Basic Properties of Solids, Liquids, and Gases Solids, Liquids, and Gases 5 th International Junior Science Olympiad (IJSO) Dr. Yu­San Cheung [email protected] Department of Chemistry The Chinese University of Hong Kong 2 1 Properties of Gases Characteristics of Gases What can we study about a gas (e.g. in a balloon)? Pressure: the gas makes the balloon expand (against ambient atmospheric pressure and tension of the balloon). No definite volume or shape: A gas fills whatever volume is available to it and is easy to compress. Temperature: if the balloon is left in a room long enough, the temperatures of the balloon and the gas are the same as that of the room. Low densities: (density = mass ÷ volume) Compared with those of liquids and solids: one mole of liquid water at 20˚C (298 K) and 1 atm pressure occupies a volume of 18.8 cm 3 , whereas the same quantity of water vapor at the same temperature and pressure has a volume of 30200 cm 3 , more than 1000 times greater. Volume: the gas fills out the whole space inside a container. (The volume of gas is taken as the container capacity, which is usually assumed to be the container volume if the wall of a container is thin.) What is the relationship between these properties? 3 4 Pressure The Pressure of a Gas The molecules of a gas, being in continuous motion, frequently strike the inner walls of their container. As they do so, they immediately bounce off without loss of kinetic energy(動能) , but the reversal of direction (acceleration) imparts a force to the container walls. This force, divided by the total surface area on which it acts, is the pressure of the gas. When a force (F) is acting on a surface with area A, a pressure (p) exerts on the surface: Downward force due to weight of mass acting on piston of area A, creating a pressure p (assuming no air outside) p = F / A The pressure of a gas is observed by measuring the pressure that must be applied externally in order to keep the gas from expanding or contracting. 5 Moveable piston (frictionless and weightless) When the piston is stationary, the gas pressure is exactly equal to p. That means the gas creates an opposing force F = p⋅A which is equal but opposite in direction to the force created by gravity acting on the weight.
6 How is pressure measured? How is pressure measured? Barometer(氣壓計) Atmospheric Pressure is measured by an instrument called barometer ( 氣 壓 計 ) , invented in the early 17th century. The barometer (氣壓計) consists of a vertical glass tube closed at the top and evacuated, and open at the bottom. liquid http://en.wikipedia.org/wiki/Barometer 7 8 Pressure Exerted by a Liquid Column Column height = h Cross­section area = A Volume = h A Mass = volume × density = h Aρ
(ρ = density) Force produced by the column = Weight = mg = h Aρ g A h (g = gravity acceleration = 9.80665 m ·s−2 ) Pressure = F / A = ρ g h p = p 0 + ∆p
∆p = pressure exerted by the height difference Note that the cross­section must be uniform, but not necessarily circular. 9 Choice of Medium
10 Mercury as the Medium ∆p = ρ g h Advantages: For the same ∆p, higher density, smaller height difference. ­ High density ­ Chemically inert (e.g., water dissolves some gases such as CO 2 , NH 3 , HCl) ­ Not sticking to the glass wall (compared with water) For mercury: 1 standard atmospheric pressure ⇒ 760 mm For water: water density = 1/13.6 of mercury density So, if water is used, 1 standard atmospheric pressure
⇒ 760 mm × 13.6 = 10.34 m Disadvantages: In practice, mercury is used. ­ Vapor is toxic ­ Difficult to deal with spillage (not absorbed by paper towel, tending to form small droplets, difficult to remove with dropper)
