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Understanding the Phases of Matter: Gas Laws and Energy Changes

Learn about the phases of matter and the relationship between energy and the different states of matter. Explore the gas laws, phase change diagrams, and thermodynamics. Understand Boyle's Law and the calculations involving pressure and volume.

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Understanding the Phases of Matter: Gas Laws and Energy Changes

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  1. The Gas Laws

  2. Phases of Matter • The Phases of Matter are related directly to the amount of energy the matter contains. • Energy occurs in 2 forms: • Kinetic: the energy of movement • Potential: stored energy • Phases are determined by how much kinetic energy the matter has. • There are 4 phases of matter: Solid, liquid, gas, and plasma. • Some say plasma is NOT a form of matter; we will count it.

  3. Phases of Matter • The first phase of matter is SOLID. • The most orderly form of matter. • Shape: Definite • Volume: Definite • Energy: Low • Entropy: Low • Intermolecular Forces: Strong

  4. Phases of Matter • Liquids are the next phase of matter. • The second most orderly form of matter. • Shape: Indefinite • Volume: Definite • Energy: Medium • Entropy: Medium • Intermolecular Forces: Medium

  5. Phases of Matter: • Gases are the third phase of matter • Shape: Indefinite • Volume: Indefinite • Energy: High • Entropy: High • Intermolecular Forces: Very weak (almost nonexistent) * Molecules that are spaced far apart and can move freely * Because of this, pressure and temperature can affect the volume of a gas in ways that it would not affect another phase of matter

  6. Phases of Matter: • Plasmas are the next phase of matter. • They are found only in stars and, in theory, fusion reactors. The least orderly form of matter. • Shape: Indefinite • Volume: Indefinite • Energy: Very High • Entropy: Very High • Intermolecular Forces: Nonexistent

  7. Phases of Matter Phase changes can be shown on a graph in an easy way: this is called the Phase Change Diagram. Temp Plasma Gas Liquid Solid E!

  8. Phase Change Diagram • Temperature does not change during a phase change– but ENERGY does. • Instead of the e! adding to temperature, e! is used to break attractions between molecules. • In ice, e! works to vibrate the molecules beyond vibrating. In water, it works to spread the molecules out far enough that their density decreases enough to escape into the air.

  9. Thermodynamics • The study of heat/energy as it moves through a system • Phase changes: Melting = Freezing? • Yes-- in one, energy is being added (melting); in the other, it’s being taken away (freezing) • Calculated using Calorimetry • Uses masses and temperature changes to determine energy flow

  10. Calorimetry • Energy can be calculated using the formula: Q = m x c x ∆T Where: Q = energy (joules, kilojoules, or calories) m = mass in g c = Specific Heat (more) ∆T = Change in Temperature (K or oC)

  11. Specific Heat • Term given for the amount of energy needed to raise 1 g of a substance a given temperature: • For water, c = 1 cal / g oC • Also c = 4.18 j / g oC • From which we can deduce: • 1 cal = 4.18 j

  12. For Ice? Steam? • Ice’s Specific Heat: • 2.09 j/g oC • Steam’s Specific Heat: • 1.84 j/g oC

  13. Energy of a Phase Change • Heat of Fusion: The amount of heat required to MELT a solid; • Also, the heat given off by freezing a liquid • For H2O: 333 j / g • Heat of Vaporization: The amount of heat needed to BOIL a liquid; • Also, the heat given off by condensing a gas • For H2O: 2230 j / g

  14. Temperature and Pressure Temperature – a measure of the kinetic energy of molecules * Kinetic energy – the energy of movement * Measured in Kelvin (K) Pressure - force exerted on a surface area *Comes from molecules colliding *Measured in atmospheres (atm)

  15. Standard Temperature and Pressure (STP) Standard pressure 1 atm Conversion Factor: 1 atm = 101.3 kPa = 760 mmHg Standard temperature 273 K Conversion Factor: K = °C + 273 Ex: Convert 25°C to Kelvin 298 K 25°C + 273 = Ex: How many kPa is 1.37 atm? 1.37 atm x (101.3 kPa/1 atm) = 139 kPa Ex: How many atm are in 231.5 mmHg? 231.5 mm Hg x (1 atm/760 mm Hg) = 0.3046 atm

  16. Boyle’s Law • As the pressure on a gas increases • As the pressure on a gas increases - the volume decreases • Pressure and volume are inversely related 1 atm 2 atm 4 Liters 2 Liters

  17. Boyle’s Law P1V1 = P2V2 P is measured in atmospheres (atm) and V in Liters (L) 1 atm = 101.3 kPa = 760 mm Hg 1 L = 1 dm3 Timberlake, Chemistry 7th Edition, page 253

  18. Boyle's Law Pressuregoes up as Volumegoes down. Why? As pressure increases, gas particles are forced closer together. This decreases the gas’ volume A bicycle pump is a good example of Boyle's law. As the volume of the air trapped in the pump is reduced, its pressure goes up, and air is forced into the tire. Robert Boyle (1627 - 1691) Son of Early of Cork, Ireland.

