critical temp & pressure = the point of no return example: water vapor temp 55° c = 118 torr...

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Page 1: Critical Temp & Pressure =  THE POINT OF NO RETURN  EXAMPLE: Water vapor Temp 55° C = 118 torr Temp 110° C = 1075 torr Temp 374° C = 1.655 x 10 5
Page 2: Critical Temp & Pressure =  THE POINT OF NO RETURN  EXAMPLE: Water vapor Temp 55° C = 118 torr Temp 110° C = 1075 torr Temp 374° C = 1.655 x 10 5

A Quick Review Critical Temp & Pressure =

THE POINT OF NO RETURN

EXAMPLE: Water vapor Temp 55° C = 118 torr Temp 110° C = 1075 torr Temp 374° C = 1.655 x 105 torr Beyond 374°C = will forever be a gas

Critical temp + pressure are the points at which we can still liquefy gas

Page 3: Critical Temp & Pressure =  THE POINT OF NO RETURN  EXAMPLE: Water vapor Temp 55° C = 118 torr Temp 110° C = 1075 torr Temp 374° C = 1.655 x 10 5

Of Br2, Ne, HCl, HBr, and N2, which is likely to have (a) the largest london dispersion forces; (b) the largest dipole-dipole attractive forces?

Page 4: Critical Temp & Pressure =  THE POINT OF NO RETURN  EXAMPLE: Water vapor Temp 55° C = 118 torr Temp 110° C = 1075 torr Temp 374° C = 1.655 x 10 5

A) Br2 = greatest molar mass B) HBr = polar molecule with greatest

mass

Page 5: Critical Temp & Pressure =  THE POINT OF NO RETURN  EXAMPLE: Water vapor Temp 55° C = 118 torr Temp 110° C = 1075 torr Temp 374° C = 1.655 x 10 5

In which of the following substances is hydrogen bonding likely to play an important role in determining physical properties:Methane (CH4)

Hydrazine (H2NNH2)

Methyl fluoride (CH3F)Hydrogen sulfide (H2S)

Page 6: Critical Temp & Pressure =  THE POINT OF NO RETURN  EXAMPLE: Water vapor Temp 55° C = 118 torr Temp 110° C = 1075 torr Temp 374° C = 1.655 x 10 5

H2NNH2 = hydrogen bonding exists between molecules of this type – not the others!

Page 7: Critical Temp & Pressure =  THE POINT OF NO RETURN  EXAMPLE: Water vapor Temp 55° C = 118 torr Temp 110° C = 1075 torr Temp 374° C = 1.655 x 10 5

In which of the following substances is significant hydrogen bonding possible:Methylene chloride (CH2Cl2)

Phosphide (PH3)Hydrogen peroxide (HOOH)Acetone (CH3COCH3)

Page 8: Critical Temp & Pressure =  THE POINT OF NO RETURN  EXAMPLE: Water vapor Temp 55° C = 118 torr Temp 110° C = 1075 torr Temp 374° C = 1.655 x 10 5

HOOH

Page 9: Critical Temp & Pressure =  THE POINT OF NO RETURN  EXAMPLE: Water vapor Temp 55° C = 118 torr Temp 110° C = 1075 torr Temp 374° C = 1.655 x 10 5

List the substances BaCl2, H2, CO, HF, and Ne in order of increasing boiling points.

Page 10: Critical Temp & Pressure =  THE POINT OF NO RETURN  EXAMPLE: Water vapor Temp 55° C = 118 torr Temp 110° C = 1075 torr Temp 374° C = 1.655 x 10 5

H2, Ne, CO, HF, BaCl2

Page 11: Critical Temp & Pressure =  THE POINT OF NO RETURN  EXAMPLE: Water vapor Temp 55° C = 118 torr Temp 110° C = 1075 torr Temp 374° C = 1.655 x 10 5

(A)Identify the intermolecular forces present in the following substances, and

(B) select the substance with the highest boiling pointCH3CH3

CH3OH

CH3CH2OH

Page 12: Critical Temp & Pressure =  THE POINT OF NO RETURN  EXAMPLE: Water vapor Temp 55° C = 118 torr Temp 110° C = 1075 torr Temp 374° C = 1.655 x 10 5

(A) CH3CH3 = dispersion forces; other substances have both dispersion forces and hydrogen bonds

(B) CH3CH2OH = hydrogen bond & molar mass

Page 13: Critical Temp & Pressure =  THE POINT OF NO RETURN  EXAMPLE: Water vapor Temp 55° C = 118 torr Temp 110° C = 1075 torr Temp 374° C = 1.655 x 10 5

Calculate the enthalpy change (in kJ) upon converting 1.00 mol of ice at -25 ° C to water vapor at 125 ° C under constant pressure of 1 atm. The specific heats are Ice = 2.09 J/g-KWater = 4.18 J/g-KSteam = 1.84 J/g-K

Heat of fusion = 6.01 kJ/mol Heat of vaporization = 40.67 kJ/mol

Page 14: Critical Temp & Pressure =  THE POINT OF NO RETURN  EXAMPLE: Water vapor Temp 55° C = 118 torr Temp 110° C = 1075 torr Temp 374° C = 1.655 x 10 5

56.0 kJ

Page 15: Critical Temp & Pressure =  THE POINT OF NO RETURN  EXAMPLE: Water vapor Temp 55° C = 118 torr Temp 110° C = 1075 torr Temp 374° C = 1.655 x 10 5

What is the enthalpy change (in kJ) during the process in which 100.0 g of water at 50 ° C is cooled to -30 ° C?

