gas laws boyle charles avogadro gay-lussac what happens to the volume of a gas when you increase the...

Gas Laws

BOYLE

CHARLES

AVOGADRO

GAY-LUSSAC

What happens to the volumeof a gas when you increase the pressure? (e.g. Press a syringe that is stoppered)

Consider This

Push!

What happens to the volumeof a gas when you increase the pressure? (e.g. Press a syringe that is stoppered)

Consider This

Push!

What happens to the Volumeof a Gas When you Increase the Pressure? (e.g. Press a syringe that is stoppered)

Consider This

Why?

There is lots of space between gas particles. Therefore, gases are compressible!

Let’s investigate the relationship between pressure and volume if the quantity of gas and temperature are held constant.100 kPa

100 kPa

Volume = 50 L

Pressure (kPa)

Volume (L)

100 50

200 kPa

Volume = 25 L

Pressure (kPa)

Volume (L)

100 50200 25

400 kPa

Volume = 12.5 L

Pressure (kPa)

Volume (L)

100 50200 25400 12.5

800 kPa

Volume = 6.25 L

Pressure (kPa)

Volume (L)

100 50200 25400 12.5800 6.25

What is the mathematicalrelationship between P and V?P x V = constant

5000500050005000

Boyle’s LawIn the 17th Century Robert Boyledescribed this property as, “the spring of air”.

Boyle showed that when temperatureand amount of gas were constant then:

P 1/VOR:

PV = k

Boyle’s LawFor a fixed quantity of gas at a constant temperature, the volume and pressure are inversely proportional.

Boyle showed that when temperatureand amount of gas were constant then:

P 1/VOR:

PV = k

Who Cares?

Scuba Divers!

At sea level air pressure = 100 kPaAt 10 m deep in water pressure = 200 kPaAt 20 m deep = 300 kPaAt 30 m deep = 400 kPa

SCUBA provides air at the same pressure

What will the volume of air in the diver’s lungs be at the surface (100 kPa)?What will happen to the diver?

A Scuba Diver goes to a depth of 90 m and takes a breath of 3 L volume from her tank.

Suddenly! A Dolphin lunges at the diver and takes the SCUBA! The Diver holds her breath and quickly returns to the surface.

A Scuba Diver goes to a depth of 90 m and takes a breath of 3 L volume from her tank.

Assuming that T is constant we can useBoyle’s Law:

PV = k

At 90 m: k = P1V1

At surface: k = P2V2

Therefore! P1V1 = P2V2

(1000 kPa)(3 L) = 100 kPa(V2)V2 = 300 L

Yikes Exploding lungs

Consider This!Two balloons are filled with equalvolumes of air.

What happens to the Volumeof each if one is heated and the other is frozen?

HEATED BALLOON

FROZEN BALLOON

50oC, V=1.18 L

FROZEN BALLOON

HEATED BALLOON

60oC, V=1.22 L 40oC, V=1.14 L

FROZEN BALLOON

HEATED BALLOON

70oC, V=1.25 L 30oC, V=1.11 L

FROZEN BALLOON

HEATED BALLOON

80oC, V=1.29 L 20oC, V=1.07 L

FROZEN BALLOON

HEATED BALLOON90oC, V=1.32 L 10oC, V=1.04 L

FROZEN BALLOON

HEATED BALLOON100oC, V=1.36 L 0oC, V=1.00 L

To study this relationship let’s look at this data in a table.

Graph this datausing temperature as the independent variable

Relationship Between Temperature and Volume for a Fixed Quantity of

Gas at a Constant Pressure

0.000.200.400.600.801.001.201.401.60

-280 -230 -180 -130 -80 -30 20 70

Temperature oC

Vo

lum

e (

L)

Relationship Between Temperature and Volume for a Fixed Quantity of

Gas at a Constant Pressure

0.000.200.400.600.801.001.201.401.60

-280 -230 -180 -130 -80 -30 20 70

Temperature oC

Vo

lum

e (

L)

If this line is extended backwards the volume of 0 L of gas is found to be

-273 oC

-273.15 oC is known as absolute zero. When using gas laws, temperature must be expressed using a temperature scale where 0 is -273.15oC. This is called the Kelvin scale of absolute temperature. 0 K = -273.15 oCWhen changing oC to K simply add 273.15What is 12.3oC in K?12.3 + 273.15 = 285.45= 285.5 K

Now let’s look at this table with temperatures in Kelvin (K).

Can you spot a mathematical relationship between T and V.

V/T in Kelvin is a constant.

0.00370.00370.00370.00370.00370.00370.00370.00370.00370.00370.0037

Charles’ LawIn the 17th Century Jacques Charles examined the relationship between Temperature and Volume

Charles showed that when Pressureand amount of gas were constant then:

V TOR:

V/T = k

Charles’ LawFor a fixed quantity of gas at a constant pressure, absolute temperature and volume are directly proportional.

Charles showed that when Pressureand amount of gas were constant then:

V TOR:

V/T = k

Combined Gas Law

If V1

T1

= V2

T2

V1P1 V2P2 and

V1P1

T1

V2P2

T2

=

=then or

V1P1 T2 =V2P2 T1

What does P2 = ?

