THREE STATES OF MATTER

Transcription

1 THREE TATE OF MATTER 1

2 Pressure is Force(N)/Area(m 2 ) Pressure I Unit is Pascal Pressure of air is measured with a BAROMETER Hg rises in tube until force of Hg (down) balances the force of atmosphere Pressure P of Hg pushing down related to Hg density column height 2

3 Always go through atm (just like moles) 1 (atm) = 760 mm Hg (or torr) 1 (atm) = Pa (I unit) 1 mm Hg = Pa Different atmospheric pressures at different altitudes Pressure Conversions I. between atmospheres and millimeters of mercury. II. One atm. equals mm Hg, so there will be a multiplication or division based on the direction of the change. A. What is 475 mm Hg expressed in atm? X 1 atm = atm 760 mm Hg A. What is 250. kpa expressed in atm? X 1 atm = 2.47 atm Pressure Column height measures Pressure of atmosphere 1 standard atmosphere (atm) = 760 mm Hg (or torr) = Pa (I unit) = inches Pressure Conversions Convert atm to mmhg. olution-multiply the atm by mmhg/atm. 665 mm Hg Convert mm Hg to atm. olution-divide the mmhg by mmhg/atm atm 3

4 Convert atm to kpa. olution - multiply the atm by kpa / atm 96.8 kpa Convert kpa to atm. olution - divide the kpa by Pa / atm atm Pressure Conversions Express in kpa Express in Atm atm 25.8 kpa kpa atm atm atm atm 15.7 kpa 56.3 kpa kpa kpa atm kpa atm kpa atm atm atm atm atm 89.8 kpa 69.8 kpa 12.5 kpa kpa kpa kpa kpa kpa atm atm atm atm Convert kpa to mmhg. Convert mmhg to kpa mm Hg atm Express in kpa 410. mm Hg 335 mm Hg 226 mm Hg 211 mm Hg 352 mm Hg 661 mm Hg 50.0 mm Hg Pressure Conversions 54.7 kpa 44.7 kpa 30.1 kpa 28.1 kpa 46.9 kpa 88.1 kpa 6.67 kpa 364 mm Hg 48.5 kpa Express in mm Hg kpa kpa 12.2 kpa 63.8 kpa 130. kpa kpa kpa 7.16 mm Hg 5.20 mm Hg 91.5 mm Hg 479 mm Hg 975 mm Hg 1897 mm Hg 1991 mm Hg Pressure Conversions Express in atm Express in mm Hg 410 mm Hg 335 mm Hg 226 mm Hg 211 mm Hg 352 mm Hg 661 mm Hg 50.0 mm Hg 364 mm Hg atm atm atm atm atm atm atm atm atm atm atm atm atm atm atm atm 194 mm Hg 118 mm Hg 422 mm Hg 751 mm Hg 475 mm Hg 421 mm Hg 93.5 mm Hg 726 mm Hg 4

5 Pressure & Kinetic Molecular Theory Gases can be compressed because most of the volume of a gas is empty space. If we compress a gas without changing its temperature, the average kinetic energy of the gas particles stays the same. There is no change in the speed with which the particles move, but the container is smaller. Thus, the particles travel from one end of the container to the other in a shorter period of time. This means that but they hit the walls more often. Any increase in the frequency of collisions with the walls must lead to an increase in the pressure of the gas. Thus, the pressure of a gas becomes larger as the volume of the gas becomes smaller. P proportional to 1/V KENETIC ENERGY 5

6 Gas Properties There is a lot of free space in a gas. Gases can be expanded infinitely. Gases fill containers uniformly and completely. Gases diffuse and mix rapidly. Gas Properties Gas properties can be modeled using math. Model depends on V = volume of the gas (L) T = temperature (K) n = amount (moles) P = pressure (atmospheres) Boyle s Law P 1/V P & V INVERELY PROPORTIONAL if n and T constant P 1 V 1 = P 2 V 2 6

7 12/airplane/Animation/frglab2.html 12/airplane/boyle.html Boyle s Law 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. OLVE What is the final volume of a 2 L balloon if the pressure changes from 101 kpa to 202 kpa? 7

8 Charles s Law OLVE If n and P are constant, then V α T V and T are directly proportional. V 1 V 2 = T 1 T 2 If one temperature goes up, the volume goes up! What is the final temp of a 2 L balloon at 100 Deg Celsius if the volume changes to 4L? Temp & Kinetic Molecular Theory The average kinetic energy of the particles in a gas is proportional to the temperature of the gas. Because the mass of these particles is constant, the particles must move faster as the gas becomes warmer. If they move faster, the particles will exert a greater force on the container each time they hit the walls, which leads to an increase in the pressure of the gas. If the walls of the container are flexible, it will expand the volume until the pressure of the gas once more balances the pressure of the atmosphere. The volume of the gas becomes larger as the temperature of the gas increases. T proportional to V 8

9 Combined Gas Law The good news is that you don t have to remember all three gas laws! ince they are all related to each other, we can combine them into a single equation. P 1 V 1 = P 2 V 2 T 1 T 2 Combined Gas Law If you should only need one of the other gas laws, you can cover up the item that is constant and you will get that gas law! P 1 V 1 T 1 = P 2 V 2 T 2 Boyle s Law Charles Law Gay-Lussac s Law Combined Gas Law (General Gas Law) Boyle s Law Charles s Law P 1 V 1 = P 2 V 2 OLVE A sample of helium gas has a volume of L, a pressure of atm and a temperature of 29 C. What is the new temperature( C) of the gas at a volume of 90.0 ml and a pressure of 3.20 atm? et up Data Table P 1 = atm V 1 =.180 L T 1 = 302 K P 2 = 3.20 atm V 2 =.0900 L T 2 =?? 9

