CHEMISTRY CHAPTER 11 TEMPERATURE A measure of the average kinetic energy of the particles of a sample. It is the property of matter that determines if heat can be transferred from one body to another, and the direction of that transfer.

COMMON TEMPERATURE SCALES English (Fahrenheit) Metric (Celsius) S.I. (Kelvin) abbreviation F C K water boils 212° 100° 373 human body temp. ~98.6° 37° 310 water freezes 32° 0° 273 absolute zero -460° -273°

Absolute zero: all motion stops A kelvin (K) is the S. I Absolute zero: all motion stops A kelvin (K) is the S.I. unit of temperature. Kelvin temperatures are written without the degree symbol (°). A temperature difference of 1 kelvin is a 1 degree difference on the Celsius scale.

CONVERSION FACTORS

EXAMPLES 1. 10° C to K: K = °C + 273.15 = 10° + 273.15 = 283.15 2. 380. K to °C °C = K – 273.15 = 380 – 273.15 = 106.85, round to 107° C

Section 1 Gases and Pressure Chapter 11. GASES Section 1 Gases and Pressure Pressure and Force Pressure (P) is defined as the force per unit area on a surface. Gas pressure is caused by collisions of the gas molecules with each other and with surfaces with which they come into contact.

The SI unit of force is the newton (N). Measuring Pressure The SI unit of force is the newton (N). The SI unit of pressure is the pascal (Pa) = 1 N/m2 Because a pascal is so small, we usually use kilopascals (kPa).

Other units: The height of a column of mercury: 1 mm Hg = 1 torr Standard atmospheric pressure at sea level and 0° C: 1 atmosphere (atm) = 760 torr = 101.3 kPa (atmospheric pressure varies, but 1 atm is defined to be exactly 760 torr)

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Pounds per square inch (psi): 1 atm = 14.7 psi 1 psi = 6.89 kPa Atmospheric pressure is measured with a barometer.

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Dalton’s Law of Partial Pressures The pressure of each gas in a mixture is called the partial pressure of that gas. John Dalton, the English chemist who proposed the atomic theory, discovered that the pressure exerted by each gas in a mixture is independent of that exerted by other gases present.

Dalton’s law of partial pressures states that the total pressure of a gas mixture is the sum of the partial pressures of the component gases.

Dalton’s Law of Partial Pressures

Equation for Dalton’s Law of Partial Pressures

Gases Collected by Water Displacement Gases produced in the laboratory are often collected over water. The gas produced by the reaction displaces the water in the reaction bottle. Dalton’s law of partial pressures can be applied to calculate the pressures of gases collected in this way.

Water molecules at the liquid surface evaporate and mix with the gas molecules. Water vapor, like other gases, exerts a pressure known as vapor pressure. To determine the pressure of a gas inside a collection bottle, you would use the following equation, which is an instance of Dalton’s law of partial pressures. Patm = Pgas +

If you raise the bottle until the water levels inside and outside the bottle are the same, the total pressure outside and inside the bottle will be the same. Reading the atmospheric pressure from a barometer and looking up the value of at the temperature of the experiment in a table, you can calculate Pgas.

Particle Model for a Gas Collected Over Water

Sample Problem Oxygen gas from the decomposition of potassium chlorate, KClO3, was collected by water displacement. The barometric pressure and the temperature during the experiment were 731.0 torr and 20.0°C. respectively. What was the partial pressure of the oxygen collected?

Solution Given: PT = Patm = 731.0 torr = 17.5 torr (vapor pressure of water at 20.0°C, from table A-8 in your book) Patm = Unknown: in torr Solution: start with the equation: rearrange algebraically to:

substitute the given values of Patm and into the equation:

Section 2. The Gas Laws Boyle’s Law: Pressure-Volume Relationship Robert Boyle discovered that doubling the pressure on a sample of gas at constant temperature reduces its volume by one-half.

This is explained by the kinetic-molecular theory: The pressure of a gas is caused by moving molecules hitting the container walls. If the volume of a gas is decreased, more collisions will occur, and the pressure will increase. If the volume of a gas is increased, less collisions will occur, and the pressure will decrease.

Boyle’s Law states that if temperature is constant, the volume of a gas is inversely proportional to pressure.

Boyle’s law can be expressed as: PV = k or V=k/P P is the pressure, V is the volume, and k is a constant. If there is a change from condition 1 to condition 2 (and temperature remains constant), then: P1V1 = P2V2

Example (p. 370): A sample of oxygen gas has a volume of 150 Example (p. 370): A sample of oxygen gas has a volume of 150.0 mL when its pressure is 0.947 atm. What will the volume of the gas be at a pressure of 0.987 atm if the temperature remains constant?

Given:V1 of O2 = P1 of O2 = P2 of O2 = Unknown: Solution: Rearrange the equation for Boyle’s law (P1V1 = P2V2) to obtain V2.

Given:V1 of O2 = 150.0 mL P1 of O2 = 0.947 atm P2 of O2 = 0.987 atm Unknown: V2 of O2 in mL Solution: Rearrange the equation for Boyle’s law (P1V1 = P2V2) to obtain V2.

Check: pressure went up, volume went down. Substitute the given values of P1, V1, and P2 into the equation to obtainV2: Check: pressure went up, volume went down.

If pressure is constant, gases expand when heated. Charles’s Law: Volume-Temperature Relationship If pressure is constant, gases expand when heated. When the temperature increases, the volume of a fixed number of gas molecules must increase if the pressure is to stay constant.