  19. One-liter flask Volume and Pressure More collisions between molecules = greater pressure Two-liter flask The average molecules hits the wall twice as often. The total number of impacts with the wall is doubled and the pressure is doubled. The molecules are closer together; the density is doubled. Bailar, Jr, Moeller, Kleinberg, Guss, Castellion, Metz, Chemistry, 1984, page 101

  20. PV Calculation (Boyle’s Law) A quantity of gas has a volume of 120 L when confined under a pressure of 0.90 atm. At what pressure will the gas be if the volume is decreased to 30 L? P1 x V1 = P2 x V2 (0.90 atm) x (120 L) = (P2) x (30 L) P2 = 3.6 atm = 4 atm

  21. V1 V2 = T1 T2 Charles’ Law Timberlake, Chemistry 7th Edition, page 259

  22. V1 V2 = T1 T2 Charles' Law Temperaturegoes up as Volumegoes up. V and T are directly related. A hot air balloon is a good example of Charles's law. Where Volume is measures in L and Temperature is measured in Kelvin (K) Jacques Charles (1746 - 1823) Isolated boron and studied gases. Balloonist. K= Celsius + 273

  23. Temperature • Raising the temperature of a gas increases the pressure if the volume is held constant. • The molecules move faster and hit the walls harder. • If the pressure is held constant, raising the temperature of a gas increases the volume

  24. 300 K • If you start with 1 liter of gas at 1 atm pressure and 300 K and heat it to 600 K

  25. 600 K 300 K • The volume will increase to 2 liters

  26. VT Calculation (Charles’ Law) At constant pressure, the volume of a gas is increased from 150 L to 300 L by heating it. If the original temperature of the gas was 20 oC, what will its final temperature be? 150 L 293 K 300 L T2 T1 = 20 oC + 273 = 293 K T2 = X K V1 = 150 L V2 = 300 L = T2 = 586 K

  27. Combined Gas Law It’s difficult to change the volume of a gas without changing the volume AND the pressure. This leads to Combined Gas Law. Note: STP = Standard Temperature and Pressure = 273 K and 1 atm

  28. The Combined Gas Law (This “gas law” comes from “combining” Boyle‘s and Charles’ law) P = pressure (any unit will work as long as it’s the same on both sides) V = volume (any unit will work as long as it’s the same on both sides) T = temperature (must be in Kelvin) 1 = initial conditions 2 = final conditions

  29. The Combined Gas Law When measured at STP, a quantity of gas has a volume of 500 L. What volume will it occupy at 10 oC and 0.90 atm? (1 atm) x (500 L) = (0.90) x (V2) 273 K 283 K 1.83 = (0.90xV2)/283 P1 = 1 atm T1 = 273 K V1 = 500 L P2 = 0.90 atm T2 = 10 oC + 273 = 283 K V2 = X L 518 = 0.90xV2 V2 = 575 L = 600 L

  30. 985 mm Hg (126 dm3) 760 mm Hg (V2) P1V1 P2V2 = = T1 T2 198 K 273 K A sample of methane occupies 126 dm3 at -75oC and 985 mm Hg. Find its volume at STP. + 273 = 198 K T1 = -75oC Cross-multiply and divide: V2 = 225 dm3 985 (126) (273) = 198 (760) V2

  31. Ideal Gas Law PV = nRT Takes into account the number of molecules of a gas (moles).

  32. Ideal Gas Equation Universal Gas Constant Volume PV = nRT Pressure Temperature No. of moles R = 0.0821 atm L / mol K These units mean we must measure: P in atm, V in L, and T in K Kelter, Carr, Scott, Chemistry A Wolrd of Choices 1999, page 366

  33. nRT V = P (500 g)(0.0821 atm . L / mol . K)(300oC) = V 740 mm Hg Ideal Gas Law What is the volume that 500 g of iodine will occupy under the conditions: Temp = 300oC and Pressure = 740 mm Hg? Step 1) Write down given information. mass = 500 g iodine T = 300oC P = 740 mm Hg R = 0.0821 atm . L / mol . K Step 2) Equation: PV = nRT Step 3) Solve for variable Step 4) Substitute in numbers and solve V = What MISTAKES did we make in this problem?

