Ratio Method of SolvingBasic Gas Laws

Boyle’s Law

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Robert Boyle

Boyle’s Law

Demonstrations that illustrate Boyle’s Law• Marshmallow or balloon in the bell jar (marshmallow or balloon

gets big when pressure gets low)

Boyle’s Law

Volume is inversely related to PressureIf volume decreases, pressure increasesIf volume increases, pressure decreases

(they move in opposite directions)

You can see this relationship when temperature and moles of gas are held constant

Boyle’s Law

Mathematically, PV = kWhere “k” is a constant

Boyle’s Law

Another way to write Boyle’s law is

P1V1 = P2V2

Temperature is constant!! If it is not, then you can not use Boyle’s Law

Boyle’s Law Problem #1

A 0.030L marshmallow undergoes a drop in pressure from 1.1 atm to 0.20 atm. What will be

the new volume?Start with a “data table” to organize what you know and what you don’t know

• V1 = 0.030L

• V2 = ?

• P1 = 1.1 atm

• P2 = 0.20 atm

Boyle’s Law Problem #1

A 0.030L marshmallow undergoes a drop in pressure from 1.1 atm to 0.20 atm. What will be

the new volume?

• V1 = 0.030L

• V2 = ?

• P1 = 1.1 atm

• P2 = 0.20 atm

V2 = 0.030L x 1.1 atm

0.20 atm= 0.17 L

Bigger number on top

makes Volume get larger

Pressuregot smallerso Volumemust getlarger

Boyle’s Law Problem #2

A 0.025L marshmallow at 0.60 atm experiences a pressure increase to 0.95 atm. What will be the

new volume?

Start with your data table

• V1 = 0.025L

• V2 = ?

• P1 = 0.60 atm

• P2 = 0.95 atm

Boyle’s Law Problem #2

A 0.025L marshmallow at 0.60 atm experiences a pressure increase to 0.95 atm. What will be the

new volume?

• V1 = 0.025L

• V2 = ?

• P1 = 0.60 atm

• P2 = 0.95 atm

V2 = 0.025L x 0.60 atm

0.95 atm= 0.016 L

Smaller number on top

Makes volume get smaller

Pressuregot larger, so Volumemust get smaller

Charles’ Law

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Charles’ Law

Demonstrations that illustrate Charles’ Law• Balloon on flask (heat the flask then the balloon expands)

Charles’ Law

Volume is directly related to TemperatureIf temperature increases, volume increases

If temperature decreases, volume decreases(they move in the same direction)

You can see this relationship when pressure and moles of gas are held constant

Charles’ Law

V = kTWhere “k” is a constant of proportionalityNote that V does NOT have to equal T, but they are

directly proportional to each other

Charles’ Law

Another way to write this is:(V1/ T1) = (V2/T2)

OrV1T2 = V2T1

Pressure must remain constant or you can not use this law!

Charles’ Law Problem #1

A 2.0L sample of air is collected at 298K then cooled to 278K. What will be the new volume?

Start with a data table to organize what you know and what you don’t know

• V1 = 2.0 L

• V2 = ?

• T1 = 298 K

• T2 = 278 K

Charles’ Law Problem #1

A 2.0L sample of air is collected at 298K then cooled to 278K. What will be the new volume?

• V1 = 2.0 L

• V2 = ?

• T1 = 298 K

• T2 = 278 K

V2 = 2.0 L x 278 K

298 K= 1.9 L

Smaller number on top

makes volume get smaller

Temperaturedecreasedso volumemustdecrease

Charles’ Law Problem #2

A 3.25L balloon at 298K changes volume to 2.50L. What temperature would cause this to happen?

Start with your data table

• V1 = 3.25 L

• V2 = 2.50 L

• T1 = 298 K

• T2 = ?

Charles’ Law Problem #2

A 3.25L balloon at 298K changes volume to 2.50L. What temperature would cause this to happen?

• V1 = 3.25 L

• V2 = 2.50 L

• T1 = 298 K

• T2 = ?

T2 = 298 K x 2.50 L

3.25 L= 229 K

Smaller number on top

makes temperature sm

aller

Volume gotsmaller so

temperaturemust getsmaller

Gay-Lussac Law

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we may continue.

