Chapter 1Chapter 1 Gases Gases

Chapter 1Chapter 1 Gases Gases

Dr. Hisham E AbdellatefDr. Hisham E Abdellatef

Professor of pharmaceutical Professor of pharmaceutical analytical chemistryanalytical chemistry

2011 - 20122011 - 2012

Characteristics of Gases

• Expand to fill and assume the shape of their container

• CompressibleNext

• Diffuse into one another and mix in all proportions. homogeneous mixtures

• Particles move from an area of high concentration to an area of low concentration.

• Next

PropertiesPropertiesPropertiesPropertiesthat determine physical behavior of a that determine physical behavior of a

gasgas

1. VOLUME1. VOLUME

2. PRESSURE2. PRESSURE

3. TEMPERATURE3. TEMPERATURE

1. Volume l x w x h πr²h

capacity of the container enclosing it. (m3), (dm3), or liter. For smaller volumes (cm3), (ml).

2. Temperature• Temperature: Three temperature

scales• Fahrenheit (ºF)• Celsius (ºC)• Kelvin K (no degree symbol) • K performing calculations with the gas

law equations.

gas

• Expand when heated

Temperature

• K = (ºC) + 273.15• Example: ºC = 20º • K = 293.15

• Absolute or Kelvin Scale: =-273.15 (ºC as its zero). (T when V= 0 )

• 1 K = 1 ºC

3. Pressure

The molecules in a gas are in constant motion. gaseous atoms that collide with each other and the walls of the container.

"Pressure" is a measure of the collisions of the atoms with the container.

Pressure

• Force per unit area

• Equation: P = F/A

F = force A = area Next

Barometer- Atmospheric pressure

is measured with a barometer.

•Height of mercury varies with atmospheric conditions and with altitude.

Mercury Barometer

Measurement of Gas Pressure

- Standard atmospheric pressure is the pressure required to support 760 mm of Hg in a column.

- There are several units used for pressure:

- Pascal (Pa), N/m2

- Millimeters of Mercury (mmHg)- Atmospheres (atm)

Manometers

• Used to compare the gas pressure with the barometric pressure.

Next

Types of Manometers

• Closed-end manometer The gas pressure is equal to

the difference in height (Dh) of the mercury column in the two arms of the manometer

Open-end Manometer

The difference in mercury levels (Dh) between the two arms of the manometer gives the difference between barometric pressure and the gas pressure

Three Possible Relationships

1. Heights of mercury in both columns are equal if gas pressure and atmospheric pressure are equal.

Pgas = Pbar

2. Gas pressure is greater than the barometric pressure.

∆P > 0

Pgas = Pbar + ∆P

3. Gas pressure is less than the barometric pressure.

∆P < 0

Pgas = Pbar + ∆P

The Simple Gas LawsThe Simple Gas LawsThe Simple Gas LawsThe Simple Gas Laws

1. Boyle’s Law1. Boyle’s Law2. Charles’ Law2. Charles’ Law

3. Gay-Lussac’s Law3. Gay-Lussac’s Law4. Combined Gas Law4. Combined Gas Law

variables required to

describe a gas• Amount of substance: moles• Volume of substance: volume• Pressures of substance: pressure• Temperature of substance:

temperature

25

The Pressure-Volume Relationship: Boyle’s Law

Boyle’s Law - The volume of a fixed quantity of gas is inversely proportional to its pressure.

2211

) and (constant 1

VPVP

TnV

P

The Pressure-Volume Relationship: Boyle’s Law

28

Example Example

An ideal gas is enclosed in a Boyle's-law apparatus. Its volume is 247 ml at a pressure of 625 mmHg. If the pressure is increased to 825 mmHg, what will be the new volume occupied by the gas if the temperature is held constant?

SolutionSolution

Method 1: P1V1 = P2V2

or, solving for V2 the final volume

2p

1V

1P

2V

ml 187 mmHg 825

ml mmHgX247 6252

V

SolutionSolution

Method 2: The pressure of the gas increases by a factor 825/625, the volume must decrease by a factor of 625/825

V2= V1 X (ratio of pressures)

ml 187 mmHg 825

mmHg 625 x 247

2V

Charles’ Law

The volume of a fixed amount of gas at constant pressure is directly proportional to the Kelvin (absolute) temperature.

V1 = V2

T1 T2

or V1T2 = V2T1

Example . Charles’ Law

A 4.50-L sample of gas is warmed at constant pressure from 300 K to

350 K. What will its final volume be?

Given: V1 = 4.50 L T1 = 300. K T2 = 350. K V2 = ?

