1 / 114

Solids, Liquids, and Gases

Solids, Liquids, and Gases. 5 th International Junior Science Olympiad (IJSO). Dr. Yu-San Cheung yscheung@cuhk.edu.hk Department of Chemistry The Chinese University of Hong Kong. Basic Properties of Solids, Liquids, and Gases. Characteristics of Gases No definite volume or shape :

lali
Télécharger la présentation

Solids, Liquids, and Gases

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Solids, Liquids, and Gases 5th International Junior Science Olympiad (IJSO) Dr. Yu-San Cheung yscheung@cuhk.edu.hk Department of Chemistry The Chinese University of Hong Kong

  2. Basic Properties of Solids, Liquids, and Gases

  3. Characteristics of Gases No definite volume or shape: A gas fills whatever volume is available to it and is easy to compress. 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 cm3, whereas the same quantity of water vapor at the same temperature and pressure has a volume of 30200 cm3, more than 1000 times greater.

  4. Properties 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). 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. 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?

  5. 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.

  6. Downward force due to weight of mass acting on piston of area A, creating a pressure p (assuming no air outside) 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. Pressure When a force (F) is acting on a surface with area A, a pressure (p) exerts on the surface: 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.

  7. 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

  8. How is pressure measured?

  9. p = p0 + p p = pressure exerted by the height difference

  10. A h Pressure Exerted by a Liquid Column Column height = h Cross-section area = A Volume = h A Mass = volume  density = h A (= density) Force produced by the column = Weight =mg = h A g (g= gravity acceleration = 9.80665 m·s2) Pressure = F / A =  g h Note that the cross-section must be uniform, but not necessarily circular.

  11. Choice of Medium p=  g h For the same p, higher density, smaller height difference. 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 In practice, mercury is used. Other advantages? But any disadvantages?

  12. Mercury as the Medium Advantages: - High density - Chemically inert (e.g., water dissolves some gases such as CO2, NH3, HCl) - Not sticking to the glass wall (compared with water) Disadvantages: - Vapor is toxic - Difficult to deal with spillage (not absorbed by paper towel, tending to form small droplets, difficult to remove with dropper)

  13. Mercury Barometer One end open to atmosphere Accuracy: depending on the ruler readout (usually 1 mm) The other end connected to closed container containing the gas

  14. Mercury Barometer (Closed-end) sealed end vacuum inside p = p

  15. Pressure Gauge Also called barometer, but the term “Pressure Gauge” (壓強計) is more commonly use in the market For measuring both “positive pressure” and “negative pressure” (relative to atmospheric pressure) “Negative pressure”: also called vacuum “Positive pressure”: can be measured up to 1380 atmospheric pressure http://en.wikipedia.org/wiki/Pressure_measurement

  16. Principle of Pressure Gauge 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. 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. Alternatively, a strain gauge circuit can be used to produce output electronically. If you are interested, see: http://www.efunda.com/DesignStandards/sensors/strain_gages/strain_gage_theory.cfm http://en.wikipedia.org/wiki/Pressure_measurement http://www.efunda.com/DesignStandards/ sensors/bourdon_tubes/bourdon_intro.cfm

  17. Principle of Thermo-conductivity Gauge • 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 • High pressure vs. electrical signal: may be complicated • Different gases: different calibration • Range: 10 – 0.001 mmHg http://www.varianinc.com.cn/ products/vacuum/measure/ transducers/531gauge/shared/ 531gauge-180.jpg

  18. Principle of Ionization Gauge • Principle: • 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 • Range: 10–3 – 10–10 mmHg http://www.varianinc.com.cn/products/vacuum/measure/shared/uhv24a-180.jpg

  19. Units of Pressure The unit of pressure in the SI system is Pascal (Pa)(帕斯卡) 1 Pa  1 N·m2  1 J·m3  1 kg·m1·s2 p = F/A 1 Pa = 1 N·m2 Energy = F ·d 1 J = 1 N·m = 1 (Pa·m2) · m F = m·a 1 N = 1 kg·m·s2

  20. Units of Pressure Other commonly used units: (i) psi = pound per square inch 1 pound = 453.59237 g = 0.45359237 kg 1 inch = 2.54 cm = 0.0254 m Exercise: 1 psi = ?? Pa (You also need: gravity acceleration g = 9.80665 m·s2)

  21. Units of Pressure (ii) 1 torr = 1 mm Hg (implying the pressure produced by a Hg column of 1 mm height) Pressure = F/A =  g h (p. 11) 1 torr = 133.32 Pa ( = 13595.1 kg·m3 for mercury) (iii) 1 atm = 760 torr 1 atm = 101325 Pa = 14.696 psi (iv) 1 bar = 105 Pa Important: 1 bar is close to 1 atm, but not exactly equal!

