Engr. Yvonne Ligaya F. Musico Chemical Engineering Department

Transcription

1 GASEOUS STATE Engr. Yvonne Ligaya F. Musico Chemical Engineering Department

2 TOPICS Objective Properties of Gases Kinetic Molecular Theory of Gases Gas Laws

3 OBJECTIVES Determine how volume, pressure and temperature are related to each other to determine the state gases Generalize the behavior of gases using different Gas Law Solve problems involving Gas Law Apply Gas Law

4 GENERAL PROPERTIES OF GASES Indefinite shape Indefinite volume Take the shape and volume of container Particles are far apart

5 GENERAL PROPERTIES OF GASES Solid Liquid Gas Particles move fast high Kinetic Energy - particles can separate and move throughout container

6 KINETIC MOLECULAR THEORY OF GASES All gases are made up of very tiny particles called molecules which are widely separated from one another in an otherwise empty space. The gas molecules move at high speed travelling in straight line but in random direction.

7 KINETIC MOLECULAR THEORY OF GASES The molecules collide with one another but the collisions are perfectly elastic. The average kinetic energy of gas molecules depends directly upon the absolute temperature of any given temperature.

8 Some Measurable Properties of Gases Volume - (also called capacity) is a quantification of how much space an object occupies Pressure - force exerted per unit area of surface by molecules in motion.

9 Some Measurable Properties of Gases Absolute Temperature - Temperature at which a thermodynamic system has the lowest energy.it corresponds to C on the Celsius temperature scale and to F on the Fahrenheit temperature scale K = o C R = o F (SI) (English)

10 GAS LAWS Boyle s Law Charles Law Gay-Lussac s Law Combined Gas Law Ideal Gas Law Dalton s Law Graham s Law

11 BOYLE S LAW

12 BOYLE S LAW If the temperature of a given quantity of gas is held constant, the volume of the gas varies inversely with the absolute pressure during the change of state.

13 BOYLE S LAW V p V C p or pv C p V p V Constant temperature Constant amount of gas

14 Sample Problem 00 ml of gas are enclosed in the cylinder under a pressure of 760 torr. What would be the volume be at a pressure of 50 torr?

15 Solution to Sample Problem Given: V = 00 ml P = 760 torr P = 50 torr Required: V Using Boyle s Law p V V V p V V p p 50 ml (00 ml)(760 torr) (50 torr)

16 Sample Problem A sample of chlorine gas occupies a volume of 946 ml at a pressure of 76 mmhg. What is the pressure of the gas (in mmhg) if the volume is reduced at constant temperature to 54 ml?

17 Solution to Sample Problem Given: V = 946 ml P = 76 mmhg V = 54 ml Required: P Using Boyle s Law P = P x V V 76 mmhg x 946 ml = 54 ml P = 4460 mmhg

18 Charles Law The Collapsing Can Activity

19 Charles Law

20 Charles Law Generalization: As the temperature of the gas increases the gas molecules will begin to move quickly. As the temperature of the gas decreases, the gas molecules will begin slow down. The molecules will hit the walls of the container with more or less force causing the volume to increase or decrease.

21 Charles Law The volume of a fixed mass of gas at constant pressure varies directly with the absolute temperature. Jaques Alaxandre Cesar Charles (746-83)

22 Charles Law Mathematically, V T V T k

23 Charles Law Applying to State and State ). (, P const at T V T V Therefore k k k T V k T V Where: V, V = volume, L, ft 3 T, T = absolute temperature, K, R

24 Sample Problem A sample of methane gas that has a volume of 3.8 L at 5.0 o C is heated to 86.0 o C at constant pressure. Calculate its new volume.

25 Solution to Problem Given: V = 3.8 L T = = 78 K T = = 359 K Required: V Using Charles Law V T V V V V T TV T (359K)(3.8L) 78K 4.9L

26 Sample Problem A mass of neon occupies 50 cm 3 at 0 o C. If the volume of the gas increases to 00 cm 3 at constant pressure, what will be the new temperature in o C?

27 Solution to Problem Given: V = 50 cm 3 T = = 73 K V = 00 cm 3 Required: T in o C Using Charles Law V T T T T T V T VT V 3 (00cm )(73K) 3 50cm 364K o 9 C

28 Sample Exercise. A mass of gas has temperature of 50 o C and volume of 0. m 3. The temperature is changed to 50 o C but the pressure is unchanged. Determine the new volume.. A gas at constant pressure is kept at 00 0 C. On decreasing the temperature to 50 0 C, the gas occupies a volume of 800 ml. Find the initial volume of the gas.

29 Gay-Lussac s Law If the volume of a particular quantity of gas is held constant, then, with any change of state, the pressure will vary directly as the absolute temperature.

30 Gay-Lussac s Law P T p CT or P C T p p T T ( V and m are const.) Temperature must be in Kelvin

31 Sample Problem An aerosol can has a pressure of.4 atm at 5 o C. What pressure would it attain at 00 o C, assuming the volume remained constant.