Equation: V1 = V2

T1 T2

or V1T2 = V2T1

(4.50 L)(350. K) = V2 (300. K)

V2 = 5.25 L

Gay-Lussac’s Law

The pressure of a sample of gas is directly proportional to the absolute temperature when volume remains constant.

P1 = P2

T1 T2

or P1T2 = P2T1

On the next slide

The amount of gas and its volume are the same in either case, but if the gas in the ice bath (0 ºC) exerts a pressure of 1 atm, the gas in the boiling-water bath (100 ºC) exerts a pressure of 1.37 atm. The frequency and the force of the molecular collisions with the container walls are greater at the higher temperature.

Combined Gas Law

Pressure and volume are inversely proportional to each other and directly proportional to temperature.

P1V1 = P2V2

T1 T2

or P1V1T2 = P2V2T1

Example. Combined Gas Law

A sample of gas is pumped from a 12.0 L vessel at

27ºC and 760 Torr pressure to a 3.5-L vessel at 52ºC. What is the final pressure?

Given:P1 = 760 Torr P2 = ?V1 = 12.0 L V2 = 3.5 LT1 = 300 K T2 = 325 K

Equation:

P1V1 = P2V2 T1 T2

or P1V1T2 = P2V2T1

(760 Torr)(12.0 L)(325 K) = ( P2)(3.5 L)(300 K)

P2 = 2.8 x 10³ Torr

Avogadro’s LawAvogadro’s LawAvogadro’s LawAvogadro’s Law

Volume & MolesVolume & Moles

Avogadro’s LawAt a fixed temperature and pressure, the

volume of a gas is directly proportional to the amount of gas.

V = c · nV = volume c = constant n= # of moles

Doubling the number of moles will cause the volume to double if T and P are constant.

Ideal Gas LawIdeal Gas LawIdeal Gas LawIdeal Gas Law

Equation

Includes all four gas variables:• Volume• Pressure• Temperature• Amount of gas Next

PV = nRT• Gas that obeys this equation if said to be an ideal gas (or perfect gas).

• No gas exactly follows the ideal gas law, although many gases come very close at low pressure and/or high temperatures.

• Ideal gas constant, R, is R = PV nT = 1 atm x 22.4 L 1 mol x 273.15 K

R = 0.082058 L·atm/mol· K

Example

• Suppose 0.176 mol of an ideal gas occupies 8.64 liters at a pressure of 0.432 atm. What is the temperature of the gas in degrees Celsius?

SolutionPV = nRT

 

To degrees Celsius we need only subtract 273 from the above result:

=258 - 273 = -15OC

K 258 )molK atmliter 1mol)(0.082 (0.176

liters) atm)(8.64 (0.432

1-1-

Example

• Suppose 5.00 g of oxygen gas, O2, at 35 °C is enclosed in a container having a capacity of 6.00 liters. Assuming ideal-gas behavior, calculate the pressure of the oxygen in millimeters of mercury. (Atomic weight: 0 = 16.0)

Solution• One mole of O2 weighs

2(16.0) = 32.0 g. 5.00 g of O2 is, therefore, 5.00 g/32.0 g mol-1, or 0.156 mol. 35 °C is 35 + 273 = 308 K

• PV = nRT atm 0.659 liters 6.00

K) 308 1)(-mol I-K atmliter 1mol)(0.082 (0.156p

Hg mm 500 atm 1

mmHg 760 659.0 atm

Molar volume of an ideal gas at STP

• The volume occupied by one mole, or molar volume, of an ideal gas at STP is

P

nRTV

liters 22.414

atm 1.0000

K) 273.15 )(molK atmliter 057mol)(0.082 (1.0000

-1 -I

Applications of the Applications of the Ideal Gas LawIdeal Gas Law

Applications of the Applications of the Ideal Gas LawIdeal Gas Law

Dalton’s Law ofDalton’s Law of Partial Pressure Partial PressureDalton’s Law ofDalton’s Law of Partial Pressure Partial Pressure

Mixture of GasesMixture of Gases

Total Pressure

The total pressure of a mixture of gases is the sum of the partial pressures of the components of the mixture.

Ptotal = PA + PB + ……

Total Pressure: Mixture of Gases

Example: Gas Mixtures & Partial Pressure

A gaseous mixture made from 6.00 g O2 and 9.00 g CH4 is placed in a 15.0 L vessel at 0ºC. What is the partial pressure of each gas, and what is the total pressure in the vessel?

