Kinetic Molecular Theory KMT THEORY THAT DISCRIBES THE PROPERTIES OF AN IDEAL GAS

Ideal Gases Particles move rapidly in constant random motion Far apart All collisions perfectly elastic

X X IDEAL GASES NEVER Stick together Or interact. Therefore can not be Solids or liquids X

Kinetic Molecular Theory KMT K: Always in motion or kinetic. The higher the temperature, the faster the particles move. M : All matter is made of very tiny molecules. Space between the piece much large than the size of molecules. T: Elastic collisions: Do NOT interact. No attractions or repulsions between.

4 Variables to describe Pressure Temperature Volume Number / amount P Atm, torr or mmHg, Kpa Celsius or Kelvin mL, L, cc, cm3 Molecules or moles

P: Pressure Collisions or Hits The more collisions, the greater the pressure

UNIT FOR PRESSURE Standard Pressure 101.3 kPa = 760 torr, Atm, torr, mmHg, KPa Standard Pressure 101.3 kPa = 760 torr, = 760 mm Hg = 1 atm

Pressure Conversions 152 kPa 0.80 kPa 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 Celsius 00 Celsius = 273 K K = 0C + 273 or 0C = K – 273 Convert 191 K to Celsius 0C = 191 – 273 = -82 0C Convert 191 C to Kelvin K = 191 + 273 = 464 K

Standard Temperature and Pressure STP Standard Temperature and Pressure ST 00C or 273 K SP 101.3 kPa, 760 torr, 760 mm Hg, or 1 atm

IDEAL GASES How do we use A, B, C, & D laws to relate Pressure, Temperature, Volume and Amount (n)?

Gas Relationships: A, B, C, & D Relationships between P, V, T, and n Avogadro’s Boyle’s Charles’s Dalton’s

A: Avogadro's Law Amount (n) is directly proportional to the space occupied (V). n1 = n2 V1 V2

n1 = n2 V1 V2 AVOGADRO’S LAW. V2 = 7.0 x 8.8 = 17.6 L 3.5 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 7.0 moles 8.8 liters V2 = 7.0 x 8.8 3.5 = 17.6 L

BOYLE’S LAW P1V1=P2V2 Pressure vs Volume

C. Charles Law T1 T2 Temperature vs Volume V1 = V2 At a constant pressure, the volume (V) of the gas varies directly with its Kelvin temperature (T).

Scuba d5 c9

5.8 atm = 3.3 atm + 0.9 atm + PHe 5.8 - 3.3 - 0.9 = PHe= 1.6 atm 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

PV = nRT Ideal Gas Law R = IDEAL GAS constant 0.08206 (L*atm)/(mol*K) or 8.31 (L*kPa)/(mol*K) n = number of moles

Mathematical relationship between P ,V, n, and T IDEAL GAS LAW Mathematical relationship between P ,V, n, and T PV = constant value “R” nT

Practice 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?

Solution First solve for n 300. g of C3H8. x 1 mole = 6.82 moles 44 grams C3H8 6.82 moles PV=nRT. T=250C +273 P= 3.0 atm? R=0.082 (L*atm)/(mol*K)

Solution V = nRT P 56 L 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 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?

Solution P = nRT V 52 atm PV=nRT. n = 2.0 moles T=30. +273 = 303 K V = .95 L P= nRT / V P = nRT V 52 atm P = 2.0(0.082) 303 atm = 0.95 R=0.082 (L*atm)/(mol*K)

PV = nRT Ideal Gas Law R = IDEAL GAS constant 0.08206 (L*atm)/(mol*K) or 8.31 (L*kPa)/(mol*K) n = number of moles