(5) What mass of NaOH is required to react exactly with 25. 0 mL of 1 (5) What mass of NaOH is required to react exactly with 25.0 mL of 1.2 M H2SO4? a) 1.2 g b) 1.8 g c) 2.4 g d) 3.5 g e) none of these

Gases The earth’s atmosphere is a mixture of gases that consists mainly of elemental nitrogen (N2) and oxygen (O2). The gases in the atmosphere also shield us from harmful radiation from the sun and keep the earth warm by reflecting heat radiation back toward the earth. Properties of a Gas: It has no fixed shape and volume. The distance between gas atoms/molecules is larger as compared to particle size. Uniformly fills any container. Mixes completely with any other gas.

PRESSURE Interparticles attraction in gas is negligible. Particles are constantly moving and colliding especially with walls of container. The collision of particles with surface result in pressure. A device to measure atmospheric pressure, the barometer, was invented in 1643 by an Italian scientist named Evangelista Torricelli (1608–1647), who had been a student of Galileo.

PRESSURE SI units of pressure = Newton/meter2 (N/m2) = 1 Pascal (Pa) 1 standard atmosphere = 101325 Pa 1 standard atmosphere = 1 atm = 760 mm Hg = 760 torr

Pressure Conversions The pressure of a gas is measured as 2.5 atm. Represent this pressure in both Torr and Pascal.

Barometer Device used to measure atmospheric pressure. Mercury flows out of the tube until the pressure of the column of mercury standing on the surface of the mercury in the dish is equal to the pressure of the air on the rest of the surface of the mercury in the dish.

Manometer Device used for measuring the pressure of a gas in a container.

Example:

Boyle’s Law An Irish chemist, Robert Boyle (1627–1691) studied the relationship between the pressure of the gas and its volume. He used a J-shaped tube closed at one end. Boyle law states; Pressure of a gas is inversely related to the volume of that gas at constant temperature (T). 1 V P 1 V P = k PV = K Boyle’s Law also can be written as;

Boyle’s Law A J-tube similar to the one used by Boyle. When mercury is added to the tube, pressure on the trapped gas is increased, resulting in a decreased volume.

Boyle’s Law EXAMPLE:

Cont.

P1 x V1 = P2 x V2 P2 = ? P1 = 726 mmHg V1 = 946 mL V2 = 154 mL P1 x V1 Example: A sample of chlorine gas occupies a volume of 946 mL at a pressure of 726 mmHg. What is the pressure of the gas (in mmHg) if the volume is reduced at constant temperature to 154 mL? P1 x V1 = P2 x V2 P2 = ? P1 = 726 mmHg V1 = 946 mL V2 = 154 mL P1 x V1 V2 726 mmHg x 946 mL 154 mL = P2 = = 4460 mmHg

Charles’s Law A French physicist, Jacques Charles (1746–1823), who was the first person to fill a balloon with hydrogen gas. Charles found in 1787 that; The volume of a gas at constant pressure increases linearly with the temperature of the gas. A Graph volume of a gas (at constant pressure) versus its temperature (K) gives a straight line. This behavior is shown for samples of several gases.

V/T = b (b is a proportionality constant) Charles’s Law V T V/T = b (b is a proportionality constant) Volume and Temperature (in Kelvin) are directly related (constant P and n). K = °C + 273 0 K is called absolute zero.

V1/T1 = V2/T2 V1 = 3.20 L V2 = 1.54 L T1 = 398.15 K T2 = ? V2 x T1 V1 A sample of carbon monoxide gas occupies 3.20 L at 125 0C. At what temperature will the gas occupy a volume of 1.54 L if the pressure remains constant? V1/T1 = V2/T2 V1 = 3.20 L V2 = 1.54 L T1 = 398.15 K T2 = ? V2 x T1 V1 1.54 L x 398.15 K 3.20 L = T2 = = 192 K

1) A balloon has a volume of 1. 20 liters at 24. 0C 1) A balloon has a volume of 1.20 liters at 24.0C. The balloon is heated to 48.0C. Calculate the new volume of the balloon. a) 1.20 L b) 1.30 L c) 1.70 L d) 2.10 L e) 2.40 L

1) A balloon has a volume of 1. 20 liters at 24. 0C 1) A balloon has a volume of 1.20 liters at 24.0C. The balloon is heated to 48.0C. Calculate the new volume of the balloon. a) 1.20 L b) 1.30 L c) 1.70 L d) 2.10 L e) 2.40 L