The specific heats are Ice = 2.09 J/g-KWater = 4.18 J/g-KSteam = 1.84 J/g-K

Heat of fusion = 6.01 kJ/mol Heat of vaporization = 40.67 kJ/mol

Page 16: Critical Temp & Pressure =  THE POINT OF NO RETURN  EXAMPLE: Water vapor Temp 55° C = 118 torr Temp 110° C = 1075 torr Temp 374° C = 1.655 x 10 5

-60.6 kJ

Page 17: Critical Temp & Pressure =  THE POINT OF NO RETURN  EXAMPLE: Water vapor Temp 55° C = 118 torr Temp 110° C = 1075 torr Temp 374° C = 1.655 x 10 5

QUESTION 8 What is this substance’s normal melting point?

Page 18: Critical Temp & Pressure =  THE POINT OF NO RETURN  EXAMPLE: Water vapor Temp 55° C = 118 torr Temp 110° C = 1075 torr Temp 374° C = 1.655 x 10 5

ANSWER 8 60 ° C

Page 19: Critical Temp & Pressure =  THE POINT OF NO RETURN  EXAMPLE: Water vapor Temp 55° C = 118 torr Temp 110° C = 1075 torr Temp 374° C = 1.655 x 10 5

Question 9 At what temperature and pressure do all

three phases coexist?

Page 20: Critical Temp & Pressure =  THE POINT OF NO RETURN  EXAMPLE: Water vapor Temp 55° C = 118 torr Temp 110° C = 1075 torr Temp 374° C = 1.655 x 10 5

ANSWER 9 45˚ C

Page 21: Critical Temp & Pressure =  THE POINT OF NO RETURN  EXAMPLE: Water vapor Temp 55° C = 118 torr Temp 110° C = 1075 torr Temp 374° C = 1.655 x 10 5

Question 10

In Denver, we live approximately 5,280 feet above sea level, which means the normal atmosphericpressure is less than 1 atm. In Denver, will water boil at a higher or lower temperature, than at 1atmosphere?

Page 22: Critical Temp & Pressure =  THE POINT OF NO RETURN  EXAMPLE: Water vapor Temp 55° C = 118 torr Temp 110° C = 1075 torr Temp 374° C = 1.655 x 10 5

Answer 10 Lower temperature. At 1atm, water boils at 100˚C.

Page 23: Critical Temp & Pressure =  THE POINT OF NO RETURN  EXAMPLE: Water vapor Temp 55° C = 118 torr Temp 110° C = 1075 torr Temp 374° C = 1.655 x 10 5

Question 11

Water is an unusual substance because the slope of theboundary between solid and liquid is negative.What happens to solid water at 0˚C if youincrease the pressure?

Page 24: Critical Temp & Pressure =  THE POINT OF NO RETURN  EXAMPLE: Water vapor Temp 55° C = 118 torr Temp 110° C = 1075 torr Temp 374° C = 1.655 x 10 5

Answer 11 It becomes a liquid! All other substances become a

solid, but water behaves differently!

Page 25: Critical Temp & Pressure =  THE POINT OF NO RETURN  EXAMPLE: Water vapor Temp 55° C = 118 torr Temp 110° C = 1075 torr Temp 374° C = 1.655 x 10 5

Question 12 Water vapor condenses on the outside

of a soda can.a. Is energy being released or absorbed by

the water?b. What phase change is the water going

through?c. If you wanted to calculate the heat

transferred, what formula would you use and why?

Page 26: Critical Temp & Pressure =  THE POINT OF NO RETURN  EXAMPLE: Water vapor Temp 55° C = 118 torr Temp 110° C = 1075 torr Temp 374° C = 1.655 x 10 5

Answer 12 Water vapor condenses on the outside

of a soda can. a. Is energy being released or

absorbed by the water? Energy is released from the water.

b. What phase change is the water going through?

Gas to Liquid. c. If you wanted to calculate the heat

transferred, what formula would you use and why?

Q = mΔHvap

Page 27: Critical Temp & Pressure =  THE POINT OF NO RETURN  EXAMPLE: Water vapor Temp 55° C = 118 torr Temp 110° C = 1075 torr Temp 374° C = 1.655 x 10 5

Question 13 How much energy in joules does 28.5g

of liquid sulfur lose when it lowers from 120°C to 115°C,then change into a solid? The specific heat of liquid sulfur is 0.71 J/g°C.

Melting point 115°Boiling point 445°Heat of fusion 54 J/gHeat of vaporization 1406 J/g

Page 28: Critical Temp & Pressure =  THE POINT OF NO RETURN  EXAMPLE: Water vapor Temp 55° C = 118 torr Temp 110° C = 1075 torr Temp 374° C = 1.655 x 10 5

Answer 13 1640.2 J Released energy.