Gases
Variable Density
Solids and liquids have a definite volume, thus they have a fixed density at a given temperature
Gases have a variable density (and thus, volume) at a given temperature
 Pressure and temperature and molar mass determine density for a gas
 Density and temperature are inversely proportional
 Density is directly proportional to pressure and to molar mass
Gas Laws Overview
Describes macroscopic behavior of gases – volume, temperature, pressure, and mol number
Combined Gas Law
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PV / T = constant
P1V1 / T1 = P2V2 / T2
P and V inversely proportional
V and T directly proportional
P and T directly proportional
Avogadro’s Law - V & n

V and n directly proportional

Applicable to stoichiometry of gas phase reactions
Ideal Gas Law
[n = mol number]
PV/T=nR
PV=nRT
PV/nT=R
R = 0.08206 L·atm·mol–1·K–1 = 62.37 L·mmHg·mol–1·K–1
Gas Law Calculations
Solving for dynamic conditions
 Initial and final values given
 Usually for two variables
 Can use combined gas law
Solving for static conditions
 Three values given
 Can use ideal gas law
Examples:
Calculate the new volume of a 5.0 L balloon when the pressure drops from 1.0 atm to 0.90 atm.
P1 V1 = P2 V2
(1.0 atm)(5.0 L) = (0.90 atm) V2
V2 = 5.6 L
Calculate the initial pressure of a piston containing 2.5 mL of air if the final volume and pressure are 7.5 mL
and 0.30 atm, respectively.
P1 V1 = P2 V2
P1 (2.5 mL) = (0.30 atm)(7.5 mL)
P1 = (0.30 atm)(7.5 mL) / (2.5 mL) = 0.90 atm
Calculate the final volume of a 5.0 L balloon if the temperature is raised from 300 K to 600 K.
V1 / T1 = V2 / T2
5.0 L / 300 K = V2 / 600 K
V2 = (5.0 L)(600 K) / (300 K) = 10 L
Calculate the final temperature if the pressure in a tank drops from 20.0 atm to 8.0 atm. Initial temperature
is 300 K.
P1 / T1 = P2 / T2
20.0 atm / 300 K = 8.0 atm / T2
T2 = (8.0 atm)(300 K) / (20.0 atm) = 120 K
Calculate the volume of a balloon containing 2.00 mol of gas with P = 1.00 atm and T = 300 K.
PV=nRT
(1.00 atm)(V) = (2.00 mol)( 0.08206 L·atm·mol–1·K–1)(300 K)
V = (2.00 mol)( 0.08206 L·atm·mol–1·K–1)(300 K) / (1.00 atm) = 49.2 L
Calculate the number of moles in a cylinder containing 20.0 L of gas at 298 K and a pressure of 3.00 atm.
PV=nRT
(3.00 atm)(20.0 L) = n (0.08206 L·atm·mol–1·K–1)(298 K)
n = (3.00 atm)(20.0 L) / (0.08206 L·atm·mol–1·K–1)(298 K) = 2.45 mol
Calculate the temperature of a cylinder holding 5.00 mol of gas with P = 6.00 atm and V = 22.5 L.
PV=nRT
T = (6.00 atm)(22.5 L) / [(5.00 mol)( 0.08206 L·atm·mol–1·K–1)] = 329 K
Calculate the molar volume of a gas with P = 0.95 atm and T = 295 K.
PV=nRT
(0.95 atm)(V) = (1.00 mol)( 0.08206 L·atm·mol–1·K–1)(295 K)
V = (1.00 mol)( 0.08206 L·atm·mol–1·K–1)(295 K) / (0.95 atm) = 25.5 L
Other Applications of the Ideal Gas Law
The ideal gas law can be rearranged to show the relationship between gas density and molar mass.
n/V=P/RT
n=m/M
d=m/V=PM/RT
M =dRT/P
M =mRT/PV
Examples:
Calculate the density of butane (C4H10) at 0ºC and 1.00 atm.
M = 58.14 gmol –1
d = P M / R T = (1.00 atm)(58.14 gmol –1) / (0.08206 Latmmol –1K –1)(273.15 K)
d = 2.59 gL–1
Calculate the molar mass of a gas with a density of 1.18 gL–1 at 25.0ºC and 1.00 atm.
M =dRT/P
M = (1.18 gL–1) (0.08206 Latmmol –1K –1)(298.15 K) / (1.00 atm)
M = 28.9 gmol –1
Dalton’s Law of Partial Pressures
PTotal = P1 + P2 + P3 + …
P1 = X1 × PTotal
Total pressure is sum of partial pressures in a mixture of gases
The partial pressure of a component in a gas mixture is the product of mole fraction and total pressure
Vapor Pressure
Small amount of liquid evaporates to form vapor over a liquid and exerts a vapor pressure.
Thus, for gases collected by bubbling through water
PTotal = Pgas + Pwater
Vapor pressure increases with temperature.
Water
T(ºC)
P(mmHg)
T(ºC)
P(mmHg)
0
4.6
60
149.4
10
9.2
70
233.7
20
17.5
80
355.1
30
31.8
90
525.8
40
55.3
100
760.0
50
92.5
Examples:
Calculate the partial pressure of oxygen in a scuba tank filled with heliox (helium and oxygen mixture). The
partial pressure of helium is 600 mmHg and the total pressure is 760 mmHg.
P = P1 + P2
760 mmHg = 600 mmHg + P2
P2 = 160 mmHg
A sample of acetylene is collected over water at a temperature of 20ºC. Total pressure is 500.0 mmHg.
Calculate the partial pressure of acetylene.
P = P1 + P2
500.0 mmHg = 17.5 mmHg + P2
P2 = 482.5 mmHg
Calculate the partial pressure (in mmHg) of O2 in air that has a total pressure of 0.985 atm and contains
20.95% O2.
PO2 = (0.2095)(0.985 atm)
PO2 = 0.206 atm
PO2 = (0.206 atm)(760 mmHgatm–1) = 157 mmHg
Calculate the partial pressure (in mmHg) of He in air that has a total pressure of 0.995 atm and contains
5.24 ppm He.
PHe = (5.24×10–6)(0.995 atm)(760 mmHgatm–1)
PHe = 3.96×10–3 mmHg
Kinetic Theory of Gases
Explains macroscopic behavior of gases – volume, temperature, and pressure.
[Quantum mechanical and relativistic effects are negligible.]
1. Gases consist of a statistically large number of particles with finite mass.
2. Particles are in constant, random motion.
[Average kinetic energy of particles directly proportional to absolute temperature].
3. Elastic collisions between particles and walls.
[Conservation of energy]
4. Intermolecular forces between particles are negligible.
5. Total particle volume negligible relative to container.
[Large intermolecular spacing]
For real gases at different temperatures and pressures:
 Intermolecular forces may not be negligible
 Particle volume may not be negligible
Thus, real gases may deviate from ideal behavior
 High pressure and/or low temperature lead to non-ideal behavior
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At extremes, gas may condense into liquid
Nitrogen boils at 77 K
[Liquid air at about T = –200 ºC]
Van der Waals Equation
Accounts for non-ideality using two constants that depend on the identity of the gas
Graham’s Laws of Effusion and Diffusion
Effusion – process by which individual gas molecules flow through holes in a container without
intermolecular collisions
Diffusion – particles moving from an area of high concentration to an area of low concentration

