The Gas Laws.

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The Gas Laws

Gases - 4 min

with the volume of gases
The Gas Laws Gas laws are concerned with the volume of gases

Gas volume can be expressed in any of the metric units

Volume 1 liter L = 3 dm 1 cubic decimeter = 1000 milliliters ml = 3 1000 cubic centimeters cm

Gas volume is determined by temperature & pressure
The Gas Laws Gas volume is determined by temperature & pressure

Changes in temperature cause changes in particle motion

Changes in particle motion cause changes in volume
Temperature Changes in particle motion cause changes in volume

Standard temperature is
zero degrees Celsius o C

Temperature All gas calculations must be done using Kelvin temperatures

Temperature o C 273 K + =

Standard atmospheric pressure is one atmosphere at sea level

Pressure 1 atm 760 mm Hg = = 101 kPa 101,300 Pascles = = 1013 mb 29.92 in Hg =

STP Standard Temperature and Pressure

Boyle's Law Charles' Law Combined Gas Law Dalton's Law

At a constant temperature, the pressure exerted by a
Boyle's Law At a constant temperature, the pressure exerted by a gas varies inversely with its volume.

Whatever pressure does, volume does the opposite.
Boyle's Law Used when only pressure changes. Whatever pressure does, volume does the opposite.

Boyle's Law

Boyle's Law V2 = V1 P1 P2

Boyle's Law Practice Problem

Boyle's Law A gas occupies 15.5 dm3 at a pressure of 30 mm Hg. What is its volume when the pressure is increased to 50 mm Hg. Assume temp does not change.

Write the proper equation.
Boyle's Law Write the proper equation. V2 = V1 P1 P2

Substitute the given numbers.
Boyle's Law Substitute the given numbers. V2 = dm3 30 mm Hg 50 mm Hg

Since pressure increases, what will happen to volume?
Boyle's Law V2 = dm3 30 mm Hg 50 mm Hg Since pressure increases, what will happen to volume?

DECREASE V2 will be less than V1.
Boyle's Law V2 = dm3 30 mm Hg 50 mm Hg DECREASE V2 will be less than V1.

Boyle's Law Do the math. V2 = dm3 30 mm Hg 50 mm Hg V2 = 9.3 dm3

All Boyle's Law problems should look like this:
V2 = V1 P1 P2 = dm3 30 mm Hg 50 mm Hg = 9.3 dm3 stop

varies directly with the Kelvin temperature.
Charles' Law At a constant pressure, the volume of a gas varies directly with the Kelvin temperature.

Whatever temperature does, volume does the same.
Charles' Law Used only when temperature changes. Whatever temperature does, volume does the same.

V2 = V1 T2 T1 K e l v i n t e m p e r a t u r e K = o C + 2 7 3
Charles' Law K e l v i n t e m p e r a t u r e K = o C V2 = V1 T2 T1

Charles' Law Practice Problem

500 ml of air at 20 oC is heated to 60 oC with no
Charles' Law 500 ml of air at 20 oC is heated to 60 oC with no pressure change. What is the new gas volume?

Write the proper equation.
Charles' Law Write the proper equation. V2 = V1 T2 T1

Substitute the given numbers.
Charles' Law Substitute the given numbers. V2 = 500 ml 60 oC 20 oC

Temps must be in Kelvins.
Charles' Law Temps must be in Kelvins. V2 = 500 ml 333 K 293 K Add 273 to Celsius temps.

Since temperature increases, what will happen to volume?
Charles' Law V2 = 500 ml 333 K 293 K Since temperature increases, what will happen to volume?

INCREASE V2 will be greater than V1.
Charles' Law V2 = 500 ml 333 K 293 K INCREASE V2 will be greater than V1.

Charles' Law Do the math. V2 = 500 ml 333 K 293 K V2 = 600 ml

All Charles' Law problems should look like this:
V2 = V1 T2 T1 = 500 ml 333 K 293 K = 600 ml stop

In the real world, pressure and temperature BOTH will change.

