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GasesGases

Chapter 4Chapter 4

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KKiinneettiicc MMoolleeccuullaarr TThheeoorryy

Particles in an Ideal Gases… have no volume. have elastic collisions. are in constant, random, straight-line

motion. don’t attract or repel each other. have an avg. KE directly related to

Kelvin temperature.

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RReeaall GGaasseess

Particles in a REAL GAS… have their own volume attract each other

Gas behavior is most ideal…at low pressures

at high temperatures

in nonpolar atoms/molecules

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CChhaarraacctteerriissttiiccss ooff GGaasseess

- Expand to fill any container. Random motion, No attraction

- Are fluids (like liquids). No attraction

- Have very low densities. No volume = lots of empty space

- Can be compressed.

- Undergo diffusion & effusion. Random motion

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AA.. TTeemmppeerraattuurree

ºF

ºC

K

-459 32 212

-273 0 100

0 273 373

32FC 95 K = ºC + 273

Always use absolute temperature (Kelvin) when working with gases.

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BB.. PPrreessssuurree

area

forcepressure

Which shoes create the most pressure?

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BB.. PPrreessssuurree ((CCoonntt..))

Barometermeasures atmospheric pressure

Mercury Barometer

Aneroid Barometer

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BB.. PPrreessssuurree ((CCoonntt..))

Manometermeasures contained gas pressure

U-tube Manometer Bourdon-tube gauge

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BB.. PPrreessssuurree ((CCoonntt..))

2m

NkPa

KEY UNITS AT SEA LEVEL

101.325 kPa (kilopascal)

1 atm

760 mm Hg

760 torr

14.7 psi

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FF.. SSTTPP

Standard Temperature & PressureStandard Temperature & Pressure

0°C 273 K

1 atm 101.325 kPa-OR-

STP

? The Gas LawsA. Boyle’s Law B. Charle’s LawC. Combined Gas LawD. Avogadro’s principleE. Ideal Gas LawF. Dalton’s Law

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AA.. BBooyyllee’’ss LLaaww

P

V

Pα 1/V or

PV = k

The pressure and volume of a gas are inversely related at constant mass & temperature

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V

T

BB.. CChhaarrlleess’’ LLaaww

Vα T or

kT

V

The volume and absolute temperature (K) of a gas are directly related at constant mass & pressure

? V α T

CC.. CCoommbbiinneedd GGaass LLaaww

P1V1

T1

=P2V2

T2

P1V1T2 = P2V2T1

1P

V α &

?GIVEN:

V1 = 473 cm3

T1 = 36°C = 309K

V2 = ?

T2 = 94°C = 367K

WORK:

P1V1T2 = P2V2T1

GGaass LLaaww PPrroobblleemmss

A gas occupies 473 cm3 at 36°C. Find its volume at 94°C.

CHARLES’ LAW

T V

(473 cm3)(367 K)=V2(309 K)

V2 = 562 cm3

?GIVEN:

V1 = 100. mL

P1 = 150. kPa

V2 = ?

P2 = 200. kPa

WORK:

P1V1T2 = P2V2T1

GGaass LLaaww PPrroobblleemmss

A gas occupies 100. mL at 150. kPa. Find its volume at 200. kPa.

BOYLE’S LAW

P V

(150.kPa)(100.mL)=(200.kPa)V2

V2 = 75.0 mL

?GIVEN:

V1 = 7.84 cm3

P1 = 71.8 kPa

T1 = 25°C = 298 K

V2 = ?

P2 = 101.325 kPa

T2 = 273 K

WORK:

P1V1T2 = P2V2T1

(71.8 kPa)(7.84 cm3)(273 K)

=(101.325 kPa) V2 (298 K)

V2 = 5.09 cm3

GGaass LLaaww PPrroobblleemmss

A gas occupies 7.84 cm3 at 71.8 kPa & 25°C. Find its volume at STP.

P T VCOMBINED GAS LAW

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kn

VV

n

DD.. AAvvooggaaddrroo’’ss PPrriinncciippllee

Equal volumes of gases contain equal numbers of moles at constant temp & pressure true for any gas.