Other advantages? But any disadvantages? 11 12 Mercury Barometer Mercury Barometer (Closed­end) One end open to atmosphere sealed end vacuum inside Accuracy: depending on the ruler readout (usually 1 mm) The other end connected to closed container containing the gas p = ∆p 13 Pressure Gauge 14 Principle of Pressure Gauge Also called barometer, but the term “Pressure Gauge” (壓強計) is more commonly use in the market A typical Bourdon tube contains a curved tube. It is open to external pressure input on one end and is connected mechanically to an indicating needle on the other end. For measuring both “positive pressure” and “negative pressure” (relative to atmospheric pressure) The external pressure is guided into the tube. Change in pressure causes the tube to flex, resulting in a change in curvature of the tube. This curvature change is linked to the dial indicator for a number readout. “Negative pressure”: also called vacuum Alternatively, a strain gauge circuit can be used to produce output electronically. If you are interested, see: “Positive pressure”: can be measured up to 1380 atmospheric pressure http://www.efunda.com/DesignS tandards/sensors/strain_gages/s train_gage_theory.cfm http://en.wikipedia.org/wiki/Pressure_measurement http://www.efunda.com/DesignStandards/ sensors/bourdon_tubes/bourdon_intro.cfm http://en.wikipedia. org/wiki/Pressure_ measurement 15 16 Principle of Ionization Gauge Principle of Thermo­conductivity Gauge Principle: Principle: ­ A filament is heated by running electrical current through it. ­ A thermocouple thermometer measures the filament temperature. ­ High pressure: more gas molecules ⇒ higher heat loss (by convection) ⇒ lower temperature ⇒ electrical signal ­ A filament is heated to emit electrons. ­ These electrons ionized gaseous molecules. ­ Ions (gaseous molecule ions) are attracted to the cathode of a electrical circuit and an electrical current is resulted. High pressure vs. electrical signal: may be complicated Different gases: different calibration High pressure vs. electrical signal: may be complicated Different gases: different calibration Range: 10 –3 – 10 –10 mmHg Range: 10 – 0.001 mmHg http://www.varianinc.com.cn/ products/vacuum/measure/ transducers/531gauge/shared/ 531gauge­180.jpg http://www.varianinc.com.cn/products/vacuum/measure/shared/uhv24a­180.jpg
17 18 Units of Pressure Units of Pressure Other commonly used units: The unit of pressure in the SI system is Pascal (Pa)(帕斯卡) (i) psi = pound per square inch 1 Pa ≡ 1 N·m− 2 ≡ 1 J·m− 3 ≡ 1 kg·m− 1 ·s− 2 1 pound = 453.59237 g = 0.45359237 kg 1 inch = 2.54 cm = 0.0254 m ⇒ 1 Pa = 1 N·m− 2 ⇒ 1 J = 1 N·m = 1 (Pa·m 2 ) · m ⇒ 1 N = 1 kg·m·s− 2 p = F / A
Energy = F ·d
F = m·a
Exercise: 1 psi = ?? Pa (You also need: gravity acceleration g = 9.80665 m ·s− 2 ) 19 Units of Pressure 20 Determination of Volume (ii) 1 torr = 1 mm Hg (implying the pressure produced by a Hg column of 1 mm height) Pressure = F / A = ρ g h (p. 11) By calculations: 1 torr = 133.32 Pa (ρ = 13595.1 kg·m−3 for mercury) Sphere:
Rectangular shape: width × depth × height (iii) 1 atm = 760 torr 4 3 π r 3 1 Pyramid:
× base area × height (including cone) 3 1 atm = 101325 Pa = 14.696 psi More complicated shape but well­defined by mathematical formulas: by calculus Irregular shape: by empirical measurements (1) Immersing the object in liquid (2) For container only: check how much liquid is needed to fill out the container (thinner wall, better result) (iv) 1 bar = 10 5 Pa Important: 1 bar is close to 1 atm, but not exactly equal! 21 Units of Volume 22 Types of thermometer (1) Glass thermometer(玻管液體溫度計) Unit of volume = (Unit of length) 3 Glass tube containing a liquid, e.g., _________, ________ High temperature: _______________ of liquid Low temperature: _______________ of liquid SI unit for volume : m 3 Units used in daily life: liter, dm 3 , ml (or mL), cm 3 (2) Resistance thermometer(電阻溫
度計) 1 m 3 = (1 dm) 3 = 1 dm 3 , also called liter (L) Consisting of a probe (e.g., platinum) and a controller (with battery, read­out, and electronic circuit). The resistance of the probe changes with temperature. By measuring the resistance, the temperature can be determined.