V2T1V2T1

RECAP

Charles’ LawAt Constant P and n:

V/T = k

Boyle’s LawAt Constant T and n:

VP = k

Combined Gas LawFor a fixed quantity of gas:

VP/T = k

If 12.5 L of a gas at a pressure of 125 kPa is placed in an elastic container at 15oC what volume would it occupy if the pressure is increased to 145 kPa?

Given:V1 = 12.5 LP1 = 125 kPaT1 = 15oCV2 = ?P2 = 145 kPaT2 = 15oC

V1P1

T1

=V2P2

T2

Since T1 = T2 cancel them to get

V1P1 V2P2=

(12.5 L)(125 kPa) = V2(145 kPa)V2 = (12.5 L)(125 kPa)/145 kPaV2 = 10.8 L

Does this answer make sense?

12.5 L @ 125 kPa

125 kPa

Does this answer make sense?

165 kPa

Does this answer make sense?

165 kPa

Does this answer make sense?

165 kPa

V reduced to10.8 L

Yes, as the pressure increases at constant temperature the volume decreases.

If 15.6 L of a gas at a pressure of 165 kPa is placed in an elastic container at 15oC what volume would it occupy if the temperature is increased to 98oC?

Given:V1 = 15.6 LP1 = 165 kPaT1 = 15oCV2 = ?T2 = 98oCP2 = 165 kPa

V1P1

T1

=V2P2

T2

Since P1 = P2 cancel them to get

V1/T1 V2 / T2=

(15.6 L)/(288 K) = V2/(371 K)V2 = (15.6 L)(371 K) / 288KV2 = 20.1 L

288 K

371 K

If 5.3 L of a gas at a pressure of 75 kPa is placed in an elastic container at 24oC what volume would it occupy if the temperature is increased to 62oC and pressure to 155 kPa?

Given:V1 = 5.3 LP1 = 75 kPaT1 = 24oCV2 = ?T2 = 62oCP2 = 155 kPa

V1P1

T1

=V2P2

T2

(5.3 L)(75 kPa)(335 K)297 K

335 K

OR V1P1T2 =V2P2 T1

(155 kPa)(297K)V2 =

V2= 2.9 L

Ideal Gas LawAn Ideal Gas is a hypothetical gas that obeys all the gas laws perfectly under all conditions.PV = nRTWhere n is the number of moles of gas, P is pressure in kPa, R = 8.313 kPaL/molK and T is temperature in K.Find the mass of helium gas which would be introduced into a 0.95 L container to produce a pressure of 125 kPa at 25oC.

Find the volume occupied by 25 g of chlorine gas at SATP.4.2 g of propane gas is introduced into a 325 mL container at 45oC. What is the pressure of the container.Propane is C3H8.What is the density of NH3 gas at STP if 1.0 mol of this gas occupies 22.4 L.At what temperature does methane gas have a density of 1.2 mg/L if its pressure is 65 kPa.Methane is CH4.

This gas exerts a pressure of 100 kPa inside this container.If another gas is injected into the same container and it exerts a pressure of 70 kPa what is the total pressure in the container?

170 kPa

This gas exerts a pressure of 100 kPa inside this container.If another gas is injected into the same container and it exerts a pressure of 70 kPa what is the total pressure in the container?

170 kPa

This gas exerts a pressure of 100 kPa inside this container.If another gas is injected into the same container and it exerts a pressure of 70 kPa what is the total pressure in the container?

170 kPa

The total pressure of a gas mixture is the sum of the partial pressures of each of the gases in the mixture.ExampleIf a container of air has a pressure of 100 kPa and the % of N2 in the container is 78%, % of O2 is 21%, what are the partial pressures of each of these gases inside this container.

PN2= 78 kPa, PO2= 21 kPa

5.0 L300 K125 kPa

6.0 L, 400 K,155 kPa

1

2

What would the total pressure be if the gas in container 1 was injected into container 2.

5.0 L300 K125 kPa

6.0 L, 400 K,155 kPa

1

2

The total pressure is the sum of the pressures of gas 1 and gas 2.Since gas 1 changed volume and temperature its pressure changed.

5.0 L300 K125 kPa

6.0 L, 400 K,155 kPa

1

2

V1=5.0 L, V2=6.0 L, T1=300 K, T2 = 400 KP1=125 kPa, P2=?V1P1T2=V2P2T1

P2 =5.0 L x 125 kPa x 400 K

6.0 L x 300 K = 139 kPa

5.0 L300 K125 kPa

6.0 L, 400 K,155 kPa

1

2

Total pressure is155 kPa + 139 kPa = 294 kPa = 2.9 x 102 kPa

139 kPa

5.0 L300 K125 kPa

6.0 L, 400 K,155 kPa

1

2

Find the total pressure in container 1 if the gas in container 2 is injected into container 1.

5.0 L300 K125 kPa

6.0 L, 400 K,155 kPa

1

2

V1=6.0 L, V2=5.0 L, T1=400 K, T2 = 300 KP1=155 kPa, P2=?V1P1T2=V2P2T1

P2 =6.0 L x 155 kPa x 300 K

5.0 L x 400 K = 139.5 kPa

5.0 L300 K125 kPa

6.0 L, 400 K,155 kPa

1

2

Total pressure is125 kPa + 139.5 kPa = 264.5 kPa = 2.6 x 102 kPa

139.5 kPa