Gay-Lussac Law

Demonstrations that illustrate Gay-Lussac’s Law

Aerosol can in the campfire (As temperature increases, pressure increases until the can explodes)

Gay-Lussac Law

Pressure is directly related to TemperatureIf temperature increases, pressure increases

If temperature decreases, pressure decreases(they move in the same direction)

You can see this relationship when volume and moles of gas are held constant

Gay-Lussac Law

P = kTWhere “k” is a constant of proportionality

Gay-Lussac Law

Another way to write it:(P1 / T1) = (P2 / T2)

OrT1P2 = P1T2

Volume stays constant!

Gay-Lussac Law Problem #1

A can of hairspray contains a gaseous propellant at 1.50 atm and 298K. What is the new pressure if

the can is heated to 500.K?

Start with a data table

• T1 = 298K

• T2 = 500.K

• P1 = 1.50 atm

• P2 = ?

Gay-Lussac Law Problem #1

A can of hairspray contains a gaseous propellant at 1.50 atm and 298K. What is the new pressure if

the can is heated to 500.K?

• T1 = 298K

• T2 = 500.K

• P1 = 1.50 atm

• P2 = ?

P2 = 1.50 atm x 500 K

298 K= 2.52 atm

Big number on top makes

pressure get bigger

Temperature

increased so

pressuremustincrease

Gay-Lussac Law Problem #2

A can of spraypaint at 298K experiences an increase in pressure from 101.3 kPa to 425 kPa. What temperature would cause such a change?

Start with a data table

• T1 = 298 K

• T2 = ?

• P1 = 101.3 kPa

• P2 = 425 kPa

Gay-Lussac Law Problem #2

A can of spraypaint at 298K experiences an increase in pressure from 101.3 kPa to 425 kPa. What temperature would cause such a change?

• T1 = 298 K

• T2 = ?

• P1 = 101.3 kPa

• P2 = 425 kPa

T2 = 298 K x 425 kPa

101.3 kPa= 1250 K

Big number on top makes

Temperature get biggerPressureincreasedsotemperaturemustincrease

Avogadro’s Law

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Avogadro’s Law

DemonstrationBlowing up a balloon

As you add moles of gas, the volume increases

Avogadro’s Law

Volume is directly related to MolesIf moles of gas increases, volume increases

If moles of gas decreases, volume decreases(they move in the same direction)

You can see this relationship when temperature and pressure of gas are held constant

Avogadro’s Law

V = knWhere “k” is a constant of proportionality

Avogadro’s Law

Another way to write this:V1/n1 = V2/n2

Or V1n2 = V2n1

Avogadro’s Law Problem #1

A 12.2 L sample of gas contains 0.500 mol of O2 at 1.00 atm and 25.0oC. If 3.00 mol of O2 are

added, what will be the new volume?

Start with a data table

• V1 = 12.2 L

• V2 = ?

• n1 = 0.500 mol

• n2 = 3.500 mol

Avogadro’s Law Problem #1

A 12.2 L sample of gas contains 0.500 mol of O2 at 1.00 atm and 25.0oC. If 3.00 mol of O2 are

added, what will be the new volume?

• V1 = 12.2 L

• V2 = ?

• n1 = 0.500 mol

• n2 = 3.500 mol

V2 = 12.2 L x 3.500 mol

0.500 mol= 85.4 L

Bigger number on top

Makes volume get biggerMolesincreasedso volumemustincrease

Avogadro’s Law Problem #2

A 2.25 L balloon contains 0.475 mol of helium. To increase the volume of the balloon to 6.50 L, how many moles of helium must it contain?

Start with a data table

• V1 = 2.25 L

• V2 = 6.50 L

• n1 = 0.475 mol

• n2 = ?

Avogadro’s Law Problem #2

A 2.25 L balloon contains 0.475 mol of helium. To increase the volume of the balloon to 6.50 L, how many moles of helium must it contain?

• V1 = 2.25 L

• V2 = 6.50 L

• n1 = 0.475 mol

• n2 = ?

n2 = 0.475 mol x 6.50 L

2.25 L= 1.37 mol

Bigger number on top makes

Moles get bigger

Volumeincreased somoles must

increase