Step 1: nO2 = 6.00 g O2 x 1 mol O2

32 g O2

= 0.188 mol O2

Next ---- >

nCH4 = 9.00 g CH4 x 1 mol CH4

16.0 g CH4

= 0.563 mol CH4

Step 2: Calculate pressure exerted by each

PO2 = nRT V = (0.188 mol O2)(0.0821 L-atm/mol-K)(273 K) 15.0 L = 0.281 atm

PCH4 = (0.563 mol)(0.0821 L-atm/mol-K)(273 K) 15.0 L = 0.841 atm

Step 3: Add pressures

Ptotal = PO2 + PCH4

= 0.281 atm + 0.841 atm

Ptotal = 1.122 atm

Try ?

• Q1, Q2, Q3 page 42(lecturer note )

Graham’s LawGraham’s LawGraham’s LawGraham’s Law

Molecular Effusion and Diffusion

Molecular Effusion and DiffusionMolecular Effusion and DiffusionGraham’s Law of Effusion

Effusion – The escape of gas through a small opening.

Diffusion – The spreading of one substance through another.

Molecular Effusion and DiffusionMolecular Effusion and Diffusion

Graham’s Law of Effusion – The rate of effusion of a gas is inversely proportional to the square root of its molecular mass.

Graham’s Law of Effusion

Molecular Effusion and DiffusionMolecular Effusion and Diffusion

Graham’s Law of Effusion – The rate of effusion of a gas is inversely proportional to the square root of its molecular mass.

Graham’s Law of Effusion

1

2

2

1MM

rr

Molecular Effusion and DiffusionMolecular Effusion and Diffusion

Graham’s Law of Effusion – The rate of effusion of a gas is inversely proportional to the square root of its molecular mass.

• Gas escaping from a balloon is a good example.

Graham’s Law of Effusion

1

2

1

2

2

1

dd

MM

r

r

Chapter 10 63

Molecular Effusion and DiffusionMolecular Effusion and DiffusionGraham’s Law of Effusion

1

2

1

2

2

1

dd

MM

r

r

Example 1. Rate of Effusion

• The rate of effusion of an unknown gas (X) through a pinhole is found to be only 0.279 times the rate of effusion of hydrogen (H2) gas through the same pinhole, if both gases are at STP. What is the molecular weight of the unknown gas?

• (Atomic weight; H = 1.01.) Next

• Solution

Mx= 26.0

• Pv = nRt n=m mass M dPv= m RT PM = m RT M V PM = dRT or d = PM RT

Example2 . Rate of Effusion

Calculate the ratio of the effusion rates of hydrogen gas (H2) and uranium hexafluoride (UF6), a gas used in the enrichment process fuel for nuclear reactors.

Known:Molar Masses H2 = 2.016 g/mol UF6 = 352.02 g/mol

(Rate of effusion)² = MU compound

MH gas

= 352.02 2.016

Rate of effusion = 13.21

Quize • 1. what is the relative rates of

diffusion of H2 and CO2 under the same condition ?

• 2. What is the density of gas which it’s diffusion is 1.414 times of the rate of diffusion of CO2 at STP ???.

ProblemCalcium hydride, CaH2, reacts with water to form hydrogen gas:CaH2(s) + 2 H2O(l) Ca(OH)2(aq) + 2 H2(g)

How many grams of CaH2 are needed to generate 10.0L of H2 gas if the pressure of H2 is 740 torr at 23oC?

Problem

CaH2(s) + 2 H2O(l) Ca(OH)2(aq) + 2 H2(g)

-Calculate moles of H2 formed

-Calculate moles of CaH2 needed

-Convert moles CaH2 to grams

ProblemCaH2(s) + 2 H2O(l) Ca(OH)2(aq) + 2 H2(g)

)(/)(08206.0

29627323

0.10

974.0760/740

KmolatmLr

KCT

LV

torrP

o

Problem

CaH2(s) + 2 H2O(l) Ca(OH)2(aq) + 2 H2(g)

)296(08206.0

)0.10(974.0

)(/)(08206.0

29627323

0.10

974.0760/740

2 K

Latmn

KmolatmLr

KCT

LV

torrP

H

o

ProblemCaH2(s) + 2 H2O(l) Ca(OH)2(aq) + 2 H2(g)

moln

K

Latmn

KmolatmLr

KCT

LV

torrP

H

H

o

401.0

)296(08206.0

)0.10(974.0

)(/)(08206.0

29627323

0.10

974.0760/740

2

2

ProblemCaH2(s) + 2 H2O(l) Ca(OH)2(aq) + 2 H2(g)

mol

x

H

CaH

moln

K

Latmn

KmolatmLr

KCT

LV

torrP

H

H

o

401.02

1

401.0

)296(08206.0

)0.10(974.0

)(/)(08206.0

29627323

0.10

974.0760/740

2

2

2

2

ProblemCaH2(s) + 2 H2O(l) Ca(OH)2(aq) + 2 H2(g)

molx

mol

x

H

CaH

moln

K

Latmn

KmolatmLr

KCT

LV

torrP

H

H

o

2005.0

401.02

1

401.0

)296(08206.0

)0.10(974.0

)(/)(08206.0

29627323

0.10

974.0760/740

2

2

2

2

ProblemCaH2(s) + 2 H2O(l) Ca(OH)2(aq) + 2 H2(g)

