Your SlideShare is downloading. ×
  • Like
Properties of Solids  & Liquids
Upcoming SlideShare
Loading in...5
×

Thanks for flagging this SlideShare!

Oops! An error has occurred.

×

Now you can save presentations on your phone or tablet

Available for both IPhone and Android

Text the download link to your phone

Standard text messaging rates apply

Properties of Solids & Liquids

  • 12,324 views
Published

Discusses macroscopic and microscopic properties of solids and liquids. …

Discusses macroscopic and microscopic properties of solids and liquids.
**More good stuff available at:
www.wsautter.com
and
http://www.youtube.com/results?search_query=wnsautter&aq=f

Published in Education , Business , Technology
  • Full Name Full Name Comment goes here.
    Are you sure you want to
    Your message goes here
  • ALSO
    I have written six books
    'Sticks - A Golfer’s Tale' - A hacker’s dream comes true.
    'Fish Farm' - Revenge of the old people.
    'Coach' - A mystery in Old school football town in a rural, bigoted, fifties town.
    'The Three Dollar Phoenix' - A mystery set in Newark, New Jersey in the 1970s.
    'The Divine Comedy MMIX' - A humorous play about Jesus returning.
    'The Blood of Judas' - A horror story of revenge set in Nazi Germany.
    All are available at www.smashwords.com
    I have video trailers for 'Coach', 'Fish Farm' and 'The Blood of Judas' at:
    http://www.youtube.com/watch?v=xXSD5Kz-fDY
    http://www.youtube.com/watch?v=a9PTRb14ldc
    http://www.youtube.com/watch?v=ToPp9k9Oq-o
    http://www.youtube.com/watch?v=3eBhMZbsP-I
    Please take a look. Thanks.
    Walt Sautter - wsautter@optonline.net
    Are you sure you want to
    Your message goes here
No Downloads

Views

Total Views
12,324
On SlideShare
0
From Embeds
0
Number of Embeds
2

Actions

Shares
Downloads
247
Comments
1
Likes
3

Embeds 0

No embeds

Report content

Flagged as inappropriate Flag as inappropriate
Flag as inappropriate

Select your reason for flagging this presentation as inappropriate.

