CHAPTER 5 GASES. Characteristics of Gases Unlike liquids and solids, gases – expand to fill their containers; – are highly compressible; – have extremely.

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Presentation transcript:

CHAPTER 5 GASES

Characteristics of Gases Unlike liquids and solids, gases – expand to fill their containers; – are highly compressible; – have extremely low densities.

Units of Pressure mm Hg or torr – These units are literally the difference in the heights measured in mm ( h ) of two connected columns of mercury. Atmosphere –1.00 atm = 760 torr

Manometer This device is used to measure the difference in pressure between atmospheric pressure and that of a gas in a vessel.

Standard Pressure Normal atmospheric pressure at sea level is referred to as standard pressure. It is equal to –1.00 atm –760 torr (760 mm Hg) – kPa

Convert 49 torr to atmospheres and pascals. – ans. 6.4 x10 -2 atm; 6.5 x 10 3 Pa

Boyle’s Law The volume of a fixed quantity of gas at constant temperature is inversely proportional to the pressure.

As P and V are inversely proportional A plot of V versus P results in a curve. Since V = k (1/P) This means a plot of V versus 1/P will be a straight line. PV = k

Consider a 1.53 L sample of gaseous SO 2 at a pressure of 5.6 x 10 3 Pa. If the pressure is changed to 1.5 x 10 4 Pa at a constant temperature, what will be the new volume of the gas? Ans. 0.57L

Charles’s Law The volume of a fixed amount of gas at constant pressure is directly proportional to its absolute temperature. A plot of V versus T will be a straight line. i.e., VTVT = k

A sample of a gas at 15°C and 1 atm has a volume of 2.58L. What volume will the gas occupy at 38°C and 1 atm? Ans. 2.79L

Avogadro’s Law The volume of a gas at constant temperature and pressure is directly proportional to the number of moles of the gas. Mathematically, this means V = kn

Suppose we have a 12.2 L sample containing 0.50 mol O 2 gas at a pressure of 1 atm and a temperature of 25°C. If all this O 2 were converted to ozone (O 3 ) at the same temperature and pressure, what would be the volume of the ozone? Ans. 8.1 L

Ideal-Gas Equation V  1/P (Boyle’s law) V  T (Charles’s law) V  n (Avogadro’s law) So far we’ve seen that Combining these, we get V V  nT P

Ideal-Gas Equation The constant of proportionality is known as R, the gas constant.

Ideal-Gas Equation The relationship then becomes nT P V V  nT P V = R or PV = nRT

A sample of hydrogen gas has a volume of 8.56 L at a temperature of 0°C and a pressure of 1.5 atm. Calculate the moles of H 2 molecules present in this gas sample. – Ans mol

Suppose we have a sample of ammonia gas with a volume of 7.0 mL at a pressure of 1.68 atm. This gas is compressed to a volume of 2.7 mL at a constant temperature. What is the final pressure? Ans. 4.4 atm

A sample of methane gas that has a volume of 3.8 L at 5°C is heated to 86°C at constant pressure. Calculate its new volume. – Ans. 4.9 L

A sample of diborane gas (B 6 H 6 ), a substance that bursts into flame when exposed to air, has a pressure of 345 torr at a temperature of -15°C and a volume of 3.48L. If conditions are changed so that the temperature is 36°C and the pressure is 468 torr, what will be the volume of the sample? Ans. 3.07L

A sample containing 0.35 mol argon gas at a temperature of 13°C and a pressure of 568 torr is heated to 56°C and a pressure of 897 torr. Calculate the change in volume that occurs. Ans. -3L

A sample of nitrogen gas has a volume of 1.75L at STP. How many moles of N 2 are present? Ans x 10 -2

Quicklime (CaO) is produced by the thermal decomposition of calcium carbonate (CaCO 3 ). Calculate the volume of CO 2 at STP produced from the decomposition of 152 g CaCO 3 by the reaction CaCO 3 (s)  CaO(s) + CO 2 (g) – Ans L

A sample of methane gas having a volume of 2.80 L at 25°C and 1.65 atm was mixed with a sample of oxygen gas having a volume of 35.0 L at 31°C and 1.25 atm. The mixture was then ignited to form carbon dioxide and water. Calculate the volume of CO 2 formed at a pressure of 2.50 atm and a temperature of 125°C. Ans L

Densities of Gases If we divide both sides of the ideal-gas equation by V and by RT, we get PV = nRT VRT nVnV P RT =

We know that – moles  molar mass = mass Densities of Gases So multiplying both sides by the molar mass (  ) gives n   = m P  RT mVmV = nM V PM RT =

