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.

• P1 = initial pressure

• V1 = initial volume

• P2 = final pressure

• V2 = final volume

• P1 & P2 can be in anything as long as they are the same

• V1 & V2 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

• P1 = 210 KPa

• P2 = ?

• V1 = 4.0 L

• V2 = 2.5 L

• P1V1 = P2V2

• (210 KPa)(4.0L) = (P2)(2.5 L)• P2 = 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

• V1 = V2

T1 T2

• V1 = initial volume

• V2 = final volume

• T1 = initial temperature

• T2 = final temperature

• V1 & V2 can be in any unit as long as they are the same

• T1 & T2 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• V1 = 2.32 L

• V2 = ?

• T1 = 40.0 °C = 313 K

• T2 = 75.0 °C = 348 K

• V1 = V2

T1 T2

• 2.32 L = V2

313K 348 K• V2 = 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:• P1 = P2

T1 T2

Gay Lussac’s Law: Pressure & Temperature

• P1 = P2

T1 T2

• P1 = initial pressure• P2 = final pressure• T1 = initial temperature• T2 = final temperature• P1 & P2 can be in any unit as long as they

are the same• T1 & T2 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• P1 = 3.20 atm• P2 = ?• T1 = 22.0 °C = 295 K• T2 = 60.0 °C = 333 K• P1 = P2

T1 T2

• 3.20 atm = P2

295K 333K• P2 = 3.61 atm

Combined Gas Law

P1V1 = P2V2

T1 T2

• 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

• P1V1 = P2V2

T1 T2

• P1 = 110.0 kPa• V1 = 2.00 L• T1 = 30.0 °C = 303 K• P2 = 440.0 kPa• V2 = ?• T2 = 80.0 °C = 353 K

Example

• P1V1 = P2V2

T1 T2

• (110.0)(2.00L) = (440.0kPa)(V2)

303K 353K• V2 = 0.583 L