1 ml = 1 milli­liter = 10 ­3 L = 10 ­6 m 3 = 10 ­6 (100 cm) 3 = 1 cm 3 Smaller unit: µL (micro­liter, 10 ­6 L) , nL (nano­liter, 10 ­9 L) Summary: fill out the blanks below with appropriate units m 3 ÷ 1000 L dm 3
÷ 1000
mL ml cm 3
÷ 1000
µL ÷ 1000 nL
http://en.wikipedia.org/wiki/Thermometer 23 24 Types of thermometer Unit of Temperature (3) Infra­red (IR) thermometer(紅外
線溫度計) All objects emit IR light. The hotter an object is, the stronger the IR light emitted. By measuring the IR light intensity, the temperature can be determined. Celsius scale: 0 °C = freezing temperature of water 100 °C = boiling temperature of water (What is its main advantage over the other types?) at 1 atm pressure for both (4) Liquid crystal thermometer Containing dots of heat­sensitive liquid crystals on a plastic strip. The dots change color at different temperatures. http://en.wikipedia.org/wiki/Thermometer 25 Pressure­volume relations: Boyle‘s law (波義耳定律) 26 Boyle's law & J­tube Robert Boyle (1627 – 91) Each time Hg is added, h 1 and h2 changes, but p 0 and h 0 remain unchanged. Volume p 0 Trapped air h 0 Pressure of trapped air = p 0 + h 1 – h 2 Volume = (h0 – h2 ) ×
cross­sectional area Pressure Exercise: p0 = 0.987 atm h1 = 45.9 cm, h 2 = 10.7 cm Modified from: http://www.grc.nasa.gov/WW W/K­12/airplane/aboyle.html h 1 h 2 p = ?? cm Hg 27 Boyle's Law p ∝ 1/V (i.e., p is inversely proportional to V ) 28 Boyle's Law p = “constant” × 1/V 1 / Volume
•
We can also say that: p is proportional to 1/V •
1 / Volume
leads to: •
••
pV = “constant” •
•
•
••
Pressure Pressure p = “constant” × 1/V “Constant” here means that: It does not change with p and V. If p changes, V also changes. V changes in such a way that pV remains the same. Plot of p vs. 1/V yields a straight line passing through the origin 29 30 Boyle's Law Boyle's Law p 1V
1 = p 2V
2 Example: Alternative form (more useful): Initial: p 1 , V 1 If p 1 = 700 torr, V 1 = 1 L, p 2 = 100 torr, V 2 = ? Final: p 2 , V 2 Ans.: V 2 = p 1V
1 / p 2 = (700 × 1) / (100) = 7 L p 1V
1 = constant = p 2V
2 Exercise: i.e., If p 1 = 700 torr, V 1 = 1 L, V 2 = 100 L, p 2 = ? p 1 V 1 = p 2 V 2 Ans.: 32 31 Effect of Temperature on p­V Curve Effect of Temperature on p­V Curve Try finding out the “constant” for each curve: Modified from: http://www.che m1.com/acad/w ebtext/gas/gas_ 2.html 500K: pV = 400K: pV = 300K: pV = 200K: pV = 100K: pV = 50K: pV = 33 34 Effect of Temperature on p­1/V Curve 1/V Exercise: In an industrial process, a gas confined to a volume of 1 L at a pressure of 20 atm is allowed to flow into a 4 L container by opening the valve that connects the two containers. What will be the final pressure of the gas? (Hint: at the end, the gas fills up both containers.) Temperature: increasing or decreasing? 20 atm at the beginning
Vacuum at the beginning p 4 L 35 1 L 36 Temperature – Volume Relationship: Put it another way, V varies linearly with T, i.e., the T­V graph is a straight line. V, liter Charles‘ Law (查理定律) p 1 When the straight lines are extrapolated, they reaches the x­axis at –273.15 ˚C. p 2 p 3 With the pressure held constant, the volume of a gas changes by the same amount for each °C change in temperature. 0 T, ˚C http://www.chem1.com/acad/webtext/gas/gas_2.html 37 38 Temperature­Volume Relationship V Kelvin Scale (K): adding 273.15 to the °C value e.g., ­273.15 °C
0.00 °C
25.00 °C
100.00 °C
Using the Kelvin scale, we have T ∝ V (T is proportional to V ) ⇒ 0.00 K (Note: the symbol “°” is not used) ⇒ 273.15 K ⇒
K ⇒
K or T / V = constant If talking about temperature difference or temperature change, change in 1 °C = change in 1 K. Diff =
°C Diff =
°C ­273.15 °C
0.00 °C
25.00 °C
100.00 °C
⇒
0.00 K ⇒ 273.15 K ⇒ 298.15 K ⇒ 373.15 K 0 °C ­273.15 °C T V or T 1 / V 1 = T 2 / V 2 Diff = K T / V = constant ⇒
(i) V = 0 if T =0 (ii) V is –ve if T is –ve Diff = K T 0 K 273.15 K 39 40 Charles’ Law
