2 Pressure What is pressure? Pressure is the amount of force - PowerPoint PPT Presentation

Gases: Chapter 9 • Characteristics: Unlike liquids and solids, they: • have large spaces between the gas molecules → not tightly packed • expand to fill the container → take the shape of the container • diffuse into one another and mix • 4 properties that affect its physical behaviour:  amount of gas (in moles)  volume  temperature 1  pressure

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Pressure What is pressure? Pressure is the amount of force applied to an area. • it is the force per unit area – (force divided by the area over which the force is applied) • The unit is N.m -2 which is one pascal (Pa) F ( N )  P ( Pa ) 2 ( ) A m 3

Units of Pressure • Pascals 1 Pa = 1 N/m 2 • Bar 1 bar = 10 5 Pa = 100 kPa • mm Hg or torr The difference in the heights in mm ( h ) of two connected columns of mercury • Atmosphere 1.00 atm = 760 torr 4

Barometric Pressure • The simplest device used to measure pressure is the Toricelli barometer • Standard atmosphere (atm) – is the pressure exerted by a mercury column of 760 mm (density of Hg = 13.5951 g.cm -3 (0 0 C and gravity = 9.80665 m.s -2 ) 5

1 atm = 760 mmHg 1 atm = 760 Torr Example: What is the height of a column of water that exerts the same pressure as a column of mercury 76.0 cm (760 mm) high? P H2O = P Hg Using equation (g x h x d) g x h H2O x 1.00 g/cm 3 = g x 76.0 cm x 13.6 g/cm 3  3 76 . 0 cm 13 . 6 g / cm    3 h 1 . 03 10 cm H O 3 2 1 . 00 g / cm  10 . 3 m 6

Common Units of Pressure Atmosphere atm Millimeter of mercury mmHg 1 atm = 760 mmHg Torr Torr = 760 Torr Pascal Pa = 101 325 Pa Kilopascal kPa = 101.325 kPa Bar bar = 1.01325 bar Millibar mb = 1013.25 mb 7

Manometer Used to measure the difference in pressure between atmospheric pressure and that of a gas in a vessel. Figure 9.2

Standard Pressure • Normal atmospheric pressure at sea level. • It is equal to: – 1.00 atm – 760 torr (760 mm Hg) – 101.325 kPa

Manometers 10

Example: When a manometer is filled with liquid mercury ( d = 13.6 g/cm 3 ), the barometric pressure is 748.2 mmHg, and the difference in mercury levels is 8.6 mmHg. What is the gas pressure P gas ? P gas = P bar + ∆P = 748.2 mmHg - 8.6 mmHg = 739.6 mmHg 11

Gas Laws Boyle’s Law: The volume of a fixed quantity of gas at constant temperature is inversely proportional to the pressure. 1  P V 12

Charles’s Law: - The volume of a fixed amount of gas at constant pressure is directly proportional to its temperature. V  T A plot of V versus T will be a straight line. 13

Standard Temperature and Pressure STP conditions: • standard temperature for gases = 0 0 C = 273.15 K • standard pressure, 1 atm = 760 mmHg Avogadro’s Law: - at a fixed temperature and pressure, the volume of a gas is directly proportional to the amount of gas V   1 22 . 4 ( ) n mol gas L gas at STP 14

Example: What is the mass of 1.00 L of cyclopropane gas, C 3 H 6 , when measured at STP?  1L =1dm 3 1 mol 22 . 4 L gas ( at STP )  x 1 . 00 L  1 . 00 L 1 mol   x 0 . 0446 mol 22 . 4 L   mass of C H n m . m 3 6    1 0 . 0446 mol 42 . 08 g . mol  1 . 88 g 15

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, if we call the proportionality constant, R: nT V = R P i.e. PV = nRT 16

The Ideal Gas Equation PV  nRT Where R = gas constant R = 8.3145 m 3 .Pa.mol -1 .K -1 (SI unit-use this) = 8.3145 J.mol -1 .K -1 (SI unit)(1m 3 Pa = 1J) = 0.082057 L.atm.mol -1 .K -1 = 8.314 kPa. dm 3 mol -1 .K -1 17

Units (1m 3 Pa =1J) R= 8.3145 J mol -1 K -1 R = 8.3145 m 3 Pa mol -1 K -1 F ( N )  P 2 A ( m ) kg m  1 N 2 s kg m 2 s  Pa 2 m kgm 1   Pa 2 2 s m   1 2  Pa kg m s  3 2 2   m Pa kg m s 1 J 18

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Volume conversion (from milliliters to litres and then to cubic meters) 3 1 L 1 m    V 305 mL 1000 1000 mL L    4 3 3 . 05 10 m 20

Pressure conversion • 1 atm → 101325 Pa • 1.37 atm → x Pa • X = (1.37 atm x 101325 Pa) / 1 atm = 138815 Pa = 1.38 x 10 5 Pa 21