  22. By calculations: Rectangular shape: width × depth × height Sphere: r3 Pyramid:× base area × height (including cone) More complicated shape but well-defined 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) 4 1 3 3 Determination of Volume

  23.  1000  1000  1000  1000 mL ml cm3 L dm3 m3 nL L Units of Volume Unit of volume = (Unit of length)3 SI unit for volume : m3 Units used in daily life: liter, dm3, ml (or mL), cm3 1 m3 = (1 dm)3 = 1 dm3, also called liter (L) 1 ml = 1 milli-liter = 10-3 L = 10-6 m3 = 10-6 (100 cm)3 = 1 cm3 Smaller unit: L (micro-liter, 10-6 L) , nL (nano-liter, 10-9 L) Summary: fill out the blanks below with appropriate units

  24. Types of thermometer (1) Glass thermometer(玻管液體溫度計) Glass tube containing a liquid, e.g., _________, ________ High temperature: _______________ of liquid Low temperature: _______________ of liquid (2) Resistance thermometer(電阻溫 度計) Consisting of a probe (e.g., platinum) and a controller (with battery, read-out, and electronic circuit). The resistance of the probe changes with temperature. By measuring the resistance, the temperature can be determined. http://en.wikipedia.org/wiki/Thermometer

  25. Types of thermometer (3) Infra-red (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. (What is its main advantage over the other types?) (4) Liquid crystal thermometer Containing dots of heat-sensitive liquid crystals on a plastic strip. The dots change color at different temperatures. http://en.wikipedia.org/wiki/Thermometer

  26. Unit of Temperature Celsius scale: 0 C = freezing temperature of water 100 C = boiling temperature of water at 1 atm pressure for both

  27. Pressure-volume relations: Boyle‘slaw (波義耳定律) Volume Pressure Robert Boyle (1627 – 91) Modified from: http://www.grc.nasa.gov/WWW/K-12/airplane/aboyle.html

  28. p0 Trapped air h0 h1 h2 Boyle'slaw & J-tube Each time Hg is added, h1 and h2 changes, but p0 and h0 remain unchanged. Pressure of trapped air = p0 + h1 – h2 Volume = (h0 – h2)  cross-sectional area Exercise: p0 = 0.987 atm h1 = 45.9 cm, h2 = 10.7 cm p = ?? cm Hg

  29. 1 / Volume      Pressure Boyle'sLaw p 1/V (i.e., p is inversely proportional to V ) We can also say that: p is proportional to 1/V p = “constant” × 1/V Plot of p vs. 1/V yields a straight line passing through the origin

  30. 1 / Volume      Pressure Boyle'sLaw p = “constant” × 1/V leads to: pV = “constant” “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.

  31. Boyle'sLaw Alternative form (more useful): Initial: p1, V1Final: p2, V2 p1V1 = constant = p2V2 i.e., p1V1 = p2V2

  32. Boyle'sLaw p1V1 = p2V2 Example: If p1 = 700 torr, V1 = 1 L, p2 = 100 torr, V2 = ? Ans.: V2 = p1V1 / p2 = (700 × 1) / (100) = 7 L Exercise: If p1 = 700 torr, V1 = 1 L, V2 = 100 L, p2 = ? Ans.:

  33. Effect of Temperature on p-V Curve Modified from: http://www.chem1.com/acad/webtext/gas/gas_2.html

  34. Try finding out the “constant” for each curve: 500K: pV = 400K: pV = 300K: pV = 200K: pV = 100K: pV = 50K: pV = Effect of Temperature on p-V Curve

  35. Effect of Temperature on p-1/V Curve 1/V Temperature: increasing or decreasing? p

  36. 20 atm at the beginning Vacuum at the beginning 4 L 1 L 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 valvethat connects the two containers. What will be the final pressure of the gas? (Hint: at the end, the gas fills up both containers.)

  37. Temperature – Volume Relationship: Charles‘ Law (查理定律) With the pressure held constant, the volume of a gas changes by the same amount for each C change in temperature.

