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## Chapter 1 Gases

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**Chapter 1 Gases**Dr. Hisham E Abdellatef Professor of pharmaceutical analytical chemistry 2009 - 2010**Characteristics of Gases**• Expand to fill and assume the shape of their container • Compressible Next**Diffuse into one another and mix in all**proportions.homogeneous mixtures • Particles move from an area of high concentration to an area of low concentration. Next**Properties**that determine physical behavior of a gas 1. VOLUME 2. PRESSURE 3. TEMPERATURE**1. Volume**l x w x h πr²h capacity of the container enclosing it. (m3), (dm3), or liter. For smaller volumes (cm3), (ml).**2. Temperature**• Temperature: Three temperature scales • Fahrenheit (ºF) • Celsius (ºC) • Kelvin K (no degree symbol) • K performing calculations with the gas law equations.**gas**• Expand when heated**Temperature**• K = (ºC) + 273.15 • Example: ºC = 20º • K = 293.15 • Absolute or Kelvin Scale: =-273.15 (ºC as its zero). (T when V= 0 ) • 1 K = 1 ºC**The molecules in a gas are in constant motion. gaseous atoms**that collide with each other and the walls of the container. "Pressure" is a measure of the collisions of the atoms with the container. 3. Pressure**Pressure**• Force per unit area • Equation: P = F/A F = force A = area Next**Atmospheric pressure is measured with a barometer.**Height of mercury varies with atmospheric conditions and with altitude. Barometer**Mercury Barometer**• Measurement of Gas Pressure**Standard atmospheric pressure is the pressure required to**support 760 mm of Hg in a column. • There are several units used for pressure: • Pascal (Pa), N/m2 • Millimeters of Mercury (mmHg) • Atmospheres (atm)**Manometers**• Used to compare the gas pressure with the barometric pressure. Next**Types of Manometers**• Closed-end manometer The gas pressure is equal to the difference in height (Dh) of the mercury column in the two arms of the manometer**Open-end Manometer**The difference in mercury levels (Dh) between the two arms of the manometer gives the difference between barometric pressure and the gas pressure**Three Possible Relationships**• Heights of mercury in both columns are equal if gas pressure and atmospheric pressure are equal. Pgas = Pbar**Gas pressure is greater than the barometric pressure.**∆P > 0 Pgas = Pbar + ∆P**Gas pressure is less than the barometric pressure.**∆P < 0 Pgas = Pbar + ∆P**The Simple Gas Laws**1. Boyle’s Law 2. Charles’ Law 3. Gay-Lussac’s Law 4. Combined Gas Law**variables required to describe a gas**• Amount of substance: moles • Volume of substance: volume • Pressures of substance: pressure • Temperature of substance: temperature**The Pressure-Volume Relationship:**Boyle’s Law Boyle’s Law - The volume of a fixed quantity of gas is inversely proportional to its pressure.**Example**An ideal gas is enclosed in a Boyle's-law apparatus. Its volume is 247 ml at a pressure of 625 mmHg. If the pressure is increased to 825 mmHg, what will be the new volume occupied by the gas if the temperature is held constant?**Solution**Method 1: P1V1 = P2V2 or, solving for V2 the final volume**Solution**Method 2: The pressure of the gas increases by a factor825/625, the volume must decrease by a factor of 625/825 V2= V1 X (ratio of pressures)**Charles’ Law**The volume of a fixed amount of gas at constant pressure is directly proportional to the Kelvin (absolute) temperature. V1= V2 T1 T2 or V1T2 = V2T1**A 4.50-L sample of gas is warmed at constant pressure from**300 K to 350 K. What will its final volume be? Given: V1 = 4.50 L T1 = 300. K T2 = 350. K V2 = ? Equation: V1= V2 T1 T2 or V1T2 = V2T1 (4.50 L)(350. K) = V2 (300. K) V2 = 5.25 L Example . Charles’ Law**Gay-Lussac’s Law**The pressure of a sample of gas is directly proportional to the absolute temperature when volume remains constant. P1 = P2 T1 T2 or P1T2 = P2T1**On the next slide**The amount of gas and its volume are the same in either case, but if the gas in the ice bath (0 ºC) exerts a pressure of 1 atm, the gas in the boiling-water bath (100 ºC) exerts a pressure of 1.37 atm. The frequency and the force of the molecular collisions with the container walls are greater at the higher temperature.**Combined Gas Law**Pressure and volume are inversely proportional to each other and directly proportional to temperature. P1V1 = P2V2 T1 T2 or P1V1T2 = P2V2T1**A sample of gas is pumped from a 12.0 L vessel at**27ºC and 760 Torr pressure to a 3.5-L vessel at 52ºC. What is the final pressure? Given: P1 = 760 Torr P2 = ? V1 = 12.0 L V2 = 3.5 L T1 = 300 K T2 = 325 K Equation: P1V1 = P2V2 T1 T2 or P1V1T2 = P2V2T1 (760 Torr)(12.0 L)(325 K) = ( P2)(3.5 L)(300 K) P2 = 2.8 x 10³ Torr Example. Combined Gas Law**Avogadro’s Law**Volume & Moles**Avogadro’s Law**At a fixed temperature and pressure, the volume of a gas is directly proportional to the amount of gas. V = c · n V = volume c = constant n= # of moles Doubling the number of moles will cause the volume to double if T and P are constant.**Equation**Includes all four gas variables: • Volume • Pressure • Temperature • Amount of gas Next**PV = nRT**• Gas that obeys this equation if said to be an ideal gas (or perfect gas). • No gas exactly follows the ideal gas law, although many gases come very close at low pressure and/or high temperatures. • Ideal gas constant, R, is R = PV nT = 1 atm x 22.4 L 1 mol x 273.15 K R = 0.082058 L·atm/mol· K**Example**• Suppose 0.176 mol of an ideal gas occupies 8.64 liters at a pressure of 0.432 atm. What is the temperature of the gas in degrees Celsius?**Solution**PV = nRT To degrees Celsius we need only subtract 273 from the above result: =258 - 273 = -15OC**Example**• Suppose 5.00 g of oxygen gas, O2, at 35 °C is enclosed in a container having a capacity of 6.00 liters. Assuming ideal-gas behavior, calculate the pressure of the oxygen in millimeters of mercury. (Atomic weight: 0 = 16.0)**Solution**• One mole of O2 weighs 2(16.0) = 32.0 g. 5.00 g of O2 is, therefore, 5.00 g/32.0 g mol-1, or 0.156 mol. 35 °C is 35 + 273 = 308 K • PV = nRT**Molar volume of an ideal gas at STP**• The volume occupied by one mole, or molar volume, of an ideal gas at STP is