Kelvin Scale: what a big deal? Fahrenheit set zero degree (0 °F = –17.78 °C) for the lowest temperature reachable at that time (1724), so as to avoid negative temperature in practical life. T 1 / V 1 = T 2 / V 2 Example: If T 1 = 200 K, V 1 = 1 L, T 2 = 100 K, V 2 = ? Celsius: more open­mined, not minding negative temperature, choosing convenient scale: freezing point of water for 0 °C. Ans.: V 2 = T 2 V 1 / T 1 = (200 × 1) / (100) = 2 L But there is no clue what the lowest temperature is in nature. Exercise: Charles’ law: 0 K is the lowest temperature in nature, otherwise we have negative volume of a gas, which is physically unacceptable. If T 1 = 200 °C , V 1 = 1 L, V 2 = 100 cm 3 , T 2 = ? 0 K: also called absolute zero Kelvin Scale: also called the absolute temperature scale Ans.: Kelvin Scale is assumed in scientific formulas (unless otherwise specified). 41 42 Boyle’s Law + Charles’ Law = What? Boyle’s Law p ∝ 1/ V or V ∝ 1/p The tires of a car were filled with air to a pressure of 30 psi at 25 °C. After the car running for several hours, the temperature raised to 100 °C and the tire volume increased by 10%. What was the pressure of the air in the tires? V ∝ T / p + Exercise: pV / T = constant Charles’ Law V ∝ T or p 1 V 1 / T 1 = p 2 V 2 / T 2 43 Amadeo Avogadro (1776­1856): "E.V.E.N principle" 44 Mole Concept 32 g of oxygen gas = 6.022 × 10 23 oxygen gas molecules = 1 mole of oxygen gas molecules Equal volumes of gases, measured at the same temperature and pressure, contain equal numbers of molecules. Combined with pV / T = constant, we have i.e., 1 mole = 6.022 x 10 23 c.f.: 1 dozen of pencils = 12 pencils 1 kB = 1024 bytes 1 catty of nails = ??? pieces of nails pV / NT = constant = k (Boltzmann constant) Avogadro Number: N A = 6.022 × 10 23 mol−1 i.e., V ∝ N (N = no. of gas molecules) Mole number (n) = numbers of moles = N / N A or pV = NkT e.g., 0.01 mol of O 2 gas molecules = 0.01 x 6.022 x 10 23 = 6.022 x 10 21 O 2 gas molecules (Ideal Gas Law) 45 Mass and Mole Molecular Mass
?? grams of a gas = 1 mole of the gas molecules 50.5 g of CH 3 Cl = 1 mole of CH 3 Cl We define: molar mass / molecular mass / molecular weight (M w ) You need to find out the atomic mass units of the atoms (from books or internet) e.g. M w for CH 3 Cl = 50.5 g mol−1 e.g. H: 1.0 amu (amu = atomic mass unit) C: 12.0 amu Cl: 35.5 amu H 2 : 2 x 1.0 = 2.0
⇒
46 e.g. Mass of 0.1 mol of CH 3 Cl = 50.5 g mol−1 x 0.1 mol = 5.05 g M w N A Mass ⇔ Mole ⇔ Number of molecules 2.0 g of H 2 = 1 mole of H 2 CH 3 Cl: 12.0 + 3 x 1.0 + 35.5 = 50.5
⇒
50.5 g of CH 3 Cl = 1 mole of CH 3 Cl 47 48 Ideal Gas Law / Perfect Gas Law pV = NkT = (N A n)kT = n (N A k)T = n (N A k)T = nRT Ideal Gas Law / Perfect Gas Law R : gas constant = 8.314 n : no. of moles (R = N A k) R: gas constant or universal gas constant Unit to be filled out Must be memorized!! pV = nRT pV = NkT Exercise: (i) What is the S.I. unit for R ? (ii) What is the S.I. unit for k ? (iii) If the experimental value for R is 8.314 in S.I. unit. What is the value for k in S.I. unit? N : no. of molecules More commonly used k : Boltzmann constant = Value to be filled out 49 Ideal Gas Law / Perfect Gas Law (a) 50 Ideal Gas Law / Perfect Gas Law (b) pV = nRT pV = nRT ⇒ pV / nT = R = constant
⇒ p 1V
1 / n 1T
1 = p 2V
2 / n 2T
2 There are five symbols in the equation. If four of them are given, the remaining one can be determined. This equation is more useful if there are changes for the variables (p, V, n, and T ). If three of p, V, n, and T are known, the remaining one can be determined because R can be found out from literature. Exercise: A box contained a gas of 1700 torr at 25 °C. A hole was punched and gas leaked out. After the hole was sealed, it was found that half of the gas remained and the temperature dropped to 0 °C. What was the final pressure of the gas? Exercise: how much gas is needed to fill an 1­L box to 700 torr at 25 °C? 51 Ideal Gas Law / Perfect Gas Law 52 Gas Mixture P : total pressure V : volume of the container N : total number of gaseous molecules T : temperature of the mixture p 1V
1 / n 1T
1 = p 2V
2 / n 2T
2 This form includes the Boyle’s Law, Charles’ Law, and EVEN Principle. You can show that by keeping some variables constant. Idea: The two gases act independently.
