CHEMISTRY 161 Chapter 11 Gases. Classification of Matter solid liquid gas.

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CHEMISTRY 161 Chapter 11 Gases

Transcript of CHEMISTRY 161 Chapter 11 Gases. Classification of Matter solid liquid gas.

Page 1: CHEMISTRY 161 Chapter 11 Gases. Classification of Matter solid liquid gas.

CHEMISTRY 161

Chapter 11

Gases

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Classification of Matter

solid liquid gas

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1. Gasessubstances that exist in the gaseous phase under

normal atmospheric conditions

T = 25oC p = 1 atm

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HF, HCl, HBr, HI

CO, CO2

CH4, NH3, H2S, PH3

NO, NO2, N2O

SO2

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Jupiter

(H2, He)

Io

(SO2)

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Helix Nebula

Orion Nebula

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2. Pressure

molecules/atoms of gas are constantly in motion

Ar

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Atmospheric Pressure

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Standard Atmospheric Pressure

pressure of the atmosphere is balanced by pressure exerted by mercury

760 mm at 273 K at sea level

1 atm = 760 mm Hg = 760 torrbarometer

Torricelli

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pressure = force area

p = F / A

[p] = Nm-2 = kg m-1 s-2 = Pa

SI units

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manometer

pressure measurement

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3. Gas Laws3.1. pressure p versus volume V

3.2. temperature T versus volume V

3.3. volume V versus amount n

p, V, T, n

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3.1. Boyle’s Law

pressure – volume

relationship

(temperature is constant)

Boyle

(1627-1691)

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p ∞ 1/V

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p ∞ 1/V

p = const/V

p × V = const

p2 × V2 = constp1 × V1 = const

p1 × V1 = p2 × V2

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3.2. Gay-Lussac’s Law

temperature – volume

relationship

(pressure is constant)

Gay-Lussac

(1778-1850)

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V ∞ T

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V ∞ T

V = const’ ×T

V/T = const’

V2 / T2 = constV1 / T1 = const’

V1 / T1 = V2 / T2

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3.3. Avogadro’s Law

amount – volume

relationship

(pressure and temperature

are constant)

Avogadro

(1776-1856)

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2 H2(g) + O2(g) → 2 H2O(l)

n ∞ V

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n ∞ V

n = const’’ × V

n/V = const’’

n2 / V2 = const’’n1 / V1 = const’’

n1 / V1 = n2 / V2

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SUMMARY3.1. Boyle’s Law

3.2. Gay-Lussac’s Law

3.3. Avogadro’s Law

p ∞ 1/V

n ∞ V

V ∞ T

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(1) p ∞ 1/V

p × V = const × n × T

(2) V ∞ T

1. IDEAL GAS EQUATION

(3) n ∞ V

V ∞ 1/p

V ∞ T

V ∞ n

V ∞ T × n / p

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p × V = const × n × T

p × V = R × n × T

p × V = n × R × T

ideal gas equation

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p × V = n × R × T

[R] = [p] × [V] / [n] / [T]

Pa = N/m2 m3 mol K

[R] = N × m / mol / K

[R] = J / mol / K

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R = 8.314 J / mol / K

[R] = J / mol / K

ideal gas constant

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p × V = n × R × T

2. MOLAR VOLUME

What is the volume of 1 mol of a gas at

273.15 K (0oC) and 1 atm (101,325 Pa)?

standard temperature and pressure

(STP)

V = 22.4 l

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p × V = n × R × T

the molar volume at standard pressure and temperature is independent on the gas type

V = 22.4 l

Vm = 22.4 l

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3. STOICHIOMETRY

NaN3(s) → Na(s) + N2(g)

How many liters of nitrogen gas are produced in the decomposition of 60.0 g sodium azide at 80oC and 823 torr?

1. Balancing

2. Mole ratios

3. Convert grams into moles

4. Convert moles into liters

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4. DENSITY CALCULATION

p × V = n × R × T

ς = m / V

relate the moles (n) to the mass (m) via the molecular weight (M)

n = m / M m = n × M

V = n × R × T / p

ς = p × M / (R × T)

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5. DALTON’S LAW

Dalton

(1801)

pure gases

gas mixtures

(atmospheres)

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DALTON’S LAW

the total pressure of a gas mixture, p, is the sum of the

pressures of the individual gases (partial pressures) at a

constant temperature and volume

p = pA + pB + pC + ….

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pA × V = nA × R × T pA = nA × R × T / V

pB × V = nB × R × T

p × V = n × R × T

pB = nB × R × T / V

p = pA + pB

p = (nA + nB) × R × T / V

p × V = n × R × T

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pA = nA × R × T / V

p × V = (nA + nB) × R × T

pA / p = nA /(nA + nB) = xA

mole fraction

x < 1

pA = xA × p

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2 KClO3 → 2 KCl + 3 O2

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SUMMARY

p × V = n × R × T

1. ideal gas equation

R = 8.314 J / mol / K

Vm = 22.4 l

2. molar volume

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ς = p × M / (R × T)

3. Density of gases

4. Dalton’s Law

p = Σ pii=1

n

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1. Kinetic Molecular Theory of Gases

Maxwell

(1831-1879)

Boltzmann

(1844-1906)

macroscopic

(gas cylinder)

microscopic

(atoms/molecules)

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Kinetic Energy of Gasesphysical properties of

gases can be described

by motion of individual

gas atoms/molecules

each macroscopic and

microscopic particle in

motion holds an energy

(kinetic energy)

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Assumptions of the Kinetic Theory of Gases

1. gases are composed of atoms/molecules which are separated from each other by a distance l much more than their

own diameter d

l

d = 10-10 m

l = 10-3 m….. few m

molecules are mass points with negligible volume

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2. gases are constantly in motion in random reactions and hold a kinetic energy

gases collide and transfer energy

(billiard ball model)

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3. gases atoms/molecules

do not exert forces on each other

(absence of intermolecular interactions)

F(inter) = 0

p(inter) = 0

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Gas Diffusion

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2. Distribution of Molecular Speeds

Maxwell-Boltzmann distribution

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3. Real Gases

p × V = n × R × T (n = 1)

deviation of ideal gas law at high pressures

p ≈ 90 atm

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p << 10-10 atm

North America Nebula

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ideal gas law

p V = n R T

real gas law

(van der Waals equation)

(p + (a n2 / V2) ) (V – n b) = n R T

corrected volume

(volume occupied by molecules)

corrected pressure

(additional pressure/force from attraction)