Cancel
    No notes for slide

Transcript

  • 1. LIQUIDS, SOLIDS, SOLUTIONS AND PHASE CHANGES Copyright Sautter 2003
  • 2. The next slide is a quick promo for my books after which the presentation will begin Thanks for your patience! Walt S. [email_address] More stuff at: www.wsautter.com
  • 3. Books available at: www. wsautter .com www.smashwords.com www.amazon.com www.bibliotastic.com www.goodreads.com Walt’s Books for Free!
  • 4. SOLIDS, LIQUIDS AND PHASE CHANGES
    • What is the difference between a solid, a liquid and a gas?
    • ENERGY STATE ! Solids are the lowest energy state, liquids are intermediate and gases are the highest energy state of matter.
    • The energy differences which separate each state are called the Heats of Phase Change. The Heat of Phase Change is the energy required to change the state of a substance without changing its temperature.
    • The Heats of Phase Change are defined by the phase change which occurs, for example, Heat of Melting (often called fusion) is the heat needed to change a solid into a liquid at its melting point. The Heat of Vaporization is the heat needed to change a liquid into a gas at its boiling point.
  • 5. SOLIDS, LIQUIDS AND PHASE CHANGES
    • When solids, liquids and gases are heated or cooled and do not undergo phase change , the heated added or removed from the substance can be calculated using the equation: q = m x c x  T
    • The letter q represents heat is calories or joules. The mass of the substance in grams is m . The letter c stands for “ specific heat ” in calories per gram degree Celsius or joules per gram degree Celsius. Specific heat indicated how easy or difficult it is to change the temperature of a substance. Large specific heats mean that it requires a large amount of heat addition or removal to change temperature. Small values mean that the substance changes temperature readily.
    • Delta T (  T ) is the Celsius temperature change of the substance
  • 6. q = M x C x T HEAT IN JOULES OR CALORIES MASS IN GRAMS SPECIFIC HEAT IN JOULES / G x 0 C OR CALORIES / G x 0 C TEMPERATURE CHANGE IN 0 C CALCULATING HEAT QUANTITIES (NO PHASE CHANGE OCCURS)
  • 7. SOLIDS, LIQUIDS AND PHASE CHANGES
    • Problem: What is the temperature of 15.0 grams of iron when 250 joules of heat are removed from it?
    • (specific of iron = 0.450 j / g x 0 C)
    • Solution: q = m x c x  T, q = 250 joules
    • m = 15.0 grams, c = 0.450 j / g x 0 C
    •  T = q / (m x c) = 250 / (15.0 x 0.450) = 37.0 0 C
    • Since heat is removed from the sample the temperature is lowered by 37.0 degrees, therefore the answer is – 37.0 0 C
  • 8. SOLIDS, LIQUIDS AND PHASE CHANGES
    • Problem: A 100 gram sample of aluminum at 100 0 C is placed in 200 grams of water at 25 0 C. What is the final temperature of the system? (c for Al = 0.900 j / g 0 C and c for water = 4.18 j / g 0 C)
    • Solution: Heat lost = Heat gained
    • (Conservation of Energy)
    • Al loses heat to the water since the Al is at the higher initial temperature.
    • m Al x c Al x  T Al = m w x c w  T w ,
    • T = temperature of system at equilibrium
    • 100g(0.900 j /g 0 C)(100 0 C – T) = 200g(4.18 j /g 0 C)(T-25 0 C)
    • 9000 – 90T = 836T – 20900
    • 926T = 29900, T = 32.3 0 C
  • 9. SOLIDS, LIQUIDS AND PHASE CHANGES
    • When a substance undergoes phase change (change in physical state) heat is added or removed from the material however the temperature of the material remains unchanged. All energy that is released or absorbed during the phase change is used to change the state of the substance. None is used to alter the temperature.
    • Only after the phase change is complete can the temperature of the system rise or fall. (Note the temperature plateaus on the next graph)
    • The heat of phase change for different processes is described by the physical change which is occurring for example: Heat of Melting (sometimes called fusion), Heat of Vaporization, Heat of Freezing, etc.
  • 10. Time of Heating T E M P E R A T U R E 0 C 110 100 0 -10 All ice Melting starts Melting completed Ice & water All water Boiling starts Water & steam Boiling completed All steam Heating Curve for Water (-10 to 110 0 C)
  • 11. CALCULATING HEAT EXCHANGES DURING PHASES CHANGES
    • During a phase change the temperature of a substance remains constant. Based on that observation, q = m c  T cannot be used to find heat gained or lost since  T = 0.
