Gas Laws

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) – mmHg – Torr

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 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 follows. • P 1 = initial pressure • V 1 = initial volume • P 2 = final pressure • V 2 = final volume • P1 & 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 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 • P 2 • V 1 • V 2 = 210 KPa = ?

= 4.0 L = 2.5 L • P 1 V 1 = P 2 V 2 • (210 KPa)(4.0L) = (P2)(2.5 L) • P 2 = 340 KPa

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 stated mathematically as follows.

Charles’ Law: Volume & Temperature • V 1 = V 2 T 1 • V 1 • V 2 • T 1 T 2 = initial volume = final volume = 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) °F = 1.8 °C + 32

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 • V 2 • T 1 • T 2 • V 1 = 2.32 L = ?

= 40.0 °C = 313 K = 75.0 °C = 348 K = V 2 T 1 T 2 • 2.32 L = V 2 313K 348 K • V 2 = 2.58 L

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 • P 1 • P 2 • T 1 T 2 = initial pressure = final pressure = 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 at 22.0 °C. If the temperature rises to 60.0 °C, what will the new pressure in the tank be?

Example

• P 1 • P 2 • T 1 • T 2 • P 1 = 3.20 atm = ?

= 22.0 = 60.0 = P 2 °C = 295 K °C = 333 K T 1 T 2 • 3.20 atm = P 2 295K 333K • P 2 = 3.61 atm

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 kPa 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 kPa, what is the new volume?

Example

• P 1 V 1 T 1 = P 2 V 2 T 2 • P 1 • V 1 • T 1 • P 2 • V 2 • T 2 = 110.0 kPa = 2.00 L = 30.0 °C = 303 K = 440.0 kPa = ?

= 80.0 °C = 353 K

Example

• P 1 V 1 = P 2 V 2 T 1 T 2 • (110.0)(2.00L) = (440.0kPa)(V 2 ) 303K 353K • V 2 = 0.583 L