10 P 1 = atm V 1 = 180 ml T 1 = 302 K P 2 = 3.20 atm V 2 = 90 ml T 2 =?? P 1 V 1 P 2 V 2 = P 1 V 1 T 2 = P 2 V 2 T 1 T 1 T 2 T 2 = P 2 V 2 T 1 P 1 V 1 T 2 = 3.20 atm x.0900 L x 302 K atm x.180 L T 2 = 604 K = 331 C = 604 K Moles & Kinetic Molecular Theory Avogadro s Hypothesis As the number of gas particles increases, the frequency of collisions with the walls of the container must increase. This, in turn, leads to an increase in the pressure of the gas. Flexible containers, such as a balloon, will expand until the pressure of the gas inside the balloon once again balances the pressure of the gas outside. Thus, the volume of the gas is proportional to the number of gas particles. n proportional to V OLVE A balloon has a volume of 675 ml at 35 C and atm pressure. What is the temperature in C when the gas has a volume of L and a pressure of 802 mm Hg? D) Avagadro s Law - Equal volumes of gas at constant temperature and pressure have equal numbers of molecules. Expressed Mathematically 179 K -94 C VOTE FOR CRIDER o for a system that changes n and V at constant P & T OLVE A balloon has a volume of 785 ml on a fall day when the temperature is 21 C. In the winter, the gas cools to 0 C. What is the new volume of the balloon? VOTE FOR CRIDER Avogadro s Hypothesis Equal volumes of gases at the same T and P have the same number of molecules. V = n (RT/P) = kn V and n are directly related..729 L twice as many molecules 10

11 Avagadro s Law And now, we pause for this commercial message from TP OK, so it s really not THI kind of TP TP in chemistry stands for tandard Temperature and Pressure tandard Pressure = 1 atm (or an equivalent) tandard Temperature = 0 deg C (273 K) TP allows us to compare amounts of gases between different pressures and temperatures III. Ideal Gas Law Experiments show that at TP, 1 mole of an ideal gas occupies L. Boyle s Law Charles s Law Avagadro s Law Expressed Mathematically IDEAL GA LAW P V = n R T 11

12 At standard temperature and pressure, how many moles of H 2 are contained in a 1.0 L container?.045 mol/l TP tandard Temperature and Pressure tandard temperature = K (0 C) tandard Pressure = 1 atm At TP 1 mole of any gas occupies L (tandard Molar Volume) R = Universal Gas Constant = Using PV = nrt P = Pressure V = Volume T = Temperature N = number of moles R is a constant, called the Ideal Gas Constant L kpa Mol K 12

13 The Gas Constant R Repeated experiments show that at standard temperature (273 K) and pressure (1 atm or N/m2 ), one mole (n = 1) of gas occupies 22.4 L volume. Using this experimental value, you can evaluate the gas constant R, P V 1 atm 22.4 L R = --- = n T 1 mol 273 K = L atm / (mol K) When I units are desirable, P = N/m2 (Pa for pascal) instead of 1 atm. The volume is m3. The numberical value and units for R is N/m m3 R = mol 273 K = J / (mol K) At standard temperature and pressure, how many grams of CO 2 is contained in a 3.00 L container? (Molar mass of CO 2 = 44.00) How many moles of N 2 are required to fill a large box with a volume of 27.0 L to 745 mm Hg at 25.0 o C? 1. Get all data into proper units V = 27.0 L T = 25 o C = 298 K P =745 mm Hg ( Kpa/760mm Hg) = Kpa R = L Kpa / mol K P V = n R T ( kpa) (3.00) = ( n ) (8.314 L Kpa / mol K) (273 K).134 moles = ( n ).134 mol x 44.00g/mol = ( n ) 5.89g = ( n ) Or magic box: 3/22.4 x = 5.89 g P V = n R T ( kpa) (27.0) = ( n ) (8.314 L Kpa / mol K) (298 K) ( kpa) (27.0 L) = ( n ) (8.314 L Kpa / mol K) (298 K) 1.08 = ( n ) What is the temperature if 1.00 mole of N 2 occupy 100. L of volume has a pressure of 20.0 kpa 13

14 P V = n R T Using PV = nrt (20.0 kpa) (100. L) = ( T ) (1.00 mol)(8.314 L Kpa / mol K) 241 K = ( T ) A 5.00 L cylinder contains oxygen gas (diatomic) at 20.0 C and 735 mm Hg. How many grams of oxygen are in the cylinder? Using PV = nrt P V = n R T Dinitrogen monoxide (N 2 O), laughing gas, is used by dentists as an anesthetic. If 2.86 mol of gas occupies a 20.0 L tank at 23 C, what is the pressure (in kpa) in the tank in the dentist office? ( kpa) (5.00) = ( n ) (8.314 L Kpa / mol K) (293 K).2011 mol = ( n ).2011 mol x 32.0g/mol = ( g ) 6.44 g O 2 = ( g ) P V = n R T Calculate the pressure exerted by moles of nitrogen gas occupying a volume of 895 ml at a temperature of 42.0 C. P = (8.314 L Kpa / mol K) (2.86 mol) (296K) (20.0 L) Answer: 24.4 atm P = kpa = 352 kpa 14

15 A gas at a pressure of 760. mm Hg and having a volume of 1024 ml is changed to a pressure of 115 mm Hg ; what is the new volume if the temperature stays constant? Answer: 6770 ml A sample of gas has a volume of 155 ml at 0 C. What will be the volume of the gas if it warmed up to a temperature of 85 C? Answer: 203 ml A sample of oxygen at 24.0 C and 745 mm Hg was found to have a volume of 455 ml. How many grams of O 2 were in the sample? Answer: g O 2 15