gases act independently. Boyle’s Law: p 1V
1 = p 2V
2 Charles’ Law: V 1 / T 1 = V 2 / T 2 EVEN Principle: V 1 / n 1 = V 2 / n 2 p = p A + p B n = n A + n B Constants:______ Constants:______ Constants:______ p AV
= n ART p BV
= n BRT
53 V & T : same for both gases Adding: (p A + p B )V = (n A + n B )RT
⇒
pV = nRT 54 Air Composition When dealing with gas mixing, the “volume %” concept is more convenient,e.g., Air is described as a gas containing 20% of O 2 and 80% of N 2 by volume. How to understand this? O 2 1 atm + N 2 O 2 1 atm O 2 Volume: 4 : 1 (Same pressure) N 2 1 atm N 2 1 atm Air 1 atm O 2 0.2 atm + N 2 0.8 atm Air Air 55 Partial Pressure Exercise: The following gases are mixed and put into a 2L box: Partial Pressure: the pressure of each gas in a gas mixture 3 L of O 2 at 0.5 atm, 2 L of N 2 at 1 atm, and 1 L of CO 2 at 1 atm If gases A and B with the same pressure are mixed in volume ratio V A :V B , then the partial pressures ratio in the mixture are p A :p B = V A :V B 56 (i) Calculate the pressure for each of the three gases before mixing if the volume of each gas is changed to 2L box. (ii) Determine the partial pressures for the three gases after mixing. (iii) Calculate the total pressure of the mixture. (according to the figure on p. 55) and the mole ratio n A :n B is also V A :V B (because p ∝ n according to the ideal gas law) 57 Molar Volume of a Gas: Standard Temperature and Pressure 58 Density of a Gas (i) Number density: no. of gas particles per unit volume = N / V = p / kT For convenient comparison it is customary to use the conditions of STP, standard temperature and pressure where T = 273.15 K and p = 1 atm. (ii) Mole density: no. of moles of gas particles per unit volume = n / V = p / RT Molar volume is the volume of one mole of a substance. Standard molar volume is the molar volume at STP. (iii) Mass density: mass of gas particles per unit volume = mass of gas / V = n × molecular weight / V = molecular weight × p / RT
Exercise: Calculate the standard molar volume of a gas. 59 60 Mass Density Change of States (iii) Mass density: Liquid molecular weight × p / RT We can conclude that: (a) hot air is lighter than cold air. (Where should we put air heater and conditioner in a room? Near Ceiling or on the floor?) Solid (b) gas with smaller molar mass is lighter than gas with larger molar mass. Sublimation Gas Deposition 61 62 63 64 Vapor Pressure When the temperature of a liquid reaches its boiling point, it becomes a gas. Why? At the boiling point, the molecules have enough energy to move into the space above the liquid. But even if the boiling point is not reached, some molecules still have enough energy to leave the liquid (surface). They become gas (though we call them vapor in our daily life). Like a pure gas, this vapor also exerts a pressure. The pressure is called vapor pressure. Condensation As long as the liquid and vapor co­exist, the vapor pressure depends on temperature only. ~ v.p. = p 0 ·exp(−∆H vap /RT )
Exercise: Why, when a person walks out a building in summer, his/her glasses get moist?
