Gas Laws Gas Pressure Pressure is defined as

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Gas Laws

Gas Laws

Gas Pressure • Pressure is defined as force per unit area. • Gas particles

Gas Pressure • Pressure is defined as force per unit area. • Gas particles exert pressure when they collide with the walls of their container. • The SI unit of pressure is the pascal (Pa). • However, there are several units of pressure – – – Pascal (Pa) Kilopascal (KPa) Atmosphere (atm) mm. Hg Torr

Boyle’s Law: Pressure and Volume • Boyle was an Irish chemist who studied the

Boyle’s Law: Pressure and Volume • Boyle was an Irish chemist who studied the relationship between volume and pressure • Boyle’s law states that the pressure and volume of a gas at constant temperature are inversely proportional.

Boyle’s Law: Pressure and Volume • At a constant temperature, the pressure exerted by

Boyle’s Law: Pressure and Volume • At a constant temperature, the pressure exerted by a gas depends on the frequency of collisions between gas particles and the container. • If the same number of particles is squeezed into a smaller space, the frequency of collisions increases, thereby increasing the pressure.

Boyle’s Law: Pressure and Volume • In mathematical terms, this law is expressed as

Boyle’s Law: Pressure and Volume • In mathematical terms, this law is expressed as follows. • P 1 = initial pressure • V 1 = initial volume • P 2 = final pressure • V 2 = final volume • P 1 & P 2 can be in anything as long as they are the same • V 1 & V 2 can be in anything as long as they are the same

Example • A sample of Helium gas is compressed from 4. 0 L to

Example • A sample of Helium gas is compressed from 4. 0 L to 2. 5 L at a constant temperature. If the pressure of the gas in the 4. 0 L volume is 210 KPa, what will the pressure be at 2. 5 L?

Example • • P 1 = 210 KPa P 2 = ? V 1

Example • • P 1 = 210 KPa P 2 = ? V 1 = 4. 0 L V 2 = 2. 5 L P 1 V 1 = P 2 V 2 (210 KPa)(4. 0 L) = (P 2)(2. 5 L) P 2 = 340 KPa

Charles’ Law: Volume & Temperature • Charles was a French physicist who looked at

Charles’ Law: Volume & Temperature • Charles was a French physicist who looked at the relationship between temperature and volume • He noted that as temperature went up, so did volume when pressure was held constant

Charles’ Law: Volume & Temperature • This observation is Charles’s law, which can be

Charles’ Law: Volume & Temperature • This observation is Charles’s law, which can be stated mathematically as follows.

Charles’ Law: Volume & Temperature • V 1 = V 2 T 1 T

Charles’ Law: Volume & Temperature • V 1 = V 2 T 1 T 2 • V 1 = initial volume • V 2 = final volume • T 1 = initial temperature • T 2 = final temperature • V 1 & V 2 can be in any unit as long as they are the same • T 1 & T 2 MUST be in Kelvin

Temperature conversions K = 273 + °C °C = 0. 56 (°F – 32)

Temperature conversions K = 273 + °C °C = 0. 56 (°F – 32) °F = 1. 8 °C + 32

Example • A sample of gas at 40. 0 °C occupies a volume of

Example • A sample of gas at 40. 0 °C occupies a volume of 2. 32 L. If the temperature is raised to 75. 0 °C what will the new volume be?

Example • • • V 1 = 2. 32 L V 2 = ?

Example • • • V 1 = 2. 32 L V 2 = ? T 1 = 40. 0 °C = 313 K T 2 = 75. 0 °C = 348 K V 1 = V 2 T 1 T 2 • 2. 32 L = V 2 313 K 348 K • V 2 = 2. 58 L

Gay Lussac’s Law: Pressure & Temperature • Gay Lussac studied the relationship between pressure

Gay Lussac’s Law: Pressure & Temperature • Gay Lussac studied the relationship between pressure and temperature • He noticed that at a constant volume a direct relationship existed between the Kelvin temperature and volume • Giving the equation: • P 1 = P 2 T 1 T 2

Gay Lussac’s Law: Pressure & Temperature • P 1 = P 2 T 1

Gay Lussac’s Law: Pressure & Temperature • P 1 = P 2 T 1 T 2 • P 1 = initial pressure • P 2 = final pressure • T 1 = initial temperature • T 2 = final temperature • P 1 & P 2 can be in any unit as long as they are the same • T 1 & T 2 MUST be in Kelvin

Example • The pressure of a gas in a tank is 3. 20 atm

Example • The pressure of a gas in a tank is 3. 20 atm at 22. 0 °C. If the temperature rises to 60. 0 °C, what will the new pressure in the tank be?

Example • • • P 1 = 3. 20 atm P 2 = ?

Example • • • P 1 = 3. 20 atm P 2 = ? T 1 = 22. 0 °C = 295 K T 2 = 60. 0 °C = 333 K P 1 = P 2 T 1 T 2 • 3. 20 atm = P 2 295 K 333 K • P 2 = 3. 61 atm

Combined Gas Law P 1 V 1 = P 2 V 2 T 1

Combined Gas Law P 1 V 1 = P 2 V 2 T 1 T 2 • Instead of memorizing all three equations, you can simply memorize this one • Just delete what you don’ t need

Example • A gas at 110. 0 k. Pa and 30. 0°C fills a

Example • A gas at 110. 0 k. Pa and 30. 0°C fills a flexible container to a volume of 2. 00 L. If the temperature was raised to 80. 0°C and the pressure was increased to 440. 0 k. Pa, what is the new volume?

Example • P 1 V 1 = P 2 V 2 T 1 T

Example • P 1 V 1 = P 2 V 2 T 1 T 2 • • • P 1 = 110. 0 k. Pa V 1 = 2. 00 L T 1 = 30. 0 °C = 303 K P 2 = 440. 0 k. Pa V 2 = ? T 2 = 80. 0 °C = 353 K

Example • P 1 V 1 = P 2 V 2 T 1 T

Example • P 1 V 1 = P 2 V 2 T 1 T 2 • (110. 0)(2. 00 L) = (440. 0 k. Pa)(V 2) 303 K 353 K • V 2 = 0. 583 L