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