~ ∆Hvap = amount of heat needed to evaporate the liquid (similar to the latent heat of vaporization) p 0 is a constant Some observations from the equation: (i) T ⇑ ⇒ v.p. ⇑
(ii) If we put the setup into a water­bath to measure v.p. ~ at different temperature, we can obtain ∆H vap . 65 66 How does the vapor pressure change in the following cases? (Temperature remains unchanged) ? ? •
•
•
•
•
Liquid
A closed container filled with 1/3 of liquid. ? ? Liquid ? The same in both cases because the temperature is the same. (i) Same volume of liquid and space above the liquid, but the surface area of the liquid is doubled. ? ? ? ? ? Liquid Answer: (i) Puzzle: the surface area of the liquid is increased, so more molecules escape from liquid? Answer: there are also more molecules returning to the liquid. (ii) Same container, but the volume of the liquid is is doubled. (ii) The vapor also obeys the ideal gas law. Though p and T are the same as before, V is halved and so is n, i.e., the amount of vapor is halved. 67 68 Vapor Pressure of Some Liquids Comparisons of Gas and Vapor In general, the weaker the intermolecular interaction, the higher vapor pressure is. (In many cases, the term “vapor” implying that it is in contact with liquid.) (i) Gas obeys the ideal gas equation. If T is fixed, two of the variables, p, V, and n are free to be varied. (The remaining one is fixed through the law.) (ii) Vapor obeys the ideal gas equation and the formula on p. 65. If T is fixed, p is also fixed. Only one of the variables, V and n can be varied. In addition, the mole density (n / V ) is constant. That is, once T is fixed, so is the mole density (and other types of density such as number density and mass density.) 69 From Vapor to Gas 70 Boiling An 1­m 3 box contains 3 mol of ethanol at 273 K. At the beginning, liquid and vapor co­exist. How does the v.p. change with increasing temperature? A liquid boils when its vapor pressure is equal to the external pressure acting on the liquid surface.
p­T curve for gas density of 3 mole/m 3 (iii) After all liquid evaporates, the vapor becomes a “pure gas” and the v. p. follows the p­T curve. (ii) At this point, all liquid evaporates. (i) At the beginning, the v. p. follow this curve, more and more liquid evaporates. 71 72 Boiling at Different External Pressures If the external pressure increases, higher vapor pressure is required for boiling, so the boiling point is higher. And vice versa. Question: from the graph on the left, what are the boiling points for ethanol at external pressure = 600 and 800 torr? Evaporation vs. Boiling Ordinary evaporation is a surface phenomenon. Below the boiling point, the vapor pressure is lower than the outside pressure and bubbles of water vapor cannot form. But at the boiling point, the vapor pressure is equal to atmospheric pressure. Bubbles form and the vaporization becomes a volume phenomena. Pressure cooker: boiling point > 100 °C At hill top, boiling point < 100 °C 73 74 Properties of Solid Why Ice Floating on Water? Ice floats on water because of its lower density. (i) Having a definite shape At 0 °C, density of ice = 0.917 g/cm 3 density of water = 0.9988 g/cm 3 (ii) Difficult to compress (as the particles are already packed closely together) (iii) Usually denser than the liquid (as the particles are packed more closely together than in liquid), except: water (according to http://en.wikipedia.org/wiki/Ice, it is the only non­metallic substance having this property) There are specific orientations between different water molecules, leaving many holes inside ice and a hollow structure is resulted. (iv) With particles vibrating around fixed locations, not moving around http://www.uwgb.edu/dutchs/PETROLGY/Ice%20Structure.HTM 75 Phase Diagram “Heavy Ice” Floating on Water However, heavy water ice (ice of the heavy water, D 2 O) sinks in water! Whether a substance exists in this phase or that phase depends on the pressure and temperature. A phase diagram indicates the phase of a substance at various combinations of pressure and temperature. The structure of a solid mainly depends of the chemical properties of the particles. We expect the two types of ices to have the same structure and the numbers of molecules per unit volume (“number densities”) are the same. But 76 p molar mass of H 2 O = 1x2+16 = 18 g/mol molar mass of D 2 O = 2x2+16 = 20 g/mol Common phases for most of the substances are: solid, liquid, and gas. Hence, there are usually three regions in a phase diagram. The following is a typical one. (Note the locations of the regions.)
Density of D 2 O ice = (20/18) x density of D 2 O ice = 1.02 g/cm 3 i.e., 2% higher than that of water! Question: can you tell which region is for solid, which one is for liquid, and so on? Would you expect that heavy water ice floats or sinks in heavy water? T 77 78 Phase Diagram Phase Diagram Answer: from left to right, temperature increases, so you go through the three phases in the order of: solid → liquid → gas Line segments: places where two phases co­exist p solid & liquid p Liquid Liquid Solid Solid Gas liquid & gas Gas T T solid & gas Triple point: three phases co­existing 79 Phase Diagram 80 Phase Diagram How melting point & boiling point change with external pressure What does it tell us? Melting point & boiling point p p p 2 p 1 Liquid p 1 Liquid Solid Solid Gas Gas T T Melting point at p 1 Melting point at p 1 Boiling point at p 1 Melting point at p 2 Boiling point at p 1 Boiling point at p 2 81 82 Decreasing Melting Point of Water With Increasing Temperature Phase Diagram The steepness of a line segment tells us the following: Hollow structure of solid
⇒ Volume decreasing on melting
⇒ M. P. decreases with increasing pressure
⇒ Negative slope of the solid­liquid segment (i) If the slope is positive, the melting point/boiling point increases with increasing temperature. - solid­gas and liquid­gas segments: positive slope - solid­liquid segment: usually positive slope (one exception: water) - solid­liquid segment: usually very steep (whether +ve or –ve) Note: the phase diagram indicates (or shows) that the m.p. of water decreases with increasing pressure. It does not prove or explain it!