At the higher temperature, the gas molecules move faster At the higher temperature, the gas molecules move faster. They collide with the walls of the container more frequently and with more force. The volume of a flexible container must then increase in order for the pressure to remain the same.

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In 1787 Charles discovered that all gases expand by about 1/273 for an increase in temperature of 1° C near 0° C. Extrapolating backwards from measurements at several temperatures gives a volume of 0 at -273.15. What is this temperature?

In 1787 Charles discovered that all gases expand by about 1/273 for an increase in temperature of 1° C near 0° C. Extrapolating backwards from measurements at several temperatures gives a volume of 0 at -273.15. What is this temperature? Absolute zero

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Charles’s law states that the volume of a fixed mass of gas at constant pressure is proportional to the Kelvin temperature.

Charles’s law can be expressed as: V is the volume, T is the Kelvin temperature, and k is a constant. For a change from condition 1 to condition 2:

Ex. (p. 372): A gas sample has a volume of 752 mL at 25° C Ex. (p. 372): A gas sample has a volume of 752 mL at 25° C. What will be the volume at 50° C if the pressure remains constant? Need to change to Kelvin temperatures: (K = °C + 273) T1 = 25 + 273 = 298 K T2 = 50 + 273 = 323 K

Check: T increased, V increased.

Gay-Lussac’s Law: Pressure-Temperature Relationship With increasing temperature, kinetic energy of gas particles increases, and they will exert more force on the sides of a container. In 1802 Gay-Lussac found that the pressure increases by about 1/273 for an increase in temperature of 1° C at about 0°. Extrapolating backward gives a pressure of 0 at -273° C.

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Gay-Lussac’s law: the pressure of a fixed mass of gas at constant volume varies directly with the Kelvin temperature.

Gay-Lussac’s law can be expressed as: P is the pressure, T is the Kelvin temperature, and k is a constant. For a change from condition 1 to condition 2:

Example (p. 373): The gas in a container is at a pressure of 3 Example (p. 373): The gas in a container is at a pressure of 3.00 atm at 25°C. Directions on the container warn the user not to keep it in a place where the temperature exceeds 52°C. What would the gas pressure in the container be at 52°C?

Given: P1 = 3.00 atm T1 = 25°C + 273 = 298 K T2 = 52°C + 273 = 325 K Unknown: P2 in atm

Check: T increased, P increased.

Summary of the Basic Gas Laws

Robert Boyle Jacques Charles Joseph Gay-Lussac crystalinks.com enciclopedia.us.es Joseph Gay-Lussac chemistryland.com

The Combined Gas Law Boyle’s law, Charles’s law, and Gay-Lussac’s law can be combined into a single equation that can be used for situations in which temperature, pressure, and volume all vary at the same time.

For a fixed amount of gas (grams or moles): For a change from situation 1 to situation 2 (temperatures in Kelvin):

The other three laws can be obtained from the combined gas law by keeping T, P, or V constant:

Example (when P, V, and T all change) (p Example (when P, V, and T all change) (p. 375): A helium-filled balloon has a volume of 50.0 L at 25°C and 1.08 atm. What volume will it have at 0.855 atm and 10.0°C?

Given: V1 = 50.0 L T1 = 25°C + 273 = 298 K T2 = 10°C + 273 = 283 K P1 = 1.08 atm P2 = 0.855 atm Unknown: V2 in L

Rearrange the equation for the combined gas law to obtain V2.

Section 3 Gas Volumes and the Ideal Gas Law Measuring and Comparing the Volumes of Reacting Gases Gay-Lussac discovered that in a reaction involving gases as reactants and products, the volumes were ratios of small whole numbers. For example: hydrogen gas + oxygen gas → water vapor (2 volumes) (1 volume) (2 volumes)

Avogadro’s Law Avogadro explained this: equal volumes of gases at the same temperature and pressure contain equal numbers of molecules.

Avogadro’s Law - 75205 Amedeo Avogadro brittanica.com

So the ratios of volumes in a reaction are the same as the ratios of molecules or the ratios of moles: 2 mol H2 + 1 mol O2 → 2 mol H2O 2H2 + O2 → 2H2O These ideas lead to the conclusion that the volume of a gas is proportional to the number of moles (n): V = kn

Ex. – what is the volume of 3 mol N2 at STP? Molar Volume of a Gas At standard temperature and pressure (STP) (0° C and 1 atm), 1 mole of a gas has a volume of 22.4 L. Ex. – what is the volume of 3 mol N2 at STP? 3 mol x (22.4 L/1 mol) = 67.2 L

Example 2: how many moles are in 5.00 L of a gas?

The Ideal Gas Law is the relationship among pressure, volume, temperature, and number of moles of a gas. PV = nRT where n = number of moles T = Kelvin temperature R = gas constant

Equation for the Ideal Gas Law - 75208

The Ideal Gas Constant (R) is expressed in different units depending on the units of pressure and volume.

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Example: What is the pressure in atmospheres exerted by a 0 Example: What is the pressure in atmospheres exerted by a 0.500 mol sample of nitrogen gas in a 10.0 L container at 298 K? Given: V = 10.0 L n = 0.500 mol T = 298 K Unknown: P in atm

Solution: Use the ideal gas law, which can be rearranged to find the pressure, as follows. Since we want P in atm, the value of 0.0821 (L•atm)/(mol•K) is used for R.