Gas Laws Problems

Boyle’s Law

Charle’s Law

Gay-Lussac’s Law

Review Kinetic Molecular Theory

• Gases consist of large numbers of tiny particles that are far apart relative to their size

• Collisions between gas particles and between particles and container walls are elastic collisions.

• Gas particles are in continuous rapid, random motion. They therefore posses kinetic energy, which is the energy of motion.

• There are no forces of attraction or repulsions between gas particles.

• The average kinetic energy of gas particles depends on the temperature of the gas

Volume

• The amount of space something occupies

• For gas molecules, the space that the gas molecules are allowed to move around in

• SI unit for volume is L or m3

Pressure• Pressure is the force exerted over an area (P

= F/A)• Pressure of a gas comes from the force of the

moving gas molecules hitting a certain area of the container they are inside

• SI units for pressure– Pa or kPa

• Other units include– Atm, torr, and mm of Hg

Temperature

• Temperature relates to the Kinetic energy of the gas molecules

• Higher temperatures have higher kinetic energies

• Si unit is Celsius

• Other units include– Farenheit and Kelvin

Temperature in Gas Laws

• When using the gas laws, we cannot use temperature in the units of Celsius

• For all gas laws, we must use the temperature unit Kelvin

• Kelvin = Celsius + 273– All temperatures used in the gas law

equations must be in Kelvin

Boyle’s Law• Relationship between

– Pressure and Volume

• If the volume is decreased, then the gas molecules have less space to move and will hit the container walls more frequently so pressure will increase

• If the volume is increased, then the gas molecules will have more space to move and will hit the container walls less frequently so pressure will decrease

Boyle’s Law• The relationship between volume and

pressure is inversely proportion– Meaning that when one increases, the other

decreases

• Mathematically, this is expressed as – Pressureinitial x Volumeinitial = Pressurefinal x Volumefinal

– Or PI x VI = PF x VF

• Note: it is always a good idea to conceptually check what should happen to a variable before solving the equation

Boyle’s Law Example• If the volume of a gas is changed from 0.5L to 1.3L and the initial

pressure was 0.4 atm, what will the final pressure be?• First, do we expect the final pressure to be less than or greater

than the initial pressure?• At the end of the problem, we need to go back and check to see

that our numbers match with what we expect to happen• Next, match the numbers in the problem to the variables in the

equation– PI = 0.4 atm– VI = 0.5L– PF = ?– VF = 1.3L

• Then, write down the equation to be used:– PI x VI = PF x VF

Boyle’s Law Example• If the volume of a gas is changed from 0.5L to 1.3L and the initial

pressure was 0.4 atm, what will the final pressure be?• After we write down the equation to be used, identify the unknown

variable– PI x VI = PF x VF

• Rearrange the equation to solve for the unknown variable– (PF x VF ) ÷ VF = (PI x VI) ÷ VF – PF = (PI x VI) ÷ VF

• Plug in the numbers to get a final answer– PI = 0.4 atm– VI = 0.5L– PF = ?– VF = 1.3L

– PF = (PI x VI) ÷ VF = (0.4 atm x 0.5 L ) ÷ 1.3 L = 0.13 atm

• Does this match with our prediction at the beginning?

Charle’s Law• Relationship between

– Temperature and Volume

• If the temperature is increased, then the molecules will be moving more rapidly and push on the walls of the container

• If the temperature is decreased, then the molecules will move more slowly and not hit the walls as frequently which will decrease the volume

Charle’s Law• The relationship between temperature and

volume is directly proportion– Meaning that when one increases, the other also

increases

• Mathematically, this is expressed as – Volumeinitial ÷ Temperatureinitial = Volumefinal ÷ Volumefinal

– Or VI ÷ TI = VF ÷ TF

• Note: it is always a good idea to conceptually check what should happen to a variable before solving the equation

Charle’s Law Example• If the volume of a gas is changed from 1.5L to 0.9L and the final temperature

was measured at 45oC, what must the initial temperature have been?• First, do we expect the initial temperature to be less than or greater than the final

temperature?• At the end of the problem, we need to go back and check to see that our

numbers match with what we expect to happen• Next, match the numbers in the problem to the variables in the equation

– VI = 1.5 L– TI = ?– VF = 0.9L– TF = 45oC + 273 = 318K– *NOTE: all temperatures must be converted to Kelvin

• Then, write down the equation to be used:– VI ÷ TI = VF ÷ TF

Charle’s Law Example• If the volume of a gas is changed from 1.5L to 0.9L and the final temperature was measured at

45oC, what must the initial temperature have been?

