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reader7 - Petrucci 6.1 Pressure Force = mass acceleration...

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Petrucci 6.1 Pressure Force = mass × acceleration Gravitational Force = mass × gravitational acceleration F g = mg ( g = 9 . 81 m/s 2 ) Pressure = Force Area P = F A Liquid Pressure P = F A = mg A = ( dV ) g A = d ( A × h ) g A = d × h × g Liquid pressure is independent of area Directionally proportional with height of liquid column Gas Pressure

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The pressure exerted by a gas is the result of the innumerable impacts of the molecules on the container walls Manometer The liquid column will be pushed up by the gas until the pressure exerted by the gas equals the pressure exerted by the liquid column plus the atmo- spheric pressure P gas = P liquid + P atm = dhg + P atm Δ P = P gas - P atm = dhg Barometer A barometer is simply a manometer open to the atmosphere on one side and has a vacuum on the other side, so instead of measuring Δ P for a gas in a container it measures the atmospheric pressure directly. It works in exactly the same way: Equilibrium when the liquid pressure of the column equals atmospheric pressure P = ghd h = P gd 2
Note that the higher the density of the liquid, the smaller h for the same gas pressure. That is why mercury is the traditional choice since it gives small manageable column heights. Density of mercury: 13.5951 g/cm 3 = 13 . 5951 10 - 3 kg (10 - 2 m) 3 = 135951 kg/m 3 Height of mercury column at 1 atm (101325 Pa) h = 101325 Pa 9 . 81 m / s 2 × 135951 kg / m 3 = 0 . 760 m = 760 mm Height of water column at same pressure ( d = 1 kg/L = 1 kg (10 - 1 m) 3 = 1000 kg/m 3 ) h = 101325 Pa 9 . 81 m / s 2 × 1000 kg / m 3 = 10 . 3 m Pressure units Pascal (Pa) SI unit for pressure 1 Pa = 1 M/m 2 = 1 kg m - 1 s - 2 Atmosphere (atm) 1 atm = 101,325 Pa Torr (Torr) 1 atm = 760 Torr 1 Torr = 1 mmHg Bar (bar) 1 bar = 100,000 Pa For practical purposes at- mospheric purposes can be set to 1 bar Millimeter mercury (mmHg) 1 atm = 760 mmHg Pounds per square inch (psi) 1 psi = 6,894.76 Pa Pressure in Davis Oct. 21 2009 (from weather.com): 1080.3 mb 1080 . 3 mb = 1 . 0803 bar = 100 , 000 Pa / bar 101325 Pa / atm × 1 . 0803 bar = 1 . 0662 atm Pressure in Paris, France Oct. 21 2009 (from weather.com): 999.0 mb 999 . 0 mb = 0 . 9990 bar = 100 , 000 Pa / bar 101325 Pa / atm × 0 . 9990 bar = 0 . 9859 atm Petrucci 6.3 Ideal Gas Law The ideal gas law is a simple relationship between pressure, volume, amount, and temperature of a gas. It was originally an empirically determined relationship, but as we shall see it can also be derived from a simple model PV = nRT 3

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R is called the gas constant R = 8 . 314472 J mol - 1 K - 1 R = 0 . 0820574 L atm mol - 1 K - 1 Petrucci 6.2 "‘Simple Gas Laws"’ Constant Relationship Boyle’s Law n , T P 1 V 1 = nRT , P 2 V 2 = nRT P 1 V 1 = P 2 V 2 V 1 P Charle’s Law n , P PV 1 = nRT 1 , PV 2 = nRT 2 PV 1 PV 2 = nRT 1 nRT 2 V 1 V 2 = T 1 T 2 V T Avogadro’s Law T , V P 1 V = n 1 RT , P 2 V = n 2 RT P 1 V P 2 V = n 1 RT n 2 RT P 1 P 2 = n 1 n 2 P n Petrucci 6.3 Solving Problems using the Ideal Gas Law PV = nRT Collect all information given in problem and identify what you need to find P 1 , V 1 , n 1 , T 1 P 2 , V 2 , n 2 , T 2 Convert units if needed TEMPERATURE ALWAYS IN KELVIN!
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