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Chapter 10 Gases

Lecture Presentation. Chapter 10 Gases. Gases. 10.1 CHARACTERISTICS OF GASES 10.2 PRESSURE 10.3 THE GAS LAWS 10.4 THE IDEAL-GAS EQUATION 10.5 FURTHER APPLICATIONS OF THEIDEAL-GAS EQUATION 10.6 GAS MIXTURES AND PARTIAL PRESSURES 10.7 THE KINETIC-MOLECULAR THEORY OF GASES

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Chapter 10 Gases

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  1. Lecture Presentation Chapter 10Gases

  2. Gases 10.1 CHARACTERISTICS OF GASES 10.2 PRESSURE 10.3 THE GAS LAWS 10.4 THE IDEAL-GAS EQUATION 10.5 FURTHER APPLICATIONS OF THEIDEAL-GAS EQUATION 10.6 GAS MIXTURES AND PARTIAL PRESSURES 10.7 THE KINETIC-MOLECULAR THEORY OF GASES 10.8 MOLECULAR EFFUSION AND DIFFUSION 10.9 REAL GASES: DEVIATIONS FROM IDEAL BEHAVIOR

  3. 10.1: Characteristics of Gases • Unlike liquids and solids, gases • Expand to fill their containers. • Are highly compressible. • Have extremely low densities.

  4. Gases • In the gas state, the particles have complete freedom of motion and are not held together • The particles are constantly flying around, bumping into each other and the container • There is a large amount of space between the particles • compared to the size of the particles • therefore the molar volume of the gas state of a material is much larger than the molar volume of the solid or liquid states

  5. Gases • Because there is a lot of empty space, the particles can be squeezed closer together – therefore gases are compressible • Because the particles are not held in close contact and are moving freely, gases expand to fill and take the shape of their container, and will flow

  6. F A P = 10.2: Pressure • Pressure is the amount of force applied to an area: • Atmospheric pressure is the weight of air per unit of area.

  7. Units of Pressure • Pascals • 1 Pa = 1 N/m2 • Bar • 1 bar = 105 Pa = 100 kPa

  8. The Pressure of a Gas • Gas pressure is a result of the constant movement of the gas molecules and their collisions with the surfaces around them • The pressure of a gas depends on several factors • number of gas particles in a given volume • volume of the container • average speed of the gas particles

  9. gravity Measuring Air Pressure • We measure air pressure with abarometer • Column of mercury supported by air pressure • Force of the air on the surface of the mercury counter balances the force of gravity on the column of mercury

  10. Units of Pressure • mmHg or torr • These units are literally the difference in the heights measured in mm (h) of two connected columns of mercury. • Atmosphere • 1.00 atm = 760 torr

  11. Manometer The manometer is used to measure the difference in pressure between atmospheric pressure and that of a gas in a vessel.

  12. Manometer for this sample, the gas has a larger pressure than the atmosphere, so

  13. Practice – What happens to the height of the column of mercury in a mercury barometer as you climb to the top of a mountain? • The height of the column increases because atmospheric pressure decreases with increasing altitude • The height of the column decreases because atmospheric pressure decreases with increasing altitude • The height of the column decreases because atmospheric pressure increases with increasing altitude • The height of the column increases because atmospheric pressure increases with increasing altitude • The height of the column increases because atmospheric pressure decreases with increasing altitude • The height of the column decreases because atmospheric pressure decreases with increasing altitude • The height of the column decreases because atmospheric pressure increases with increasing altitude • The height of the column increases because atmospheric pressure increases with increasing altitude

  14. Standard Pressure • Normal atmospheric pressure at sea level is referred to as standard pressure. • It is equal to • 1.00 atm • 760 torr (760 mmHg) • 101.325 kPa • 1 atm = 14.7 psi (pounds per square inch

  15. psi atm mmHg Example1: A high-performance bicycle tire has a pressure of 132 psi. What is the pressure in mmHg? Given: Find: 132 psi mmHg Conceptual Plan: Relationships: 1 atm = 14.7 psi, 1 atm = 760 mmHg Solution: Check: because mmHg are smaller than psi, the answer makes sense

