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Physical Properties Unit 5: Gases

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Unit 5: Gases. Physical Properties. Standards. 4a. Students know the random motion of molecules and their collisions with a surface create the observable pressure on the surface 4b. Students know the random motion of molecules explains the diffusion of gases. - PowerPoint PPT Presentation

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Page 1: Physical  Properties

Physical Properties

Unit 5: Gases

Page 2: Physical  Properties

Standards4a. Students know the random

motion of molecules and their collisions with a surface create the observable pressure on the surface

4b. Students know the random motion of molecules explains the diffusion of gases

Page 3: Physical  Properties

A. Kinetic Molecular TheoryUsed to predict and explain the

behavior of a theoretical gas or ‘ideal gas’

Particles in an ideal gas…• have no volume or elastic

collisions• in constant, rapid, random,

straight-line motion• don’t attract or repel each other

Page 4: Physical  Properties

‘Ideal gases’- are elastic (do not lose energy upon collision)

Cannot be compressed given a change in temperature

Can be measured using the eq. KE= 1/2mv2

Page 5: Physical  Properties

B. Real GasesParticles in a REAL gas…

• have their own volume• attract each other

Gas behavior is most ideal…• at low pressures• at high temperatures• in nonpolar atoms/molecules

Page 6: Physical  Properties

C. Characteristics of GasesGases expand to fill any

container• Random constant motion, no

attractionvery low densities

Page 7: Physical  Properties

C. Characteristics of Gasescan be compressed given a change

in Temp/Pressure

State Changes

Page 8: Physical  Properties

DiffusionThe movement of one material

through another.The rate depends on the mass of

the particlesLighter = rapid diffusion

Page 9: Physical  Properties

EffusionWhen a gas escapes through a tiny

openingRate of effusion can be calculated

according to Graham’s law of effusion:

Rate of effusion = 1/SQRT MM

Page 10: Physical  Properties

EffusionUsing Graham’s Law, you can also

set up a proportion to compare the diffusion rates for two gases

** see eq on board.

Page 11: Physical  Properties

Ammonia has a molar mass of 17.0 g/mol; hydrogen chloride has a molar mass of 36.5 g/mol. What is the ratio of their diffusion?

Page 12: Physical  Properties

D. Describing GasesGases can be described by their:

• Temperature• Pressure• Volume• Number of molecules/moles

• K• atm• L• #

Page 13: Physical  Properties

E. Temperature

ºF

ºC

K

-459 32 212

-273 0 100

0 273 373

32FC 95 K = ºC + 273

Always use absolute temperature (Kelvin) when working with gases!

Page 14: Physical  Properties

F. Pressure

areaforcepressure

Which shoes create the most pressure?

Page 15: Physical  Properties

F. PressureBarometer

• measures atmospheric pressure

exact height of the Hg depends on atmospheric pressure

usually measured in mm Hg

Page 16: Physical  Properties

F. PressureManometer

• measures contained gas pressure Difference in

height in two arms of U-tube is measure of pressure of gas sample

measured in various different units

Page 17: Physical  Properties

F. Pressure

2mNPa

KEY EQUIVALENT UNITS 101.325 kPa (kilopascal)

1 atm

760 mm Hg

760 torr

14.7 psi

Page 18: Physical  Properties

G. STP

Standard Temperature & Pressure

0°C 273 K

1 atm 101.325 kPa-OR-

STP

Page 19: Physical  Properties

H. Pressure Problem 1The average pressure in Denver,

Colorado, is 0.830 atm. Express this in (a) mm Hg and (b) kPa.

(a) 0.830 atm1 atm

760 mm Hg = 631 mm Hg

(b) 0.830 atm1 atm

101.325 kPa= 84.1 kPa

Page 20: Physical  Properties

H. Pressure Problem 2Convert a pressure of 1.75 atm to

kPa and mm Hg.

(a) 1.75 atm1 atm

101.325 kPa= 177 kPa

(b) 1.75 atm1 atm

760 mm Hg = 1330 mm Hg

Page 21: Physical  Properties

H. Pressure Problem 3Convert a pressure of 570. torr to

atmospheres and kPa.

(a) 570 torr760 torr

1 atm = .750 atm

(b) 570 torr760 torr

101.325 kPa= 76.0 kPa