)/10.42(2005.0

2005.0

401.02

1

401.0

)296(08206.0

)0.10(974.0

)(/)(08206.0

29627323

0.10

974.0760/740

2

2

2

2

2

molgmolgCaH

molx

mol

x

H

CaH

moln

K

Latmn

KmolatmLr

KCT

LV

torrP

H

H

o

ProblemCaH2(s) + 2 H2O(l) Ca(OH)2(aq) + 2 H2(g)

g

molgmolgCaH

molx

mol

x

H

CaH

moln

K

Latmn

KmolatmLr

KCT

LV

torrP

H

H

o

44.8

)/10.42(2005.0

2005.0

401.02

1

401.0

)296(08206.0

)0.10(974.0

)(/)(08206.0

29627323

0.10

974.0760/740

2

2

2

2

2

Kinetic-Molecular Kinetic-Molecular Theory of GasesTheory of Gases

Kinetic-Molecular Kinetic-Molecular Theory of GasesTheory of Gases

Kinetic-Molecular TheoryKinetic-Molecular Theory- Theory developed to explain gas behavior- To describe the behavior of a gas, we must first

describe what a gas is:– Gases consist of a large number of molecules in constant

random motion.– Volume of individual molecules negligible compared to

volume of container.– Intermolecular forces (forces between gas molecules)

negligible.– Energy can be transferred between molecules, but total

kinetic energy is constant at constant temperature.– Average kinetic energy of molecules is proportional to

temperature.

Nonideal (Real) GasesNonideal (Real) GasesNonideal (Real) GasesNonideal (Real) Gases

Gases may be1.  Ideal gas on obeys the gas law

PV = constant (at constant T)

2. Real gas Deviate (not obey to gas law PV ≠ Constant)

• Real gas deviate from ideal gases when (T is very low and P is very high).

• two factors :• Real gases posse’s attractive forces

between molecules.• Every molecule in a real has a real

volume.

True volume (Real) =• V container – non copressiable

volume (b) = V- b for 1 moleActual volume = V – nb ( for n

mole)

force of attraction between the molecules

of gases: (pressure)

2V

ap

2

2

Van

p

1 mole

n moles

van der Waals van der Waals EquationEquation

van der Waals van der Waals EquationEquation

van der Waals EquationEquation corrects for volume and intermolecular forces

(P + n²a/V²)(V-nb) = nRT• n²a/V² = related to intermolecular forces of attraction

• n²a/V² is added to P = measured pressure is lower than expected

• a & b have specific values for particular gases

• V - nb = free volume within the gas

EXAMPLE Calculate the pressure exerted by 10.0 g of methane, CH4, when enclosed in a 1.00-liter container at 25 °C by using (a) the ideal-gas law and (b) the van der Waals equation.

Gas a (L2.atm/mol2) b (L/mol)

CO2 3.658 0.04286

Ethane C2H6 5.570 0.06499

Methane CH4 2.25 0.0428

Helium He 0.0346 0.0238

Hydrogen H2 0.2453 0.02651

Oxygen O2 1.382 0.03186

Sulfur dioxide SO2 6.865 0.05679

The molecular weight of CH4 is 16.0; so n, the number of moles of methane, is 10.0 g/16.0 g mol-1, or 0.625 mol.

(a) Considering the gas to be ideal and solving for P, we obtain

(b) Treating the gas as a Van der Waals gas and solving for P, we have

 

 

 

 

atm 15.3liter 1.00

)(298K)molK atmliter 1mol)(0.082 (0.625

V

nRT P

1-1

2

2

V

an

nbV

nRTP

atm8.41

liter) (1.00

)mol atmliters (2.25mol) (0.625

)molliter 0.248 ( mol) (0.625 -liter 1.00

)(298K)molK atmliter 1mol)(0.082 (0.625 P

2

222

1-

1-1

Try ?

• Q1, Q2, Q3 page 50 (lecturer note )