(ii) If the line is steep, the melting point/boiling point is insensitive to pressure change. - solid­liquid segment: usually very steep (whether +ve or –ve) - solid­gas and liquid­gas segments: usually not as steep 83 84 From Solid to Gas: Sublimation From Solid to Gas: Sublimation There are not many substances subliming at 1 atm. Examples are CO 2 and iodine. But under high­enough external pressure, these substances can go through the stages from solid to liquid to gas. Substances not subliming at 1 atm can sublime if pressure is lowered. http://userpages.umbc.edu/~neumann/Chem102/Notes/ch11/c1306d.html http://userpages.umbc.edu/~neumann/Chem102/Notes/ch11/c1306d.html 85 Significance of Triple­point 86 Critical Point External pressure > triple­point pressure: solid ⇔ liquid ⇔ gas External pressure < triple­point pressure: solid
⇔
gas In a phase diagram, the liquid­gas boundary stops somewhere. 87 88 Critical Point Critical Point But, if the temperature is too high (above the “critical temperature” (臨界溫度)), the gas cannot be liquefied, however high the pressure is applied. A gas can usually be liquefied by increasing pressure and/or decreasing temperature. As the path crosses the liquid­gas boundary, the gas changes to liquid. In order to achieve liquefaction, the temperature must be lowered to a value below the critical temperature.
89 90 Critical Point Supercritical Fluid How to rationalize the existence of critical temperature(臨界
溫度)? The substance at the rectangular region with T
> T c and p > p c is called supercritical fluid. If the temperature is too high, the motions of particles are too vigorous. Attraction between the particles is not strong enough to hold the particles together, even though they are brought very closely together (by high pressure). Therefore, the gas must be cooled down to slow down the particle motions. The molecular attraction may overcome the particle motions. The fluid is not a liquid, but more like a gas. It may have high density (e.g., comparable to that of liquid). p c : Critical pressure T c : Critical temperature 91 Formation of Supercritical Fluids 92 Properties of Supercritical Fluid http://www.chem.leeds.ac.uk/People/CMR/criticalpics.html http://sunny.vemt.bme.hu/sfe/angol/supercritical.html 93 94 Cleansing of Electronic Circuit Boards Green Chemistry in Electronic Industry coating Silicon wafer etching Using Chlorofluorocarbons (CFCs) Reasons for using CFCs as solvent in electronics cleaning: 1. 2. 3. 4. 5. washing Chips soldering Inert Volatile Low surface tension Non­flammable Non­corrosive Cl
Cl Cl C F C F F 1,1,2­trichloro­1,2,2­trifluoroethane (CFC­113)
Circuit board 95 96 Ozone Depletion Caused By CFCs Supercritical CO 2 Cleaning System Enormous ozone “hole” over Antarctica http://www1.boc.com/eco­snow/index.htm Increased damages to • • • Human immune system Skin Ecology 97 98 Supercritical CO 2 in Extraction Graph Plotting – A tool for data analysis Also used in extraction (of components in herbs) • Data: collected observations and facts Advantages over boiling water: (1) (2) (3) • Regularities in observations can be found through careful analysis of data. • Graphing is a way to present data that shows relationships among data analyzed. Commercial product: http://elchem.kaist.ac.kr/vt /chem­ed/sep/sf/sfe.htm Home­made apparatus at CUHK →
99 100 Graph Plotting What Should Be Included in a Graph 101 • Title • Axis, legend, number, tick, unit • Data point • If you fit the data to an equation, the best­fit line (or “trendline” in some computer softwares), the equation and the square of correlation coefficient (R2 ) obtained should also be included. • If your purpose is to present a formula with a graph, connect the data points with lines.