Densities of Gases Mass  volume = density So, Note: One only needs to know the molar mass, the pressure, and the temperature to calculate the density of a gas. P  RT mVmV = d =

Molar Mass We can manipulate the density equation to enable us to find the molar mass of a gas: Becomes P  RT d = dRT P  =

The density of a gas was measured at 1.50 atm and 27°C and found to be 1.95 g/L. Calculate the molar mass of the gas. – Ans g/mol

Dalton’s Law of Partial Pressures The total pressure of a mixture of gases equals the sum of the pressures that each would exert if it were present alone. In other words, P total = P 1 + P 2 + P 3 + …

Mixtures of helium and oxygen can be used in scuba diving tanks to help prevent “the bends.” For a particular dive, 46 L He at 25°C and 1.0 atm and 12 L O 2 at 25°C and 1.0 atm were pumped into a tank with a volume of 5.0 L. Calculate the partial pressure of each gas and the total pressure in the tank at 25°C. – Ans. P He = 9.3 atm P O 2 = 2.4 atm P total = 11.7 atm

MOLE FRACTION Mole fraction is the ratio of the number of moles of a given component in a mixture to the total number of moles in the mixture. The mole fraction of each component in a mixture of ideal gases is directly related to its partial pressure. Since n 1 =P 1 (V/RT) and n 2 =P 2 (V/RT) and V/RT is constant, the mole fraction n 1 /n total = P 1 /P total

The partial pressure of oxygen was observed to be 156 torr in air with a total atmospheric pressure of 743 torr. Calculate the mole fraction of O 2 present. Ans

The mole fraction of nitrogen in the air is Calculate the partial pressure of N 2 in air when the atmospheric pressure is 760 torr. – Ans. 593 torr

Partial Pressures When one collects a gas over water, there is water vapor mixed in with the gas. To find only the pressure of the desired gas, one must subtract the vapor pressure of water from the total pressure.

A sample of solid potassium chlorate (KClO 3 ) was heated in a test tube and decomposed by the following reaction: 2KClO 3 (s)  2KCl(s) + 3O 2 (g) The oxygen produced was collected by displacement of water at 22°C at a total pressure of 754 torr. The volume of the gas collected was L, and the vapor pressure of water at 22°C is 21 torr. Calculate the partial pressure of O 2 in the gas collected and the mass of KClO 3 in the sample that was decomposed. Ans. P O 2 =733 torr2.12 g KClO 3

Kinetic-Molecular Theory This is a model that aids in our understanding of what happens to gas particles as environmental conditions change.

Kinetic-Molecular Theory The particles are so small compared with the distances between them that the volume of the individual particles can be assumed to be negligible (zero).

Kinetic-Molecular Theory 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.

Kinetic-Molecular Theory Attractive and repulsive forces between gas molecules are negligible.

Kinetic-Molecular Theory Energy can be transferred between molecules during collisions, but the average kinetic energy of the molecules does not change with time (elastic collisions.)

Kinetic-Molecular Theory The average kinetic energy of the molecules is proportional to the absolute temperature (K).

REAL GASES The molecules of a real gas have finite volumes and do exert forces on each other.

Why will decreasing the volume of a gas increase its pressure?

Why will increasing the temperature of a gas increase its pressure?

Why does increasing the volume of a gas at constant pressure mean that its temperature must also increase?

Effusion Effusion is the escape of gas molecules through a tiny hole into an evacuated space.

Effusion The difference in the rates of effusion for helium and nitrogen, for example, explains a helium balloon would deflate faster.

Graham’s Law of Effusion Rate of effusion for gas 1 M 2 Rate of effusion for gas 2 M 1

Freon-12 is used as a refrigerant in central home air conditioners. The rate of effusion of Freon-12 to Freon-11 (molar mass = 137.4) is 1.07:1. The formula of Freon-12 is one of the following: CF 4, CF 3 Cl, CF 2 Cl 2, CFCl 3, or CCl 4. Which formula is correct for Freon-12?

Diffusion Diffusion is the spread of one substance throughout a space or throughout a second substance.

Real Gases In the real world, the behavior of gases only conforms to the ideal-gas equation at relatively high temperature and low pressure.

Deviations from Ideal Behavior The assumptions made in the kinetic-molecular model (negligible volume of gas molecules themselves, no attractive forces between gas molecules, etc.) break down at high pressure and/or low temperature.

Corrections for Nonideal Behavior The ideal-gas equation can be adjusted to take these deviations from ideal behavior into account. The corrected ideal-gas equation is known as the van der Waals equation.

The van der Waals Equation ) (V − nb) = nRT n2aV2n2aV2 (P +