  16. psi atm kPa Practice—Convert 45.5 psi into kPa Given: Find: 645.5 psi kPa Conceptual Plan: Relationships: 1 atm = 14.7 psi, 1 atm = 101.325 kPa Solution: Check: because kPa are smaller than psi, the answer makes sense

  17. 10.3 :The Gas Laws • Four Variables are need to define state of a gas • temperature (T) • pressure (P) • volume (V) • number of moles(n) • We examine several gas laws, which are empirical relationships among these four variables

  18. Standard Conditions • Because the volume of a gas varies with pressure and temperature, chemists have agreed on a set of conditions to report our measurements so that comparison is easy – we call these standard conditions • STP • Standard pressure = 1 atm • Standard volume 22.4 L • Standard temperature = 273 K • 0 °C

  19. Boyle’s Law Robert Boyle (1627–1691) The volume of a fixed quantity of gas at constant temperature is inversely proportional to the pressure.

  20. Boyle’s Experiment • Added Hg to a J-tube with air trapped inside • Used length of air column as a measure of volume

  21. Since V = k (1/P) This means a plot of V versus 1/P will be a straight line. PV = k P and V are Inversely Proportional • P1V1= P2V2 A plot of V versus P results in a curve.

  22. Boyle’s Law Robert Boyle (1627–1691) • Pressure of a gas is inversely proportional to its volume • constant T and amount of gas • graph P vs V is curve • graph P vs 1/V is straight line • As P increases, V decreases by the same factor • P x V = constant • P1 x V1 = P2 x V2

  23. V1, P1, P2 V2 Example 2: A cylinder with a movable piston has a volume of 7.25 L at 4.52 atm. What is the volume at 1.21 atm? Given: Find: V1 =7.25 L, P1 = 4.52 atm, P2 = 1.21 atm V2, L Conceptual Plan: Relationships: P1∙ V1 = P2∙ V2 Solution: Check: because P and V are inversely proportional, when the pressure decreases ~4x, the volume should increase ~4x, and it does

  24. Practice – A balloon is put in a bell jar and the pressure is reduced from 782 torr to 0.500 atm. If the volume of the balloon is now 2.78 x 103 mL, what was it originally?

  25. V2, P1, P2 V1 A balloon is put in a bell jar and the pressure is reduced from 782 torr to 0.500 atm. If the volume of the balloon is now 2.78x 103 mL, what was it originally? Given: Find: V2 =2780 mL, P1 = 762 torr, P2 = 0.500 atm V1, mL Conceptual Plan: Relationships: P1∙ V1 = P2∙ V2 , 1 atm = 760 torr (exactly) Solution: Check: because P and V are inversely proportional, when the pressure decreases ~2x, the volume should increase ~2x, and it does

  26. Charles’s Law • The volume of a fixed amount of gas at constant pressure is directly proportional to its absolute temperature.

  27. V T = k • So, • A plot of V versus T will be a straight line. Charles’s Law

  28. Charles’s Law Jacques Charles (1746–1823) • Volume is directly proportional to temperature • constant P and amount of gas • graph of V vs. T is straight line • As T increases, V also increases • Kelvin T = Celsius T + 273 • V = constant x T • if T measured in Kelvin

  29. V1, V2, T2 T1 Example 3: A gas has a volume of 2.57 L at 0.00 °C. What was the temperature at 2.80 L? Given: Find: V1 =2.57 L, V2 = 2.80 L, t2 = 0.00 °C t1, K and °C Conceptual Plan: Relationships: Solution: Check: because T and V are directly proportional, when the volume decreases, the temperature should decrease, and it does

  30. Practice – The temperature inside a balloon is raised from 25.0 °C to 250.0 °C. If the volume of cold air was 10.0 L, what is the volume of hot air?

  31. The temperature inside a balloon is raised from 25.0 °C to 250.0 °C. If the volume of cold air was 10.0 L, what is the volume of hot air? V1, T1, T2 V2 Given: Find: V1 =10.0 L, t1 = 25.0 °C L, t2 = 250.0 °C V2, L Conceptual Plan: Relationships: Solution: Check: when the temperature increases, the volume should increase, and it does

  32. V = kn • Mathematically, this means Avogadro’s Law • The volume of a gas at constant temperature and pressure is directly proportional to the number of moles of the gas.

