2 V1

V2

AVOGADRO’S LAW. • At room temperature 3.5 moles of gas

occupy a volume of 8.8 liters. If the amount is doubled (7.0 moles), what volume is needed to keep the same pressure?

n1 = n2 V1

V2

3.5 moles

8.8 liters

7.0 moles

V2 = 7.0 x 8.8 3.5

= 17.6 L

BOYLE’S LAWBOYLE’S LAW

•P1V1=P2V2• Pressure vs Volume

C. C. Charles LawCharles Law• Temperature vs Volume

V1 = V2

T1 T2

At a constant pressure, the volume (V) of the gas varies directly with its

Kelvin temperature (T).

Scuba d5 c9Scuba d5 c9

• DALTON’S LAW. Solve the following problems. SHOW ALL WORK

• A scuba tank contains oxygen, nitrogen and helium gas. If the oxygen pressure is 0.9 atm and the nitrogen pressure is 3.3 atm, what pressure is needed for the helium gas to get a total pressure of 5.8 atm?

5.8 atm = 3.3 atm + 0.9 atm + PHe

5.8 - 3.3 - 0.9 = PHe= 1.6 atm

PV = nRT Rap

Ideal Gas LawIdeal Gas Law

•PV = nRT•R = IDEAL GAS constant 0.08206 (L*atm)/(mol*K)

or 8.31 (L*kPa)/(mol*K)

•n = number of moles

IDEAL GAS LAWIDEAL GAS LAW• Mathematical relationship

between P ,V, n, and T

• PV = constant value “R” nT

Practice ProblemPractice Problem• A camping stove uses a propane tank

that holds 300. g of C3H8. How larger a tank would be needed to hold the propane as a gas at 250C and a pressure of 3.0 atm?

SolutionSolution• First solve for n

• 300. g of C3H8. x 1 mole =

44 grams C3H8

6.82 moles

•PV=nRT.

•T=250C +273•P= 3.0 atm?

R=0.082 (L*atm)/(mol*K)

SolutionSolution

V = nRT P

•T=298 K•P= 3.0 atm

•n=6.82 moles

R=0.082 (L*atm)/(mol*K)

V = 6.82(0.082) 298 L = 3.0 atm

56 L

Practice ProblemPractice Problem

•2.0 moles of N2 is kept in a volume of 0.95 L and a temperature of 30.00C. What is the pressure of the gas under these conditions?

SolutionSolution

• PV=nRT.• n = 2.0 moles• T=30. +273 = 303 K • V = .95 L• P= nRT / V

R=0.082 (L*atm)/(mol*K)

P = nRT V

P = 2.0(0.082) 303 atm = 0.95

52 atm

Ideal Gas LawIdeal Gas Law

PV = nRT•R = IDEAL GAS constant

0.08206 (L*atm)/(mol*K)or 8.31 (L*kPa)/(mol*K)

•n = number of moles