102 Choosing x and y for Plotting Choosing x and y for Plotting A straight line has the form y = mx + c (m: slope, c: y­intercept) A straight line has the form y = mx + c (m: slope, c: y­intercept) Sometimes we have to rearrange the equation Sometimes we have to rearrange the equation Constants (known/unknown) e.g., A particle falls freely. The distance at different time is measured. We have a series of data (t, x). According to the classical mechanics, t and x obeys the equation: x = x 0 – (1/2)gt2 . x x = –(1/2)g t 2 + x 0 cf. y = m x + c y = m x + c To obtain a straight line, we plot x against t2 . So, slope = –(1/2)g, y­intercept = x 0 Line­fitting gives the value of the slope and y­ intercept. We get the value of g (gravity acceleration) from the slope. Known (variables/constants) 103 Example 104 Line­fitting The mass, A, of a radioactive substance is: Some computer software (such as Excel) provide fitting other than straight line: polynomial, logarithmic, exponential, and power. A = A 0 e−kt where A 0 is the amount at t = 0 and k is an unknown constant Take ln on both sides: ln(A) = ln(A 0 e−kt ) = ln(A 0 ) + ln(e−kt ) = ln(A 0 ) − kt ln(A) = −kt + ln(A 0 ) y = , x = , m = , c = . 105 106 Reading a Graph Reading a Graph
M ass of Beaker vs. Volume for W ater (273 K) Experiment measuring the density of water: A beaker containing different volumes of water is weighed. The mass of the beaker, M, is related to the volume of the water, V, in the following manner: M = ρV + M 0 Plot: M against V Also note the followings: Slope = ρ = 0.9933 y­intercept = M 0 = 30.283 • Unit: (i) Unit for y­intercept = (unit for y) = ________ (ii) Slope = ∆y / ∆x Unit for slope = (unit for y) ÷ (unit for x) = ________ So, ρ = and M 0 = The R2 ­value indicates how good the data fit the trendline. 0 < R2 < 1. For a perfect fit (e.g., fitting a straight line to two data points), R2 = 1. The closer to 1 it is, the better is the fitting. Mass of Beaker (g) The axes do not have to start from zero. In the previous graph, if both axes start at zero, the graph is squeezed and space is wasted. 50 40 30 y = 0.9933x + 30.283 R 2 = 0.9787 20 10 0 0 5 10 15 20 25 Volume (ml) Mass of Beaker vs. Volume for Water (273 K) 47 Mass of Beaker (g) • 60 46.5 46 45.5 R 2 = 1.00000000000 45 15 15.5 16 16.5 Volume (ml) 107 108 Exercise Linear Scale vs. Log­Scale In a second­order reaction, two molecules of A react to form molecule B. The concentration of A, a, changes with time as: a = The following graph shows the pH of a weak acid at different concentration, x 0 . We can tell the pH at x 0 = 0.01 M, 0.02 M, … But it is hard to tell the pH below 0.01 M, where pH ≈ 3 – 5. a0 1 + kta 0 where a = concentration of A (variable) a 0 = initial concentration of A (i.e., a at t = 0, a constant) k = “rate constant” (constant) t = time (variable) pH The following is a set of experimental data: t (s) 199 246 367 686 1200 a (mol·dm –3 ) 0.0094 0.0079 0.056 0.0027 0.0015 Rearrange the formula in the form of y = mx + c, make a plot, and determine k and a 0 (with appropriate units). Ans: k = 0.42 dm 3 ·mol –1 ·s –1 a 0 = 0.041 mol·dm –3 109 Linear Scale vs. Log­Scale 110 x 0 (M) Comparison The part for small x 0 can be expanded with “log­scale” as shown below. We can tell easily that pH = 5.0 at x 0 = 0.00001 M and pH = 3.45 at x 0 = 0.001 M. Linear scale: equal distance, equal increment. Linear scale: equal distance, equal multiplication factor. pH 0.00009 0.00002 0.00003 0.09 0.02 0.03 111 x 0 (M) Reading Log­Scale 112 Exercise http://electrochem.usask.ca/Chem112/Notes/L22.pdf
Shown on the right is the phase diagram for carbon. Find out: = 10 −5 = 10 −4 = 10 −3 1/3 0.00001 = 10 −5 = 10 −2 = 10 −1 (a)Transition pressure for: graphite →
diamond at 1000 ˚C = 10 0 (b)Minimum temperature and pressure for melting graphite into liquid 1/10 = 10 −4.666667 = 0.000022 (c)Boiling point of carbon liquid at 500 atm 0.0001 = 10 −4 = 10 −4.1 = 0.000079 Ans. (a) 25000 atm (b) 3600 ˚C, 52 atm (c) 3700 ˚C 113 114