    • In order to find heat values during phase change we must multiply the heat of phase change by the quantity of substance involved. Heat of phase change is the quantity of heat which is gained or lost when a substance changes state without a temperature change.
    • The heat of phase change may be expressed in calories or joules per gram of substance or calories or joules per mole of substance.
    • Q = Heat of Phase Change x Amount of Substance
  • 12. Heat of Melting Heat of Vaporization Heat of Condensation Heat of Freezing T E M P E R A T U R E Time of Heating HEATS OF PHASE CHANGE DURING HEATING & COOLING
  • 13. Heat = Heat of Phase Change X Amount HEAT REQUIRED FOR PHASE CHANGE HEAT OF MELTING (FUSION) HEAT OF FREEZING HEAT OF VAPORIZATION HEAT OF CONDENSATION HEAT OF SUBLIMATION (ALL IN JOULES / GRAM, CALORIES/ GRAM OR PER MOLE AMOUNT OF SUBSTANCE CHANGING PHASE (GRAMS OR MOLES) CALCULATING HEAT GAINED OR LOST DURING A PHASE CHANGE
  • 14. CALCULATING HEAT EXCHANGES DURING PHASES CHANGES
    • Problem: How much heat is needed to melt 200 grams of ice at its melt point (0 0 C) ? The heat of fusion for ice is 6019 joules per mole.
    • Solution: Since a phase change is occurring without a temperature change:
    • q melting = Heat of Melting x moles of ice melted
    • Moles = grams / grams per mole, moles = 200 / 18 = 11.1
    • q melting = 6019 joules / mole x 11.1 moles = 66,878 joules or 66.9 kilojoules.
  • 15. CALCULATING HEAT EXCHANGES DURING PHASES CHANGES
    • Problem: How much heat is needed change 100 grams of ice at –10 0 C to steam at 110 0 C. (c for ice = 2.06 j / g 0 C, c for water = 4.18 j / g 0 C, c for steam = 2.02 j / g 0 C, Heat of Fusion for ice = 6.02 Kj / mole, Heat of Vaporization = 40.7 Kj / mole)
    • Solution: q heating ice = m x c x  T
    • q heating ice = 100 x 2.06 x (0 – (-10)) = 2060 j
    • q melting = Heat of Melting x moles of ice melted
    • q melting = ( 6020 j / mole)(100 / 18 ) mole = 3.34 x 10 4 j
    • q heating water = 100 x 4.18 x (100 – 0) = 4180 j
    • q vaporization = (40.7 x 10 3 j / mole)(100 / 18) mole = 2.26 x 10 5 j
    • q heating steam = 100 x 2.02 x (110 – 100) = 2020 j
    • q total = 2060 + 3.34 x10 4 + 4018 + 2.26 x 10 5 + 2020 = 2.67 x 10 5 j
  • 16. CALCULATING HEAT EXCHANGES DURING PHASES CHANGES
    • IMPORTANT POINTS IN CALCULATING HEAT QUANTITIES:
    • (1) The specific heat of a substance is different for the same substance in a different physical state. For example, as we have seen, the specific heat of ice (2.06 j /g 0 C), that for water (4.18 j / g 0 C) and steam (2.02 j / g 0 C) are different.
    • (2) Heats of phase change are different for different changes involving the same material. For example Heat of Fusion for ice is 6.02 Kj / mole while Heat of Vaporization for water is 40.7 Jj / mole. Note that the Heat of Vaporization for a substance is always noticeably larger than its Heat of Fusion.
    • (3) Heat of phase change for a substance in the heating process are equal but opposite in sign as compared for the reverse phase change in the cooling process . For example, the Heat of Melting for ice is 6.02 Kj / mole while the Heat of Freezing for water is – 6.02 Kj / mole.
  • 17. TEMPERATURE AND PRESSURE EFFECTS ON PHASE CHANGE
    • Phase changes occur as:
    • Melting or Fusion – solid to liquid (endothermic)
    • Freezing or Solidification – liquid to solid (exothermic)
    • Vaporization – liquid to gas (endothermic)
    • Condensation – gas to liquid (exothermic)
    • Sublimation – solid to gas (endothermic)
    • Deposition – gas to solid (exothermic)
    • All changes of physical state are affected by both temperatures and pressures. Gas state is favored by low pressures and high temperatures both allowing for free, random motion. Solids are favored by low temperatures and high pressures both contributing to an ordered low energy state. A phase diagram shows the relationship between temperature, pressure and physical state under a variety of conditions.
  • 18. TEMPERATURE AND PRESSURE EFFECTS ON PHASE CHANGE
    • After a certain temperature (depending on the substance) no amount of pressure can cause a gas to form a liquid. This temperature is called the critical temperature of the substance. The pressure which the gas exhibits at this temperature is called the critical pressure.This temperature – pressure point is called the critical point.