Vα n or

? PV

T

PVnT

EE.. IIddeeaall GGaass LLaaww

= k

UNIVERSAL GAS CONSTANT

R=0.0821 Latm/molKR=8.315 dm3kPa/molK

= R

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EE.. IIddeeaall GGaass LLaaww

UNIVERSAL GAS CONSTANTR=0.0821 Latm/molKR=8.315 dm3kPa/molK

PV=nRT

?GIVEN:

P = ? atm

n = 0.412 mol

T = 16°C = 289 K

V = 3.25 LR = 0.0821Latm/molK

WORK:

PV = nRT

P(3.25)=(0.412)(0.0821)(289) L mol Latm/molK K

P = 3.01 atm

EE.. IIddeeaall GGaass LLaaww

Calculate the pressure in atmospheres of 0.412 mol of He at 16°C & occupying 3.25 L.

IDEAL GAS LAW

?GIVEN:

V = ?

n = 85 g

T = 25°C = 298 K

P = 104.5 kPaR = 8.315 dm3kPa/molK

EE.. IIddeeaall GGaass LLaaww

Find the volume of 85 g of O2 at 25°C and 104.5 kPa.

= 2.7 mol

WORK:

85 g 1 mol = 2.7 mol

32.00 g

PV = nRT(104.5)V=(2.7) (8.315) (298) kPa mol dm3kPa/molK K

V = 64 dm3

IDEAL GAS LAW

? IV. Gas Stoichiometry at Non-STP Conditions

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GGaass SSttooiicchhiioommeettrryy

Moles Moles Liters of a Gas Liters of a GasSTP - use 22.4 L/mol Non-STP - use ideal gas law

Non-Non-STP ProblemsSTP ProblemsGiven liters of gas?

i. start with ideal gas law

Looking for liters of gas?i. start with stoichiometry conv.

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FF.. DDaallttoonn’’ss LLaaww

The total pressure of a mixture of gases equals the sum of the partial pressures of the individual gases.

Ptotal = P1 + P2 + ...

Patm = PH2 + PH2O

?GIVEN:

PH2 = ?

Ptotal = 94.4 kPa

PH2O = 2.72 kPa

WORK:

Ptotal = PH2 + PH2O

94.4 kPa = PH2 + 2.72 kPa

PH2 = 91.7 kPa

FF.. DDaallttoonn’’ss LLaaww

Hydrogen gas is collected over water at 22.5°C. Find the pressure of the dry gas if the atmospheric pressure is 94.4 kPa.

Look up water-vapor pressure on p.899 for 22.5°C.

Sig Figs: Round to least number of decimal places.

The total pressure in the collection bottle is equal to atmospheric pressure and is a mixture of H2 and water vapor.

?GIVEN:

Pgas = ?

Ptotal = 742.0 torr

PH2O = 42.2 torr

WORK:

Ptotal = Pgas + PH2O

742.0 torr = PH2 + 42.2 torr

Pgas = 699.8 torr

A gas is collected over water at a temp of 35.0°C when the barometric pressure is 742.0 torr. What is the partial pressure of the dry gas?

DALTON’S LAW

Look up water-vapor pressure on p.899 for 35.0°C.

Sig Figs: Round to least number of decimal places.

FF.. DDaallttoonn’’ss LLaaww

The total pressure in the collection bottle is equal to barometric pressure and is a mixture of the “gas” and water vapor.

?Deviations from Ideal Gas Law

Real gas: Deviate (Not Obey Gas law

PV ≠ Constant

Repulsive Forces

Attractive Forces

When:

1- T is very low and P is very high.

This deviation is due to 2 factors which were ignored by kinetic theory:

1- Real gases posses attractive forces between molecules.

2- Every molecule in a real gas has a real volume.

?There are 2 correction factors should be taken into consideration:

I- Volume of gas molecules is not negligible:

True volume (Real) =

V container – non compressible volume (b)

= V – b (1 mole)

Actual volume = V – nb (for n moles)

II- There is an attraction force between the molecules of gases (pressure): Actual Pressure = Measured pressure + Pressure due to attraction force.

= P + (1 mole)

Actual pressure = P + (for n moles)

a

V2 an2

V2

?Van der Waals Equation

By introducing actual volume and actual pressure:

The last equation is known as “Van der Waal’s equation” where a,b are Van der Waal’s constants depend on the nature of the gas.

a

V2

For ideal gases: PV = nRTBut in case of Real gas: PV ≠ nRT

(P + ) (V – b) = RT 1 mole

(P + an2

V2

)(V – b) = nRT n mole

According to Van der Waal’s, the pressure of real gas will be lower than that of ideal gas because attraction of neighbouring molecules tends to decrease the impact of a real molecule that it make with the wall of the container, this can be expressed as: PReal < PIdeal (Same rule but replace +ve sign with –ve sign)