  38. Put it another way, V varies linearly with T, i.e., the T-V graph is a straight line. V, liter p1 When the straight lines are extrapolated, they reaches the x-axis at –273.15°C. p2 p3 0 T, °C http://www.chem1.com/acad/webtext/gas/gas_2.html

  39. -273.15 C  0.00 K 0.00 C  273.15 K 25.00 C  298.15 K 100.00 C  373.15 K Diff = K Diff = °C Diff = K Diff = °C Kelvin Scale (K): adding 273.15 to the C value e.g., -273.15 C  0.00 K (Note: the symbol “” is not used) 0.00 C  273.15 K 25.00 C  K 100.00 C  K If talking about temperature difference or temperature change, change in 1 C = change in 1 K.

  40. V T 0 °C -273.15 °C V T 273.15 K 0 K Temperature-Volume Relationship Using the Kelvin scale, we have T V (T is proportional to V ) or T / V = constant or T1/V1 = T2 / V2 T/V = constant  (i) V = 0 if T =0 (ii) V is –ve if T is –ve

  41. 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. Celsius: more open-mined, not minding negative temperature, choosing convenient scale: freezing point of water for 0 C. But there is no clue what the lowest temperature is in nature. Charles’ law: 0 K is the lowest temperature in nature, otherwise we have negative volume of a gas, which is physically unacceptable. 0 K: also called absolute zero Kelvin Scale: also called the absolute temperature scale Kelvin Scale is assumed in scientific formulas (unless otherwise specified).

  42. Charles’Law T1 / V1 = T2 / V2 Example: If T1 = 200 K, V1 = 1 L, T2 = 100 K, V2 = ? Ans.: V2 = T2V1 / T1 = (200 × 1) / (100) = 2 L Exercise: If T1 = 200 C , V1 = 1 L, V2 = 100 cm3, T2 = ? Ans.:

  43. Boyle’s Law + Charles’ Law = What? Boyle’s Law p 1/ V or V 1/p V T / p + pV / T = constant Charles’ Law V T or p1V1 / T1 = p2V2 / T2

  44. Exercise: 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?

  45. Amadeo Avogadro (1776-1856): "E.V.E.N principle" Equal volumes of gases, measured at the same temperature and pressure, contain equal numbers of molecules. i.e., V N (N = no. of gas molecules) Combined with pV / T = constant, we have pV / NT = constant = k (Boltzmann constant) or pV = NkT (Ideal Gas Law)

  46. Mole Concept 32 g of oxygen gas = 6.022  1023 oxygen gas molecules = 1 mole of oxygen gas molecules i.e., 1 mole = 6.022 x 1023 c.f.: 1 dozen of pencils = 12 pencils 1 kB = 1024 bytes 1 catty of nails = ??? pieces of nails Avogadro Number: NA= 6.022  1023 mol1 Mole number (n) = numbers of moles = N / NA e.g., 0.01 mol of O2 gas molecules = 0.01 x 6.022 x 1023 = 6.022 x 1021 O2 gas molecules

  47. Mass and Mole ?? grams of a gas = 1 mole of the gas molecules You need to find out the atomic mass units of the atoms (from books or internet) e.g. H: 1.0 amu (amu = atomic mass unit) C: 12.0 amu Cl: 35.5 amu H2: 2 x 1.0 = 2.0  2.0 g of H2 = 1 mole of H2 CH3Cl: 12.0 + 3 x 1.0 + 35.5 = 50.5  50.5 g of CH3Cl = 1 mole of CH3Cl

  48. Molecular Mass 50.5 g of CH3Cl = 1 mole of CH3Cl We define: molar mass / molecular mass / molecular weight (Mw) e.g. Mw for CH3Cl = 50.5 g mol1 e.g. Mass of 0.1 mol of CH3Cl = 50.5 g mol1 x 0.1 mol = 5.05 g Mw NA Mass  Mole  Number of molecules

  49. Ideal Gas Law / Perfect Gas Law pV = NkT = (NAn)kT = n (NAk)T = n (NAk)T = nRT (R = NAk) R: gas constant or universal gas constant • Exercise: • What is the S.I. unit for R ? • What is the S.I. unit for k ? • If the experimental value for R is 8.314 in S.I. unit. What is the value for k in S.I. unit?

  50. Ideal Gas Law / Perfect Gas Law R : gas constant = 8.314 n : no. of moles Unit to be filled out pV = nRT Must be memorized!! More commonly used pV = NkT N : no. of molecules k : Boltzmann constant = Value to be filled out

More Related