    • At a specific temperature and pressure all three phases of a substance can coexist (solid, liquid and gas). This point is called the triple point. At this point a very slight change in temperature and / or pressure can cause a solid to melt, a liquid to vaporize, a solid to sublime, a liquid to freeze, etc.
  • 19. P R E S S U R E TEMPERATURE GAS SOLID LIQUID PHASE DIAGRAM TRIPLE POINT a b c d e f g a = freezing b = melting c = vaporization d = condensation e= deposition f = sublimation g = critical point
  • 20. P R E S S U R E GAS SOLID LIQUID PHASE DIAGRAM (for water) a b Notice that moving along the line a-b, as pressure rises, the solid melts without a temperature increase TEMPERATURE This behavior is typical for water but not most substances
  • 21. P R E S S U R E GAS SOLID LIQUID PHASE DIAGRAM (for most substances) a b Notice that moving along the line a-b, as pressure rises, the solid Will not melt Increasing pressure without a temperature change will not melt the substance TEMPERATURE This behavior is typical for most substances
  • 22. VAPOR PRESSURE OF LIQUIDS
    • The measure of the tendency of a liquid to form a gas is indicated by its vapor pressure . Liquids which evaporate readily have high vapor pressures. The tendency of a liquid to enter the gaseous state varies based on several factors:
    • (1) the intermolecular forces which exist between the liquid molecules. Factors such as polar character, London Forces and hydrogen bonding (all of which have been discussed in other programs) affect vaporization. As the strength of intermolecular forces increases the vapor pressure of the liquid decreases.
    • (2) the Heat of Vaporization of a substance. Materials with large  H vap have low vapor pressures .
    • (3) the temperature of the liquid. As temperature rises the liquid molecules increase their kinetic energies and the tendency to form a gas increases .
  • 23. LIQUID A HIGH VAPOR PRESSURE RAPID VAPORIZATION LIQUID B LOW VAPOR PRESSURE SLOW VAPORIZATION
    • VAPOR PRESSURE OF LIQUID A IS GREATER
    • THAN LIQUID B BECAUSE OF:
    • WEAK INTERMOLECULAR BONDS
    • LOWER MOLECULAR WEIGHT AND / OR
    • HIGHER TEMPERATURE
    • LOWER  H vaporization
  • 24. RELATING VAPOR PRESSURE, TEMPERATURE &  H VAPORIZATION
    • Vapor pressure of liquids is directly related to the temperature of the liquid and inversely related to the Heat of Vaporization of that liquid. As temperature rises, the vapor pressure rises. As the Heat of Vaporization becomes larger the tendency to evaporate decreases as shown by a reduced vapor pressure.
    • The temperature at which the vapor pressure of a liquid reaches that of the surroundings (generally atmospheric pressure) is called the boiling point of the liquid. Liquids, therefore, can be boiled by increasing the temperature or reducing the pressure on the liquid. (Boiling refers to the spontaneous vaporization of the liquid at all point within the liquid not just at the surface of the liquid which is evaporation.)
  • 25. KINETIC ENERGY DISTRIBUTION CURVE FOR A LIQUID TEMPERATURE = T 1 TEMPERATURE = T 2 KINETIC ENERGY N U M B E R O F M O L E S HEAT OF VAPORIZATION MOLECULES WITH SUFFICIENT ENERGY TO VAPORIZE AT T 1 AS TEMPERATURE , VAPORIZATION RATE MOLECULES WITH SUFFICIENT ENERGY TO VAPORIZE AT T 2
  • 26. VAPOR PRESSURE VS TEMPERATURE FOR VARIOUS LIQUIDS V A P O R P R E S S U R E TEMPERATURE 0 10 20 30 40 50 60 70 80 90 100 H 2 O C 2 H 5 OH DIETHYL ETHER NORMAL BOILING OCCURS WHEN VAPOR PRESSURE EQUALS 1 ATM 1.0
  • 27. RELATING VAPOR PRESSURE, TEMPERATURE &  H VAPORIZATION
    • The vapor pressure of a liquid can be calculated at a particular temperature, if its Heat of Vaporization is known, using the Clausius-Clapeyron Equation.
    • In its basic form the equation is:
    • P = A x e -(  H / RT)
    • Or in expanded form:
    • ln P 2 – ln P 1 = (  H vap / R) ((1 / T 1 ) – (1 / T 2 ))
    • Where: P 1 = vapor pressure at T 1 in Kelvin
    • P 2 = vapor pressure at T 2 in Kelvin
    •  H vap = Heat of Vaporization in joules
    • R = 8.31 joules / mole Kelvin
    *
  • 28. ln P ln P = Hvap 1 1 T T R 1 2 ( ) _ _ _ 1 2 _ _ R = 8.31 JOULES / MOLE K H VAP IN JOULES T = KELVIN TEMPERATURE PRESSURE IN TORRS OR ATM THE CLAUSIUS – CLAPEYRON EQUATION *
  • 29. RELATING VAPOR PRESSURE, TEMPERATURE &  H VAPORIZATION
    • Problem: Hexane, C 6 H 14 , has a heat of vaporization of 30.1 Kj / mole. At 25 0 C its vapor pressure is 148 torrs. What is the normal boiling point of hexane?
    • Solution:
    • ln P 2 – ln P 1 = (  H vap / R) ((1 / T 1 ) – (1 / T 2 ))