What do we do then??

Combined Gas Law Used when both temperature and pressure change

Combined Gas Law V2 = V1 P1 T 2 P2 T1

Combined Gas Law Practice Problem

would be the gas volume at 227 oC and 650 mm Hg?
Combined Gas Law A gas has a volume of 810 ml at 44 oC and 325 mm Hg. What would be the gas volume at 227 oC and 650 mm Hg?

Combined Gas Law Write the proper equation. V2 = V1 P1 T 2 P2 T1

Substitute the given numbers.
Combined Gas Law Substitute the given numbers. V2 = 810 ml 325 mm Hg 500 K 650 mm Hg K

Do the math. = 640 ml V2 = 810 ml 325 mm Hg 500 K 650 mm Hg 317 K
Combined Gas Law Do the math. V2 = 810 ml 325 mm Hg 500 K 650 mm Hg K = 640 ml

Calculate the volume of
a gas at STP if 5.05 dm3 of the gas are collected at 27.5 oC and 95.0 kPa.

V1 P1 T2 V2 = P2 T1 5.05 dm kPa K = 101 kPa K = 4.32 dm3

The total pressure in a container is the sum of the partial
Dalton's Law The total pressure in a container is the sum of the partial pressures of all the gases in the container

calculations with gases collected over water.
Dalton's Law This law is used most often when doing calculations with gases collected over water.

Dalton's Law Practice Problem #1

A container holds three gases: O2, CO2, and N2. The partial
Dalton's Law A container holds three gases: O2, CO2, and N2. The partial pressures are 2 atm, 3 atm, and 4 atm respectively. What is the total pressure inside the container?

Dalton's Law O atm CO2 - 3 atm N atm Total Pressure = 9 atm

Collecting a Gas Over Water

As gas bubbles through the water ...

water vapor mixes with the gas being collected.

Gas pressure in container is equal to air pressure ...

when water levels are the SAME.

Dalton's Law Practice Problem #2

If 60 liters of nitrogen gas is collected over water when the
Dalton's Law If 60 liters of nitrogen gas is collected over water when the temperature is 40 oC and atmospheric pressure is 760 mm Hg, what is the partial pressure of the nitrogen?

From research, the vapor pressure of H2O at 40 oC is 7.4 kPa.
Dalton's Law From research, the vapor pressure of H2O at 40 oC is 7.4 kPa. 101.0 kPa Total kPa H2O 93.6 kPa N2

Gas Diffusion

At equal temperature and pressure, equal volumes
Avogadro's Principle At equal temperature and pressure, equal volumes of gases contain the same number of molecules.

At STP, 22.4 dm3 of any gas contains one mole of molecules,
Molar Volume At STP, dm3 of any gas contains one mole of molecules, 6.02 X 1023

Ideal Gas Theoretical gas molecules that have mass, but no volume

Ideal Gas Equation P V = n R T

Ideal Gas Equation P V = n R T P - standard pressure

Ideal Gas Equation P V = n R T V - molar volume

Ideal Gas Equation P V = n R T n - number of moles

Ideal Gas Equation P V = n R T R dm3 . kPa mole . K

Ideal Gas Equation P V = n R T T - standard temp (K)

How many moles of gas are found in a 500 dm3 container if the conditions inside the container are 25 oC and 200 kPa?

PV PV = nRT n = RT = 40.4 moles 200 kPa 500 dm3 mole K
298 K dm3 kPa = 40.4 moles

Substituting for n: moles = mass (m) molecular mass (M)
Ideal Gas Equation Modification Substituting for n: moles = mass (m) molecular mass (M)

Ideal Gas Equation Modification P V = m R T M

The relative rates at which two gases diffuse under
Graham's Law The relative rates at which two gases diffuse under identical conditions vary inversely as the square roots of their molecular masses.

Fifteenth Lab Fuel in a Butane Lighter

END Gas Laws