32 Solution to Sample Problem Given: P =.4 atm T = 5 o C T = 00 o C Required: P Using Gay-Lussac s Law P P p T p T PT T 6.9 atm (.4 atm)( ) K (5 73.5) K

33 Sample Problem Sulfur dioxide at a temperature of 04 o C occupies a volume of 0.3 m 3. (a) If the volume increased to 0.90 m 3 while the pressure is maintained constant (ideal gas behavior), what is the final temperature in K? In o C? (b) If the initial volume is maintained constant and the pressure tripled, what is the final temperature in K? in o C?

34 Combined Gas Law The expression indicates that the volume of a given mass of gas varies directly and inversely with the pressure and temperature respectively.

35 Combined Gas Law p V pv ( const. for fixed mass of gas ) T T

36 Sample Problem A sample oxygen collected over water has a volume of 400 ml. at 7 o C and mm Hg pressure. What volume would this sample of oxygen occupy dry at STP?

37 Solution to Problem Given V P T T P V Applying Combined Gas Law pv T V 400ml mm Hg K 73.5K 760 mm Hg? pv T PV T P T mm Hg400.0ml73.5K 760 mm Hg300.5K 37.9ml

38 Sample Problem A sample of carbon dioxide occupies 4.5 L at 30 o C and 650 mm Hg. What would it occupy at 800 mm Hg and 00 o C?

39 Ideal Gas Law Ideal gas law describe the relationship among the four variable temperature, pressure, volume and mole or mass for gaseous substance.

40 Ideal Gas Law pv nrt Where: n = number of moles of gas V = volume in liter p = pressure in atmosphere T = absolute temperature, K R = molar gas constant = 0.08 liter-atm.k -.mole -

41 Sample Problem An experiment calls for 3.50 moles of chlorine, Cl. What volume would this be if the gas volume is measured at 34 o C and.45 atm.

42 Solution to Problem L atm K K mol L atm moles p nrt V nrt pv Law Gas Ideal Applying V K mol L atm R moles n K T atm p Given ) )( )(0.08 (3.5?

43 Sample Problem A 0.76 grams sample of sulfur dioxide (SO ) occupies a volume of 0.70 liters at 739 torr and 98 o C. Calculate the molecular weight of gaseous SO.

44 Sample Problem 3 Calculate the density of CO in g/l at. atm pressure and 35 o C.

45 Solution to Problem 3 Given : P. atm T R MW of Applying pv but n K 0.08L. atm. mol? nrt m MW CO. K 44 g / mol Ideal Gas Law,

46 Solution to Problem 3 L g K K mol L atm mol g atm RT pmw Then RT pmw Therefore V m where RT V m pmw arranging RT MW m pv so that /.04 ) ( ) / )(44 (.,,,, Re,

47 Dalton s Law of Partial Pressure Dalton's Law The total pressure of a mixture of gases is equal to the sum of the partial pressures of the gases in a mixture P total = P A + P B + P C +..P i P A =X A P T Where, X A =(n A )/(n T )

48 Sample Problem Given a mixture of gases in the atmosphere at 760 torr, what is the partial pressure of N (x = ) at 5 o C

49 Solution to Sample Problem P N = X N P tot P N = (0.7808)(760 torr) P N = 593 torr

50 Sample Problem Suppose a 56 ml sample of H gas collected over water at 9 o C and 769 mm Hg. What is the mass of H collected? The vapor pressure of water at 9 o C is 6.5 mm Hg.

51 Solution to Sample Problem P total = P H + P HO P H = P total P HO P H = 769 mm Hg 6.5 mm Hg P H = 75 mm Hg P H = 75 mm Hg x atm = atm V = 56 ml = 0.56 L n =? 760 mm Hg

52 Continuation of Solution to Sample Problem Using Ideal Gas Law PV = nrt n = PV = ( atm ) ( 0.56 L ) RT (0.08 L-atm/mol.K)(9 K) n = mole H Next convert moles of H to grams mol H x.0 g H = g H mol H

53 Grahams Law: Diffusion and Effusion of Gases Diffusion the process whereby a gas spreads out through another gas to occupy the space with uniform partial pressure. Effusion the process in which a gas flows through a small hole in a container.

54 Grahams Law: Diffusion and Effusion of Gases Graham s law of Effusion the rate of effusion of gas molecules through a hole is inversely proportional to the square root of the molecular mass of the gas at constant temperature and pressure. Rate k MW

55 Grahams Law: Diffusion and Effusion of Gases R A = M B = D B R B M A D A Where: R is the rate of diffusion or effusion M is the molecular weight of the gas D is the density of the gas

56 Sample Problem A gas diffuses.5 times faster than SO gas. Calculate the molecular weight of the gas AB.

57 Solution to Sample Problem mol g M R R R R M M M M R R AB SO SO AB SO SO AB AB SO SO AB / )() (.5 ) 64)( ( ) (

58 Sample Exercise. Determine the molecular mass of an unknown compound if it effused through a small orifice if it effused 3.55 times slower than CH 4.. A compound with a molecular mass of 3.0 g/mol effused through a small opening in 35 s; determine the effusion time for the same amount of a compound with a molecular mass of 6.0.

59 Thanks for Listening.