    • P 2 = 760 torr (normal air pressure) recall the definition of boiling previously mentioned
    • P 1 = 148 torr, T 1 = (25 + 273) = 298 K
    •  H vap = 30.1 x 10 3 joules / mole, T 2 = ?
    • ln(760) – ln(148) = (30.1 x 10 3 / 8.31)((1/298) –(1/T 2 ))
    • 1.64 = 3622(0.00336 – (1/ T 2 )), (1.64 / 3622) = 0.00336 – (1/T 2 )
    • (1/T 2 ) = 0.00336 – 0.000453 , T 2 = 344 K = (344 – 273) = 71 0 C
    *
  • 30. COLLIGATIVE PROPERTIES OF SOLUTIONS
    • Colligative means “collective”. Recall that “The Gas Laws” were a set of equations that applied to all gases generally irrelevant of their individual nature. Boyle’s Law, Charles Law, Gay-Lussac’s Law, etc. were applied to hydrogen, oxygen, neon, etc. equally well. (In actuality all gases were assume to exhibit ideal behavior which is not entirely true.)
    • In the case of solutions we can also discuss the behavior of solutions in general without regard to the individual nature of the particular solutes and solvents used. Again we will assume ideal behavior which is not always entirely true.
  • 31. Vapor Pressure of Solutions
    • A solution is a solute dissolved in a solvent. The vaporization of a solution differs from that of the solvent from which it is made. The differences are based on the molar composition of the solution and the nature of the solute and solvent used.
    • The vapor pressure of a solution is related to its molar composition by Raoult’s Law. It tells us that the vapor pressure is directly related to the mole fraction of the solvent(s). Mole fraction refers to the number of moles of a particular component out of the total moles present.
    • Mole fraction (X) = (moles of substance / total moles)
    • Raoult’s Law:
    • P A = (mole fraction of A) x (vapor pressure of A)
    • And if more than one liquid components are present:
    • P SOLUTION = P A + P B + P C + …
    *
  • 32. Vapor Pressure of Solutions
    • Problem: What is the vapor pressure of an aqueous solution containing 250 grams of water and 25.0 grams of sucrose (C 12 H 22 O 11 ) at 35 0 C? The vapor pressure of water at 35 0 C is 41.2 torr.
    • Solution: Find the mole fraction of the water.
    • Mole fraction (X) = (moles of substance / total moles)
    • X water = (250 / 18) / ((250/18) + (25/342)) = 13.89 / 13.95
    • X water = 0.996
    • P water = (mole fraction of water) x (vapor pressure of water)
    • P water = 0.996 x 41.2 = 41.0 torrs
    • Note – the vapor pressure of a solution is always less than that of the pure solvent from which it is made.
    *
  • 33. Vapor Pressure of Solutions
    • When a solution consists of two or more liquid components, the vapor pressure of the solution requires that the vapor pressure of each component be determined using Raoult’s Law and then added together.
    • Problem: Find the vapor pressure of a solution consisting of 29 grams of acetone (C 3 H 6 O) and 9.0 grams of water at 20 0 C.(v.p. or acetone = 162 torrs and v.p. of water = 17.5 torrs)
    • Solution: Find the moles fractions of each
    • Mole fraction (X) = (moles of substance / total moles)
    • Moles of acetone = 29 / 58 = 0.500 moles
    • Moles of water = 9.0 / 18.0 = 0.500 moles
    • P A = (mole fraction of A) x (vapor pressure of A)
    • X acetone = 0.500 / (0.500 + 0.500) = 0.500, V.P. = 162 x 0.500 = 81.0 t
    • X water = 0.500 / (0.500 + 0.500) = 0.500, V.P. = 17.5 x 0.500 = 8.75 t
    • P SOLUTION = P A + P B + P C + …
    • P solution = 81.0 torrs + 8.75 torrs = 89.8 torrs
    *
  • 34. Freezing point Depression & Boiling point Elevation of Solutions
    • Solutions always have lower freezing points and higher boiling points than the solvents from which they are made.
    • The exact freezing point depressions and boiling point elevations can be calculated for solutions using the equation:
    •  T = k x m x i
    • k is called the freezing point depression constant or the boiling point elevation constant depending upon the calculation
    • m is the molali t y of the solution (not molarity !)
    • molality = moles of solute / kilograms of solvent
    • i is called the Van’t Hoff factor = the number of particles (ions or molecules) in solution formed by each dissolved solute component
  • 35. Sample Calculations for Freezing Point Depression & Boiling Point Elevation of Solutions
    • Problem: What is the freezing point of a solution made with 20.0 grams of sucrose (C 12 H 22 O 11 ) in 100 grams of water? (k freezing = 1.86 0 C / m)
    • Solution:  T = k x m x i