• After we write down the equation to be used, identify the unknown variable– VI ÷ TI = VF ÷ TF

• Rearrange the equation to solve for the unknown variable– (VI ÷ TI ) x TI = (VF ÷ TF ) x TI

– VI ÷ (VF ÷ TF ) =(VF ÷ TF ) x TI ÷ (VF ÷ TF ) – TI = VI ÷ (VF ÷ TF )

• Plug in the numbers to get a final answer– VI = 1.5 L– TI = ?– VF = 0.9L– TF = 45oC + 273 = 318K

– TI = VI ÷ (VF ÷ TF )= 1.5L ÷ (0.9L ÷ 318K)= 530 K • Does this match with our prediction at the beginning?

Gay-Lussac’s Law• Relationship between

– Temperature and Pressure

• If the temperature is increased, then the molecules will be moving more rapidly and push on the walls of the container so the pressure will increase

• If the temperature is decreased, then the molecules will move more slowly and not hit the walls as frequently so the pressure will decrease

Gay-Lussac’s Law• The relationship between temperature and pressure is directly

proportion– Meaning that when one increases, the other also increases

• Mathematically, this is expressed as

– Pressureinitial ÷ Temperatureinitial = Pressurefinal ÷ Volumefinal

– Or PI ÷ TI = PF ÷ TF

• Note: it is always a good idea to conceptually check what should happen to a variable before solving the equation

Gay-Lussac’s Law Example• If the pressure of a gas is changed from 145torr to 450torr and the initial

temperature was measured at 23oC, what must the final temperature be?• First, do we expect the final temperature to be less than or greater than the initial

temperature?• At the end of the problem, we need to go back and check to see that our

numbers match with what we expect to happen• Next, match the numbers in the problem to the variables in the equation

– PI = 145 torr– TI = 23oC + 273 = 296K– PF = 450 torr– TF = ?– *NOTE: all temperatures must be converted to Kelvin

• Then, write down the equation to be used:– PI ÷ TI = PF ÷ TF

Gay-Lussac’s Law Example• If the pressure of a gas is changed from 145torr to 450torr and the initial

temperature was measured at 23oC, what must the final temperature be?• After we write down the equation to be used, identify the unknown variable

– PI ÷ TI = PF ÷ TF

• Rearrange the equation to solve for the unknown variable– (PI ÷ TI) x TF = PF ÷ TF x TF

– (PI ÷ TI) x TF ÷ (PI ÷ TI) = PF ÷ (PI ÷ TI) – TF= PF ÷ (PI ÷ TI)

• Plug in the numbers to get a final answer– PI = 145 torr– TI = 23oC + 273 = 296K– PF = 450 torr– TF = ?– TF= PF ÷ (PI ÷ TI) = 450 torr ÷ (145torr ÷ 296K) = 918

• Does this match with our prediction at the beginning?

Steps For Solving Gas Law Problems:

• Predict whether the variable will increase or decrease

• Label all known and identify unknown– Change all temperatures to Kelvin

• Write the equation to be used• Identify the unknown in the equation• Rearrange equation to solve for unkown• Plug in numbers and solve equation• Check that number matches prediction

Distinguishing Laws• Identify which two variable change, then match that

to the law• Pressure and Volume => Boyle’s Law• Temperature and Volume => Charle’s Law• Temperature and Pressure => Gay-Lussac’s Law

Temperature

Volume Pressure

DEc INC

INC

INC

DEC

DEC

Temperature,Pressure, and

Volume changes

Temp Pressure

Volume

mols