Daniel L. Reger Scott R. Goode David W. Ball Chapter 6 The Gaseous State

A solid has fixed shape and volume. Solid Phase Solid Br 2 at low temperature

A liquid has fixed volume but no definite shape. The density of a solid or a liquid is given in g/mL. Liquid Phase Liquid Br 2

A gas has no fixed volume or definite shape. The density of a gas is given in g/L whereas liquids and solids are in g/mL. Gas Phase Gaseous Br 2

Pressure is the force per unit area exerted on a surface. The pressure of the atmosphere is measured with a barometer. Pressure of a Gas

Both open and closed end manometers measure pressure differences. Manometers

Units of Pressure One atmosphere of pressure (1 atm) is the normal pressure at sea level. The SI unit of pressure is the pascal (Pa), but is a very small unit and is not used frequently by chemists. 1 atm = 760 mm Hg 1 atm = kPa 1 atm = 760 torr 1 atm = 1.01 bar 1 torr = Pa 1 atm = 29.9 in Hg 1 atm = 14.7 psi

Increasing the pressure on a gas sample, by addition of mercury to an open ended manometer, causes the volume to decrease. Boyle’s Law

A plot of volume versus 1/P is a straight line. V = k 1 x Boyle’s Law

A sample of a gas occupies 5.00 L at atm. Calculate the volume of the gas at 1.00 atm, when the temperature held is constant. Example: Changing P and V

A plot of volume versus temperature is a straight line. Extrapolation to zero volume yields absolute zero in temperature: -273 o C. V = k 2 x T, where T is given in units of kelvin. Charles’s Law

Avogadro’s Hypothesis Equal volumes of gases at constant T and P contain the same number of particles. The pressure in both containers is the same, but the mass of the gases is different.

A plot of the volume of all gas samples, at constant T and P, vs. the number of moles (n) of gas is a straight line. V = k 3 x n Avogadro’s Law

Changing P, T and V Example: Changing P, T and V A sample of a gas occupies 4.0 L at 25 o C and 2.0 atm of pressure. Calculate the volume at STP (T = 0 o C, P = 1 atm).

Test Your Skill A sample of a gas occupies 200 mL at 100 o C. If the pressure is held constant, calculate the volume of the gas at 0 o C.

The ideal gas law combines the three gas laws into a single equation: PV = nRT where: R = L. atm/mol. K The volume of one mole of an ideal gas at STP is 22.4 L Ideal Gas Law

Ideal Gas Law Calculation Calculate the number of moles of argon gas in a 30 L container at a pressure of 10 atm and temperature of 298 K.

Ideal Gas Law Calculation Calculate the number of moles of argon gas in a 30 L container at a pressure of 10 atm and temperature of 298 K. PV = nRT n = n = = 12 mol

Molar Mass and Density The ideal gas law can be used to calculate density (mass/volume) and molar mass (mass/moles) of a gas. At constant pressure and temperature the density of a gas is proportional to its molar mass, so the higher the molar mass, the greater the density of the gas.

Example: Molar Mass Calculate the molar mass of a gas if a 1.02 g sample occupies 220 mL at 95 o C and a pressure of 750 torr.

Gases and Chemical Equations The ideal gas law can be used to determine the number of moles, n, for use in problems involving reactions. The ideal gas law relates n to the volume of gas just as molar mass is used with masses of solids and molarity is used with volumes of solutions.

Example: Gases with Equations Calculate the volume of O 2 gas formed in the decomposition of 2.21 g of KClO 3 at STP. 2KClO 3 (s)  2KCl(s) + 3O 2 (g)

Gas Volumes in Reactions

Calculate the volume of NH 3 gas produced in the reaction of 4.23 L of H 2 with excess N 2 gas. Assume the volumes are measured at the same temperature and pressure. Example: Gas Volumes in Reactions

The pressure exerted by each gas in a mixture is called its partial pressure. For a mixture of two gases A and B, the total pressure, P T, is P T = P A + P B Dalton’s Law of Partial Pressure

Pressure of a Mixture of Gases

Example: Partial Pressures Calculate the pressure in a container that contains O 2 gas at a pressure of 3.22 atm and N 2 gas at a pressure of 1.29 atm.

Mole Fraction Mole fraction ( , chi) is the number of moles of one component of a mixture divided by the total number of moles of all substances present in the mixture.  A  +  B +  C = 1 The partial pressure of any gas, A, in a mixture is given by: P A =  A x P T

Mole fraction of the yellow gas is 3/12 = 0.25 and the mole fraction of the red gas is 9/12 = 0.75 Mole Fraction

Example: Partial Pressure Calculate the partial pressure of Ar gas in a container that contains 2.3 mol of Ar and 1.1 mol of Ne and is at a total pressure of 1.4 atm.

Water vapor is also present in a sample of O 2 gas collected over water. Collecting Gases over Water

Example: Collecting Gases Sodium metal is added to excess water, and H 2 gas produced in the reaction is collected over water with the gas volume of 1.2 L. If the pressure is 745 torr and the temperature 26 o C, what was the mass of the sodium? The vapor pressure of water at 26 o C is 25 torr. 2Na(s) + 2H 2 O( l )  H 2 (g) + 2NaOH(aq)

1. Gases consist of small particles that are in constant and random motion. 2. Gas particles are very small compared to the average distance that separates them. 3. Collisions of gas particles with each other and the walls of the container are elastic. 4. The average kinetic energy of gas particles is proportional to the temperature on the Kelvin scale. Kinetic Molecular Theory of Gases

Gas particles move at different speeds. Average speed is called the root mean square (rms) speed, u rms, and is the square root of the average squared speed. Maxwell-Boltzmann distribution curves Average Speed of a Gas

R = J/mol. K; molar mass in kilograms per mole Average Speed of a Gas

Effusion - the passage of a gas through a small hole into an evacuated space. Gases with low molar masses effuse more rapidly. Diffusion is the mixing of particles due to motion. Effusion and Diffusion

Gases deviate from the ideal gas law at high pressures. Deviations from Ideal Behavior

The assumption that gas particles are small compared to the distances separating them fails at high pressures. The observed value of PV/nRT will be greater than 1 under these conditions. Deviations from Ideal Behavior

The forces of attraction between closely spaced gas molecules reduce the impact of wall collisions. These attractive forces cause the observed value of PV/nRT to decrease below the expected value of 1 at moderate pressures. Forces of Attraction in Gases

A gas (O 2 below) deviates from ideal gas behavior at low temperatures (near the condensation point ) and high pressures. Ideal Gases

The van der Waals equation corrects for attractive forces and the volume occupied by the gas molecules. a is a constant related to the strength of the attractive forces. b is a constant that depends on the size of the gas particles. a and b are determined experimentally for each gas. van der Waals Equation