Avogadro’s Law V n V1/n1 = V2/n2 In 1811 the Italian chemist Avogadro postulated that equal volumes of gases at the same temperature and pressure contain the same number of “particles.” This observation is called Avogadro’s law Mathematically, V = an (a = proportionality constant) Avogadro’s law is where V is the volume of the gas, n is the number of moles of gas particles, and a is a proportionality constant at constant T and P. V n V1/n1 = V2/n2

2) A gas sample is held at constant pressure. The gas occupies 3 2) A gas sample is held at constant pressure. The gas occupies 3.62 L of volume when the temperature is 21.6C. Determine the temperature at which the volume of the gas is 3.45 L. a) 309 K b) 281 K c) 20.6 K d) 294 K e) 326 K

2) A gas sample is held at constant pressure. The gas occupies 3 2) A gas sample is held at constant pressure. The gas occupies 3.62 L of volume when the temperature is 21.6C. Determine the temperature at which the volume of the gas is 3.45 L. a) 309 K b) 281 K c) 20.6 K d) 294 K e) 326 K

Avogadro’s Law V a number of moles (n) V = constant x n V1/n1 = V2/n2 Constant temperature Constant pressure V = constant x n V1/n1 = V2/n2

Ammonia burns in oxygen to form nitric oxide (NO) and water vapor Ammonia burns in oxygen to form nitric oxide (NO) and water vapor. How many volumes of NO are obtained from one volume of ammonia at the same temperature and pressure? 4NH3 + 5O2 4NO + 6H2O 1 mole NH3 1 mole NO At constant T and P 1 volume NH3 1 volume NO

3) Gaseous chlorine is held in two separate containers at identical temperature and pressure. The volume of container 1 is 1.30 L and it contains 6.70 mol of the gas. The volume of container 2 is 2.20 L. How many moles of the gas are in container 2? a) 11.3 mol b) 19.2 mol c) 0.427 mol d) 3.96 mol e) none of these

3) Gaseous chlorine is held in two separate containers at identical temperature and pressure. The volume of container 1 is 1.30 L and it contains 6.70 mol of the gas. The volume of container 2 is 2.20 L. How many moles of the gas are in container 2? a) 11.3 mol b) 19.2 mol c) 0.427 mol d) 3.96 mol e) none of these

Ideal Gas Equation 1 Boyle’s law: V a (at constant n and T) P Charles’ law: V a T (at constant n and P) Avogadro’s law: V a n (at constant P and T) V a nT P V = constant x = R nT P R= is the gas constant PV = nRT OR universal gas constant

PV = nRT PV (1 atm)(22.414L) R = = nT (1 mol)(273.15 K) The conditions 0 0C and 1 atm are called standard temperature and pressure (STP). Experiments show that at STP, 1 mole of an ideal gas occupies 22.414 L. PV = nRT R = PV nT = (1 atm)(22.414L) (1 mol)(273.15 K) R = 0.082057 atm • L / (mol • K)

where R is the combined proportionality constant called the universal gas constant. When the pressure is expressed in atmospheres and volume in liters, R has the value 0.08206 L atm/K mol. The preceding equation can be rearranged to the more familiar form of the ideal gas law: PV = nRT

PV = nRT V = nRT P V = 1 atm 1.37 mol x 0.0821 x 273.15 K What is the volume (in liters) occupied by 49.8 g of HCl at STP? T = 0 0C = 273.15 K P = 1 atm n = 49.8 g x 1 mol HCl 36.45 g HCl = 1.37 mol PV = nRT V = nRT P V = 1 atm 1.37 mol x 0.0821 x 273.15 K L•atm mol•K V = 30.6 L

PV = nRT n, V and R are constant nR V = P T = constant P1 T1 P2 T2 = Argon is an inert gas used in light bulbs to retard the vaporization of the filament. A certain light bulb containing argon at 1.20 atm and 18 0C is heated to 85 0C at constant volume. What is the final pressure of argon in the lightbulb (in atm)? PV = nRT n, V and R are constant nR V = P T P1 = 1.20 atm T1 = 291 K P2 = ? T2 = 358 K = constant P1 T1 P2 T2 = P2 = P1 x T2 T1 = 1.20 atm x 358 K 291 K = 1.48 atm

Example: A sample of hydrogen gas (H2) has a volume of 8 Example: A sample of hydrogen gas (H2) has a volume of 8.56 L at a temperature of 0 oC and a pressure of 1.5 atm. Calculate the moles of H2 molecules present in this gas sample.

