CHEMISTRY 161 Chapter 5. Classification of Matter solid liquid gas.
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Transcript of CHEMISTRY 161 Chapter 5. Classification of Matter solid liquid gas.
CHEMISTRY 161
Chapter 5
Classification of Matter
solid liquid gas
1. Gasessubstances that exist in the gaseous phase under
normal atmospheric conditions
T = 25oC p = 1 atm
HF, HCl, HBr, HI
CO, CO2
CH4, NH3, H2S, PH3
NO, NO2, N2O
SO2
Jupiter
(H2, He)
Io
(SO2)
Helix Nebula
Orion Nebula
2. Pressure
molecules/atoms of gas are constantly in motion
Ar
EXP I
Atmospheric Pressure
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
pressure = force area
p = F / A
[p] = Nm-2 = kg m-1 s-2 = Pa
SI units
manometer
pressure measurement
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
3.1. Boyle’s Law
pressure – volume
relationship
(temperature is constant)
Boyle
(1627-1691)
p ∞ 1/V EXP II
p ∞ 1/V
p = const/V
p × V = const
p2 × V2 = constp1 × V1 = const
p1 × V1 = p2 × V2
3.2. Gay-Lussac’s Law
temperature – volume
relationship
(pressure is constant)
Gay-Lussac
(1778-1850)
V ∞ T EXP III
V ∞ T
V = const’ ×T
V/T = const’
V2 / T2 = constV1 / T1 = const’
V1 / T1 = V2 / T2
3.3. Avogadro’s Law
amount – volume
relationship
(pressure and temperature
are constant)
Avogadro
(1776-1856)
2 H2(g) + O2(g) → 2 H2O(l)
n ∞ V
n ∞ V
n = const’’ × V
n/V = const’’
n2 / V2 = const’’n1 / V1 = const’’
n1 / V1 = n2 / V2
SUMMARY3.1. Boyle’s Law
3.2. Gay-Lussac’s Law
3.3. Avogadro’s Law
p ∞ 1/V
n ∞ V
V ∞ T
(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
p × V = const × n × T
p × V = R × n × T
p × V = n × R × T
ideal gas equation
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
R = 8.314 J / mol / K
[R] = J / mol / K
ideal gas constant
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 EXP IV
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
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
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)
5. DALTON’S LAW
Dalton
(1801)
pure gases
gas mixtures
(atmospheres)
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 + ….
EXP V
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
pA = nA × R × T / V
p × V = (nA + nB) × R × T
pA / p = nA /(nA + nB) = xA
mole fraction
x < 1
pA = xA × p
2 KClO3 → 2 KCl + 3 O2
EXP VI/VII
SUMMARY
p × V = n × R × T
1. ideal gas equation
R = 8.314 J / mol / K
Vm = 22.4 l
2. molar volume
ς = p × M / (R × T)
3. Density of gases
4. Dalton’s Law
p = Σ pii=1
n
1. Kinetic Molecular Theory of Gases
Maxwell
(1831-1879)
Boltzmann
(1844-1906)
macroscopic
(gas cylinder)
microscopic
(atoms/molecules)
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)
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
2. gases are constantly in motion in random reactions and hold a kinetic energy
gases collide and transfer energy
(billiard ball model)
3. gases atoms/molecules
do not exert forces on each other
(absence of intermolecular interactions)
F(inter) = 0
p(inter) = 0
Gas Diffusion
2. Distribution of Molecular Speeds
Maxwell-Boltzmann distribution
3. Real Gases
p × V = n × R × T (n = 1)
deviation of ideal gas law at high pressures
p ≈ 90 atm
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)