  34. What mistakes did we make in this problem? What is the volume that 500 g of iodine will occupy under the conditions: Temp = 300oC and Pressure = 740 mm Hg? Step 1) Write down given information. mass = 500 g iodine Convert mass to gram; recall iodine is diatomic (I2) x mol I2 = 500 g I2(1mol I2 / 254 g I2) n = 1.9685 mol I2 T = 300oC Temperature must be converted to Kelvin T = 300oC + 273 T = 573 K P = 740 mm Hg Pressure needs to have same unit as R; therefore, convert pressure from mm Hg to atm. x atm = 740 mm Hg (1 atm / 760 mm Hg) P = 0.8 atm R = 0.0821 atm . L / mol . K

  35. nRT V = P (1.9685 mol)(0.0821 atm . L / mol . K)(573 K) = V 0.9737 atm Ideal Gas Law What is the volume that 500 g of iodine will occupy under the conditions: Temp = 300oC and Pressure = 740 mm Hg? Step 1) Write down given information. mass = 500 g iodine n = 1.9685 mol I2 T =573 K (300oC) P =0.9737 atm(740 mm Hg) R = 0.0821 atm . L / mol . K V = ? L Step 2) Equation: PV = nRT Step 3) Solve for variable Step 4) Substitute in numbers and solve V = 95.1 L I2 = 100 L I2

  36. Dalton’s Law of Partial Pressures

  37. Ptotal = P1 + P2 + ... Dalton’s Law The total pressure of a mixture of gases equals the sum of the partial pressures of the individual gases when the gases are at equal volume. Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

  38. + + = Partial Pressures In a gaseous mixture, a gas’s partial pressure is the one the gas would exert if it were by itself in the container. ? kPa 200 kPa 500 kPa 400 kPa 1100 kPa

  39. A B Z Two 1.0 L containers, A and B, contain gases under 2.0 and 4.0 atm, respectively. Both gases are forced into Container Z (w/vol. 2.0 L). Find total pres. of mixture in Z. Use BOYLE’S LAW to find the new volume if the volumes are not the same. PAVA = PZVZ 2.0 atm (1.0 L) = X atm (2.0 L) 1.0 L 1.0 atm 2.0 atm X = 1.0 atm 2.0 L PBVB = PZVZ 2.0 atm 4.0 atm 1.0 L 4.0 atm (1.0 L) = X atm (2.0 L) Total = 3.0 atm

  40. Find total pressure of mixture in Container Z. A B C Z 1.3 L 2.6 L 3.8 L 2.3 L 3.2 atm 1.4 atm 2.7 atm X atm PAVA = PZVZ 3.2 atm (1.3 L) = X atm (2.3 L) 1.8 atm 3.2 atm 1.3 L X = 1.8 atm PBVB = PZVZ 2.6 L 2.3 L 1.4 atm 1.6 atm 1.4 atm (2.6 L) = X atm (2.3 L) PCVC = PZVZ 2.7 atm 4.5 atm 3.8 L 2.7 atm (3.8 L) = X atm (2.3 L) Total = 7.9 atm

  41. Diffusion

  42. Diffusion vs. Effusion Diffusion - The tendency of the molecules of a given substance to move from regions of higher concentration to regions of lower concentration Examples: A scent spreading throughout a room or people entering a theme park Effusion - The process by which gas particles under pressure pass through a tiny hole Examples: Air slowly leaking out of a tire or helium leaking out of a balloon

  43. Effusion Particles in regions of high concentration spread out into regions of low concentration, filling the space available to them.

  44. Graham’s Law of Diffusion

  45. Gas Diffusion and Effusion Graham's law governs effusion and diffusion of gas molecules. Rate of effusion is inversely proportional to its molar mass. (Heavier gases move slower) Thomas Graham (1805 - 1869)

  46. NET MOVEMENT To use Graham’s Law, both gases must be at same temperature. diffusion: particle movement from high to low concentration effusion: diffusion of gas particles through an opening For gases, rates of diffusion & effusion obey Graham’s law: more massive = slow; less massive = fast

  47. Graham’s Law • The rate of diffusion/effusion is proportional to the mass of the molecules 80 g F I N I S H 250 g S T A R T Large molecules move slower than small molecules

  48. On average, carbon dioxide travels at 410 m/s at 25oC. Find the average speed of chlorine at 25oC. **Hint: Put whatever you’re looking for in the numerator.

  49. CH4 moves 1.58 times faster than which noble gas?

  50. 2 17 He Cl 4.0026 35.453 Find the relative rate of diffusion of helium and chlorine gas Step 1) Write given information GAS 1 = helium GAS 2 = chlorine He Cl2 M1 = 4.0 g M2 = 71.0 g v1 = x v2 = x Step 2) Equation Step 3) Substitute into equation and solve v1 71.0 g 4.21 = v2 4.0 g 1 He diffuses 4.21 times faster than Cl2

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