Example: What is the pressure, in kPa, exerted by 1.00 x 10 20 molecules of N 2 in a 305 mL flask at 175 0 C?   23 1 mol 6 . 02 10 molecules   20 x 1 . 00 10 molecules   20 1 . 00 10 molecules 1 mol  x  23 6 . 022 10 molecules  0 . 000166 mol N 2     3 1 1 nRT 0 . 000166 mol 8 . 3145 m . Pa . mol . K 448 K   P   4 3 V 3 . 05 10 m   3 2 . 03 10 Pa  2 . 03 kPa 22

General Gas Equation P V P V  f f i i n T n T i i f f For a fixed mass of gas, P V P V  f f i i T T i f 23

Example: If a fixed amount of gas held at constant volume goes from STP conditions to boiling at 100 0 C, what is the new pressure? P V P V  f f i i n T n T i i f f  P  P P T  f i f i P f T T T i i f  1 . 00 atm 373 K  273 K  1 . 37 atm 24

Applications of the Ideal Gas Equation – Molar mass Molar mass determination PV = nRT m n  M mRT PV  M 25

Example: Propylene is used in the production of plastics. A glass vessel weighs 40.1305 g when clean, dry and evacuated; it weighs 138.2410 g when filled with water at 25 0 C(density of water = 0.9970 g/mL) and 40.2959 g when filled with propylene gas at 740.3 mmHg and 24.0 0 C. What is the molar mass of propylene? Mass of water to fill vessel = 138.2410 g – 40.1305 g = 98.1105 g Density = mass/volume Volume = mass/density Volume of water (vol. of vessel) = 98.1105 g/0.9970 g/ml = 98.41 mL = 0.09841 L 26

Mass of gas = 40.2959 g – 40.1305 g = 0.1654 g Temperature (K) = (24 0 C + 273.15 0 C) x 1K/1 0 C = 297.15 K R = 8.3145 m 3 Pa mol -1 K -1 Pressure = 740.3 mmHg/760 mmHg = 0.9741 atm (convert to Pa) = 98700 Pa mRT  M PV     3 1 1 0 . 1654 g 8 . 3145 m Pa mol K 297 . 15 K     5 3 98700 Pa 9 . 841 10 m  42 . 08 g / mol 27

Gas Densities and Molar Mass If we divide both sides of the ideal-gas equation by V and by RT , we get: PV = nRT n P = V RT 28

Applications of the Ideal Gas Equation - Gas Densities  m n M n     d M V V V We can then replace n/V with P/RT m MP   d V RT The density of a gas at STP is calculated by dividing its molar mass by the molar volume (22.4 L/mol) 29

Example: What is the density of O 2 at 298 K and 0.947 atm? (Convert 0.947 atm to Pa) R = 8.3145 m 3 Pa mol -1 K -1 m MP   d V RT   1 32 . 00 g mol 95954 Pa     3 1 1 8 . 3145 m Pa mol K 298 K 3 1239 . 5 g 1 m   3 m 1000 L  1 . 24 g / L 30

Gases in Chemical Reactions Air bags use the decomposition of sodium azide, NaN 3 to produce N 2 (g) that inflates the air bag Example: From the following equation determine what volume of N 2 is produced when 70.0 g of NaN 2 is decomposed at 735 mmHg and 26 0 C? 2NaN 3 (s) →2Na(l) + 3N 2 (g) 70 . 0 g   n ( NaN ) 1 . 08 mol 3 65 . 01 g / mol  1 . 08 mol 3   n ( N ) 1 . 62 mol 31 2 2

P = 735 mmHg/760 mmHg = 0.967 atm (convert to Pa) n = 1.62 mol R = 8.3145 m 3 Pa mol -1 K -1 T = (26 0 C + 273 0 C) x 1K/1 0 C = 299 K nRT  V P     3 1 1 1 . 62 mol 8 . 3145 m Pa mol K 299 K  97981 Pa 1000 L   3 0 . 0411 m 3 1 m  41 . 1 L 32

Law of Combining Volumes Example: For the following reaction, determine what volume of SO 2 is produced per liter of O 2 consumed? Both gases are measured at 25 0 C and 745 mmHg.    2 ZnS ( s ) 3 O 2 ZnO ( s ) 2 SO ( g )  2 2  1 L 3 mol O ( g ) 2  x 2 mol SO ( g ) 2  2 mol 1 L  x 3 mol  0 . 667 L SO ( g ) 33 2

Mixtures of Gases 34

Dalton’s law of partial pressures: • The total pressure of a mixture of gases is the sum of the partial pressures of the components of the mixture P tot = P A + P B + …… n RT n RT V  V  B A B A P P tot tot   V V V tot A B 35

P n ( RT / V ) n V n ( RT / P ) n     A tot A tot A A A A and P n ( RT / V ) n V n ( RT / P ) n tot tot tot tot tot tot tot tot n P V    A A A x A n P V tot tot tot Where X A is the mole fraction (fraction of all the molecules in a mixture contributed by that component) 36