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Gases and Pressure Section 11.1. Vocabulary Pressure: the force per unit area on a surface Atmospheric pressure: the force per unit area exerted against.

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Presentation on theme: "Gases and Pressure Section 11.1. Vocabulary Pressure: the force per unit area on a surface Atmospheric pressure: the force per unit area exerted against."— Presentation transcript:

1 Gases and Pressure Section 11.1

2 Vocabulary Pressure: the force per unit area on a surface Atmospheric pressure: the force per unit area exerted against a surface by the weight of the air above that surface.

3 Measuring Pressure Barometer: device used to measure atmospheric pressure First introduced by Evangelista Torricelli during the early 1600s. Discovered that atmospheric pressure supports a column of mercury about 760 mm above the surface of Hg in a dish

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5 Units of Pressure There are several units for pressure mm Hg, also called torr in honor of Torricelli Average atmospheric pressure at sea level at 0°C is 760 mm Hg, or 760 torr Atmospheres: one atm is defined as being exactly equivalent to 760 mm Hg

6 More about Units The SI unit of measurement for pressure is the pascal One pascal (Pa) is the pressure exerted by a force of one newton acting on an area of one square meter Usually expressed as kPa 1 atm = 101.325 kPa

7 STP STP = standard temperature and pressure STP has conditions of exactly 0°C and 1 atm of pressure

8 Dalton’s Law of Partial Pressures The pressure of each gas in a mixture is called the partial pressure Dalton’s law states that the total pressure of a gas mixture is the sum of the partial pressures of the component gases P T = P 1 + P 2 + P 3 + …

9 Problem 1 A mixture of O 2, CO 2, and N 2 has a total pressure of 0.97 atm. What is the partial pressure of O 2 if the partial pressure of CO 2 is 0.70 atm and the partial pressure of N 2 is 0.12 atm? P T = P oxygen + P carbon dioxide + P nitrogen P oxygen = P T – (P carbon dioxide + P nitrogen ) P oxygen = 0.97 atm – (0.70 atm + 0.12 atm) P oxygen = 0.15 atm

10 Problem 2 Find the total pressure for a mixture that contains four gases with partial pressure of 5.00 kPa, 4.56 kPa, 3.02 kPa, and 1.20 kPa P T = P 1 + P 2 + P 3 + P 4 P T = (5.00 + 4.56 + 3.02 + 1.20) kPa P T = 13.78 kPa

11 Water Displacement Gas can be collected by water displacement, but some water vapor will be present The total pressure will be the sum of the partial pressures of the gas collected and water vapor The partial pressure of water vapor is found on a table (dependent on temperature)

12 Gas Collected by Water Displacement P gas = P atm – P water vapor

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14 Problem 3 O 2(g),from the decomposition of KClO 3, was collected by water displacement. The barometric (atmospheric) pressure and the temperature during the experiment were 97.5 kPa and 20.0° C. What was the partial pressure of the O 2 collected?

15 Solution P T = P atm = 97.5 kPa P water vapor = 2.34 kPa Unknown = P oxygen gas P oxygen gas = P T – P water vapor P oxygen gas = 97.5 kPa – 2.34 kPa P oxygen gas = 95.2 kPa


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