  33. Avogadro’s Law Amedeo Avogadro (1776–1856) • Volume directly proportional to the number of gas molecules • V = constant x n • constant P and T • more gas molecules = larger volume • Count number of gas molecules by moles • Equal volumes of gases contain equal numbers of molecules • the gas doesn’t matter

  34. V1, V2, n1 n2 Example 4:A 0.225 mol sample of He has a volume of 4.65 L. How many moles must be added to give 6.48 L? Given: Find: V1 = 4.65 L, V2 = 6.48 L, n1 = 0.225 mol n2, and added moles Conceptual Plan: Relationships: Solution: Check: because n and V are directly proportional, when the volume increases, the moles should increase, and they do

  35. Practice — If 1.00 mole of a gas occupies 22.4 L at STP, what volume would 0.750 moles occupy?

  36. Practice — If 1.00 mole of a gas occupies 22.4 L at STP, what volume would 0.750 moles occupy? V1, n1, n2 V2 Given: Find: V1 =22.4 L, n1 = 1.00 mol, n2 = 0.750 mol V2 Conceptual Plan: Relationships: Solution: Check: because n and V are directly proportional, when the moles decrease, the volume should decrease, and it does

  37. nT P V 10.4: Ideal-Gas Equation • So far we’ve seen that V 1/P (Boyle’s law) VT (Charles’s law) Vn (Avogadro’s law) • Combining these, we get

  38. Ideal-Gas Equation The constant of proportionality is known as R, the gas constant. • The value of R depends on the units of P and V • we will use 0.08206 L-atm/mol-K and convert P to atm and V to L

  39. nT P V nT P V= R Ideal-Gas Equation The relationship We replace the proportionality sign with by inserting the R constant  = PV = nRT then becomes or The value of R depends on the units of P and V. However we usually will use 0.08206 and convert P to atm and V to L

  40. P, V, T, R n Example 6: How many moles of gas are in a basketball with total pressure 24.3 psi, volume of 3.24 L at 25°C? Given: Find: V = 3.24 L, P = 24.3 psi, t = 25 °C n, mol Conceptual Plan: Relationships: Solution: 1 mole at STP occupies 22.4 L, because there is a much smaller volume than 22.4 L, we expect less than 1 mole of gas Check:

  41. Practice – A gas occupies 10.0 L at 44.1 psi and 27 °C. What volume will it occupy at standard conditions?

  42. P1, V1, T1, R n P2, n, T2, R V2 A gas occupies 10.0 L at 44.1 psi and 27 °C. What volume will it occupy at standard conditions? Given: Find: V1 = 10.0L, P1 = 44.1 psi, t1 = 27 °C, P2 = 1.00 atm, t2 = 0 °C V2, L Conceptual Plan: Relationships: Solution: Check: 1 mole at STP occupies 22.4 L, because there is more than 1 mole, we expect more than 22.4 L of gas

  43. Practice — Calculate the volume occupied by 637 g of SO2 (MM 64.07) at 6.08 x 104 mmHg and –23 °C

  44. Practice—Calculate the volume occupied by 637 g of SO2 (MM 64.07) at 6.08 x 104 mmHg and –23 °C. P, n, T, R g V n Given: Find: mSO2 = 637 g, P = 6.08 x 104 mmHg, t= −23 °C, V, L Conceptual Plan: Relationships: Solution:

  45. 10.5: Further application of the Ideal-Gas Equation • We use ideal-gas equation to • define The relationship between the density of a gas and its molar mass • calculate the volume of gasses formed or consumed in chemical reactions

  46. n V P RT = Densities of Gases If we divide both sides of the ideal-gas equation by V and by RT, we get → PV = nRT

  47. n V n V P RT P RT m n = = = m V P RT = Densities of Gases • We know that • Moles  molecular mass = mass  n  = m • So multiplying both sides by the molecular mass () or m/n gives )  *( )* (

  48. m V P RT d = = Densities of Gases • Mass  volume = density • So, Note: One needs to know only the molecular mass (M), the pressure(P), and the temperature (T) to calculate the density of a gas.

  49. P RT d = dRT P  = Molecular Mass We can manipulate the density equation to enable us to find the molecular mass of a gas: becomes • Density is directly proportional to molar mass

  50. Practice – Calculate the density of N2(g) at STP

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