Gas Laws

Kinetic Theory

• Particles of matter are ALWAYS in motion.

• The volume of individual particles is approximately zero.

• Collision of particles with container wall causes pressure.

• Particles exert no forces on each other.

• The average kinetic energy is approximately equal to the Kelvin temperature of gases.

Measuring Pressure

• The first device for measuring atmospheric pressure was developed by Evangelista Torricelli during the 17th century

• The device was called a “barometer”– baro = weight

– meter = measure

The Early Barometer

• The normal pressure due to the atmosphere at sea level can support a column of mercury that is 760 mm high

Modern Day Barometers

• Caused by the collisions of molecules colliding with the wall of the container

• equal to force/unit area

• SI unit = Newton/meter2 = 1 pascal (Pa)

• 1 standard atmosphere = 101.31 kPa

• 1 standard atmosphere = 1 atm = 760 mmHg = 760 torr

Pressure Units

Pressure Units

Unit Symbol Definition/Relationship

Pascal Pa SI pressure unit1 Pa = Newton/meter2

Millimeters of mercury

mmHg Pressure that supports a1 millimeter column ofmercury in a barometer

Atmosphere Atm Average atmosphericpressure at sea leveland 0° C

Torr Torr 1 torr = 1 mmHg

Converting Celsius to Kelvin

• For every 1 °C you cool a gas, the volume decreases by 1/273.

• –273 °C is absolute zero (0 K) – the coldest temperature possible; all motion ceases.• K = °C + 273

• °C = K - 273

Standard Temperature and Pressure (STP)

• Pressure:• 1 atm or 760 mmHg or 101.3 kPa

• Temperature:• 0° C or 273 K

The Nature of Gases

• Gases expand to fill their container

• Gases are fluid – they flow

• Gases have low density

• Gases are compressible

• Gases effuse and diffuse

Boyle’s Law

• Robert Boyle (1627-1691), Irish chemist

• The volume of a given amount of gas held at a constant temperature varies inversely with the pressure. As pressure increases, volume decreases.

• P1V1 = P2V2

Boyle’s Law• Solve the following:

825 Torr = _____ kPa

• Solve the following:The volume of oxygen at 120 kPa is 3.20 L.

What is the volume of oxygen at 101.3 kPa?

101.3 kPa / 760 Torr = x kPa / 825 Torr

109.96 kPa = x

P1V1 = P2V2

(120 kPa)(3.20L) = (101.3 kPa)V2

3.79 L = V2

Charles’ Law

• Jacques Charles (1746-1823), French physicist

• The volume of a gas varies directly with its absolute temperature. As temperature increases, volume increases.

• V1 = V2

T1 T2

Charles’ Law Cont.

• Solve the following:– The volume of a gas at 40.0 °C is 4.50 L. Find

the volume of the gas at 80 °C .

40°C + 273 = 313 K 80°C + 273 = 353 KV1/T1 = V2/T2

4.5L / 313 K = V2 / 353 K

9 L = V2

Gay-Lussac’s Law

• Joseph Gay-Lussac • The pressure of a given mass of gas varies

directly with the Kelvin temperature when the volume remains constant.

• P1 = P2

T1 T2

*** Remember – temperature must be in

Kelvin

• Solve the following:– 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 be the gas pressure in the tank?

Gay-Lussac’s Law

Temp must be in Kelvin, so convert °C to K

T1 = 22.0 °C + 273 = 295K T2 = 60°C + 273 = 333K

P1/T1 = P2/T2

3.2 atm / 295K = x atm / 333K

3.61 atm = P2

Combined Gas Law

• Boyle’s, Charles’s, and Gay Lussac’s laws can be combined into a single law.

• The combined gas law states the relationship among pressure, volume, and temperature of a fixed amount of gas.

• P1V1 = P2V2

T1 T2

*** Remember – temperature must be in

Kelvin

Combined Gas Law• Solve the following:

– A gas at 110 kPa and 30.0 °C fills a flexible container with an initial volume of 2.00 L. If the temperature is raised to 80.0 °C and the pressure increased to 440 kPa, what is the new volume? Convert temp to Kelvin:

T1 = 30°C + 273 = 303K T2 = 80°C + 273 = 353K

P1V1/T1 = P2V2/T2

(110 kPa)(2.00L) / 303K = (440 kPa)V2 / 353K

0.582 L = V2

Ideal Gas Law• In all the other gas laws, the relationships hold true

for a “fixed mass” or “given amount” of a gas sample.

• Because pressure, volume, temperature, and the number of moles present are all interrelated, one equation is used to describe their relationship.– PV = nRT

• n = moles • R = ideal gas law constant• V must be in liter• T must be in Kelvin

Ideal Gas Law

• The value of R depends on pressure unit given.– If pressure unit is mm Hg orTorr :

• R = 62.4 L • mm Hg or Torr mol • K

– If pressure is atm:• R = 0.0821 L • atm mol • K

– If pressure is kPa:• R = 8.31 L • kPa mol • K

Ideal Gas Law

• Solve the following:

44.01 g of CO2 occupies a certain volume at

STP. Find that volume. n = sample mass

molar mass

n = 44.01 g 44.0098g/mol

n = 1 mol

(1atm)(V) = (1mol)(0.0821) (273 K)V = 22.4 L

Ideal Gas Law

• Solve the following:– Find the molar mass of a gas that weights 0.7155 g

and occupies a volume of 250. mL at STP.(1 atm)(0.250 L) = (n)(0.0821)(273 K)

n = 0.011 mol

0.011 mol = 0.755 g

X

X = 68.64 g/mol