4) Gaseous C2H4 reacts with O2 according to the following equation: C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(g) What volume of oxygen at STP is needed to react with 1.50 mol of C2H4? a) 4.50 L b) 33.6 L c) 101 L d) 67.2 L e) Not enough information is given to solve the problem.

4) Gaseous C2H4 reacts with O2 according to the following equation: C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(g) What volume of oxygen at STP is needed to react with 1.50 mol of C2H4? a) 4.50 L b) 33.6 L c) 101 L d) 67.2 L e) Not enough information is given to solve the problem.

Dalton’s law of partial pressure In 1803 Dalton stated: For a mixture of gases in a container, the total pressure exerted is the sum of the pressures that each gas would exert if it were alone. This statement, known as Dalton’s law of partial pressures, Mathematically, The symbols P1, P2, P3, and so on represent each partial pressure, the pressure that a particular gas would exert if it were alone in the container.

Dalton’s Law of Partial Pressure V and T are constant P1 P2 Ptotal = P1 + P2

Cont. The total pressure of the mixture Ptotal can be represented as; Example: The mole fraction of nitrogen in the air is 0.7808. Calculate the partial pressure of N2 in air when the atmospheric pressure is 760 torr. Solution The partial pressure of N2 can be calculated as follows:

Postulates of the Kinetic Molecular Theory The gas particles are so small compared with the distances between them that the volume of the individual particles can be assumed to be negligible (zero). 2. The particles are in constant motion. The collisions of the particles with the walls of the container are the cause of the pressure exerted by the gas. 3. The particles are assumed to exert no forces on each other; they are assumed neither to attract nor to repel each other. 4. The average kinetic energy of the particles is proportional to the temperature of the gas in kelvins. Any two gases at the same temperature will have the same average kinetic energy.

(a) (b) (c) (a) A balloon filled with air at room temperature. (b) The balloon is dipped into liquid nitrogen at 77 K. (c) The balloon collapses as the molecules inside slow down due to the decreased temperature. Slower molecules produce a lower pressure.

Diffusion and Effusion Diffusion is the term used to describe the mixing of gases. When a small amount of pungent-smelling ammonia is released in a classroom, it takes some time before everyone in the room can smell it, because time is required for the ammonia to mix with the air. The rate of diffusion is the rate of the mixing of gases. Effusion is the term used to describe the passage of a gas through a tiny orifice into an evacuated chamber, as shown in Fig 1. The rate of effusion measures the speed at which the gas is transferred into the chamber. Thomas Graham (1805–1869), a Scottish chemist, found that the rate of effusion of a gas is inversely proportional to the square root of the mass of its particles.

OR OTHER DEFINITION Diffusion: Molecules moving from areas of high concentration to low concentration. Example: perfume molecules spreading across the room. Effusion: Gas escaping through a tiny hole in a container. Both of these depend on the molar mass of the particle, which determines the speed.

Diffusion: describes the mixing of gases Diffusion: describes the mixing of gases. The rate of diffusion is the rate of gas mixing. Molecules move from areas of high concentration to low concentration.

Graham’s law of effusion Figure 1. The effusion of a gas into an evacuated chamber. The rate of effusion (the rate at which the gas is transferred across the barrier through the pin hole) is inversely proportional to the square root of the mass of the gas molecules. The relative rates of effusion of two gases at the same temperature and pressure are given by the inverse ratio of the square roots of the masses of the gas particles: where M1 and M2 represent the molar masses of the gases. This equation is called Graham’s law of effusion.

Graham’s law of effusion Example: Calculate the ratio of the effusion rates of hydrogen gas (H2) and uranium hexafluoride (UF6). To calculate the molar masses: Molar mass of H2 2.016 g/mol, and molar mass of UF6 352.02 g/mol. Using Graham’s law, The effusion rate of the very light H2 molecules is about 13 times that of the massive UF6 molecules.

Graham’s law of Diffusion Two cotton plugs soaked in ammonia and hydrochloric acid are simultaneously placed at the ends of a long tube. A white ring of ammonium chloride (NH4Cl) forms where the NH3 and HCl molecules (vapors) meet several minutes later. White ring of NH4Cl(s) forms where the NH3 and HCl meet.