Rate of effusion (diffusion) proportional to inverse square root of molar mass
Denser gases (higher molar mass) diffuse and effuse more slowly
Henry’s Law and Gas Solubility
 Gas solubility is directly proportional to the partial pressure of the gas
 Henry’s law constant – depends on solute, solvent, and temperature
[Not universal like ideal gas constant.]
Background and Reference Information
DRY AIR
Pressure = Force / Area
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Pressure is omni-directional
Atmospheric pressure – referenced at sea level
Common Units
1 atm
78.08% N2
0.93% Ar
20.95% O2
0.03% CO2
= 760 mmHg
= 14.7 psi
= 101.325 kPa
= 29.92 inHg
= 1013.25 mb (millibar)
Temperature – Average Kinetic Energy
Absolute temperature is directly proportional to average kinetic energy of the gas particles
TK = TC + 273.15
generally, can use: TK = TC + 273
Hemoglobin and Gas Transport
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Partial pressure of O2 in air ~160 mmHg and in alveoli ~100 mmHg
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Partial pressure of CO2 in air ~40 mmHg and in alveoli ~45 mmHg
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CO2 present in blood is primarily in form of bicarbonate ion, HCO3–
Historical development of gas laws and kinetic molecular theory
NB: many discoveries were made independently by two scientists; attribution can be controversial
Barometer invented
Air pump invented
Manometer invented
Mercury thermometer invented
Boyle’s Law
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[1662, Pressure and Volume]
Volume of a gas is inversely proportional to its pressure (at constant temperature)
Amonton’s Law
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[1702, Pressure and Temperature]
Pressure of a gas is directly proportional to its absolute temperature
Charles’ Law
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[1787-1802, Volume and Temperature]
Volume of a gas is directly proportional to its absolute temperature
Applicable to physical and chemical processes
Dalton’s Law
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[1801, Partial Pressures]
Total pressure of a mixture of gases is the sum of the partial pressures of the components
Henry’s Law
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[1643, Torricelli]
[1652, Otto van Guericke]
[1661, Christian Huygens]
[1714, Daniel Fahrenheit]
[1803, Solubility of Gases]
At constant temperature, the solubility of a gas in some liquid is directly proportional to the partial
pressure of that gas in equilibrium with that liquid
Gay-Lussac’s Law [1809, Volume and Temperature]
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Law of combining volumes – volumes of reactants and products are ratio of small integers
Avogadro’s Law
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Graham’s Laws
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[1831, Diffusion or Effusion]
Rate of effusion inversely proportional to the square root of molar mass
Ideal gas law
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[1811, Volume and Mol Number]
Two gas samples with equal volume contain equal number of particles (same mol number)
[1834, Clapeyron; 1857, Clausius (from kinetic theory)]
Pressure, volume, temperature, and mol number (number of particles) related by constant
Combined gas law (for constant mol number) useful variant for closed systems
Van der Waals’ equation

Correction to pressure term and volume term
[1873, corrections for real gases]