    • molality = moles of solute / kilograms of solvent
    • Moles = 20.0 g / 342 grams per mole = 0.0585
    • Molality = 0.0585 / 0.100 kg of water = 0.585 m
    •  T = 1.86 0 C/m (0.585 m) 1 = 1.09 0 C
    • i = 1 because sucrose does not make ions. One molecule dissolves to make only 1 dissolved molecule!
    • The normal freezing point of water is 0 0 C and it is lowered by 1.09 0 C
    • F.P. solution = normal F.P. -  T
    • F.P. solution = 0 0 C – 1.09 0 C = -1.09 0 C
  • 36. Sample Calculations for Freezing Point Depression & Boiling Point Elevation of Solutions
    • Problem: Find the molecular mass of a substance if 5.65 g when dissolved in 110 g of benzene gives the solution a freezing point of 4.39 0 C. (k f for benzene is 5.07 0 C/m and normal f.p. is 5.45 0 C)
    • Solution:
    • If we find molality we can then calculate molar mass
    •  T = k x m x i, benzene does not form ions, i = 1
    • (5.45 0 C – 4.39 0 C) = 5.07 0 C/m x m x 1, molality = 0.209 m
    • molality = moles of solute / kilograms of solvent
    • 0.209 m = moles of solute / 0.110 kg, moles = 0.0230 moles
    • Moles = grams / molar mass, 0.0230 mole = 5.65 g / molar mass
    • Molar mass = 5.65 g / 0.0230 moles = 246 grams per mole
  • 37. Sample Calculations for Freezing Point Depression & Boiling Point Elevation of Solutions
    • Problem: Find the boiling point of a 0.0500 molal solution of calcium chloride. Assume the CaCl 2 dissociates 100%. The K b for water is 0.512 0 C/m.
    • Solution:  T = k x m x i
    • CaCl 2(s)  Ca +2 (aq) + 2 Cl - (aq) , three particles (ions) are formed from each CaCl 2 dissolved, i = 3.
    •  T = 0.512 0 C/m (0.0500) x 3 = 0.0768 0 C
    • The normal boiling point for water = 100.000 0 C
    • B.P. solution = normal B.P. +  T = 100.000 + 0.0768
    • B.P. solution = 100.0768 0 C
  • 38. Osmotic Pressure
    • Osmosis is defined as the diffusion (moving) of water across a semi-permeable membrane from an area of higher concentration to an area of lower concentration.
    • A semi-permeable membrane is one that a allows some materials to pass through while blocking others. As water moves through a membrane at unequal rates, a higher concentration and pressure builds up on the side of the membrane with the larger quantity of water.
    • Osmotic pressure can be calculated using the equation:
    •  = M x R x T ,  is used to represent osmotic pressure
    • M = Molarity (not molality)
    • R = 0.0821 atm liters / moles K
    • T = Kelvin temperature
    • This equation is derived from: PV = nRT, P = (n/V) RT
    • Since moles / liters = molarity, Pressure = M x R x T
    *
  • 39. Osmotic Pressure water sugar Semi permeable membrane Allows water to pass through but not sugar molecules More water moves left and pressure rises in the left chamber *
  • 40. Osmotic Pressure
    • Problem: Find the osmotic pressure in the system shown on the previous slide if the sugar concentration is 0.0010 M and the temperature is 25 0 C.
    • Solution:  = M x R x T
    •  = 0.0010 M x (0.0821 atm liter / mol K) x 298 K
    •  = 0.024 atm or 0.24 x 760 = 18 torrs
    *
  • 41. Final Note
    • All solutions used in the preceding problems are assumed to exhibit ideal behavior meaning that they conform to the laws and equations discussed and that no solute particle interaction occurs.
    • In reality, no solution is ideal just as no gas was ideal when the gas laws where discussed.
    • Like gases, some solutions are more ideal than others. Recall that gases of low molecular weight which there non polar, at high temperatures and low pressures were most ideal. Solutions with solutes which are non polar and non ionic and which are very dilute approach ideal behavior most closely.
    • Like the gas laws, the relationships for solutions give good approximations for many systems.
  • 42. SUMMARY
    • (1) q = m x c x  T (Heat to warm or cool a material)
    • (2) Heat lost = Heat gained (Conservation of Energy)
    • (3) Q = Heat of Phase Change x Amount of Substance
    • (Heat lost of gained during a phase change)
    • (4) ln P 2 – ln P 1 = (  H vap / R) ((1 / T 1 ) – (1 / T 2 ))
    • ( Clausius-Clapeyron Equation for vapor pressures)
    • (5) P A = (mole fraction of A) x (vapor pressure of A)
    • (6) P SOLUTION = P A + P B + P C + …
    • (Equations (5) & (6) are Raoult’s Law)
    • (7)  T = k x m x i (Changing F.P. or B.P. for solutions)
    • (8)  = M x R x T (Osmotic Pressure)
  • 43. The End