The three 'historical' gas laws are relationships between
two physical (state) properties of a gas, with two other properties constant.
(Why does it take just four properties to define the state of a gas?):
These relationships can be combined into a single relationship
to make a more general gas law:
If the proportionality constant is called "R",
then we have:
Rearranging to a more familiar form:
This equation is known as the IdealGas Equation of State
Units for the Gas Constant, R 

Units 
Numerical Value 
L ^{.} atm / mol ^{.} K 
0.08206 
cal / mol ^{.} K 
1.987 
J / mol ^{.} K 
8.314 
m^{3} ^{.} Pa / mol ^{.} K 
8.314 
L ^{.} torr / mol ^{.} K 
62.36 
Use of the Ideal Gas Equation
Numerical Example:
1.00 mol of gas at 1.00 atm of pressure at 0.00°C
(273.15 K) occupies what Volume?
pV = nRT
V = nRT/p
V = (1.00 mol)(0.0821 L atm/mol K)(273.15 K) / (1.00 atm)
Therefore: V = 22.4 L
 The MOLAR VOLUME of any ideal gas is 22.4 liters at STP 
Another Numerical Example:
Hydrogen peroxide, H_{2}O_{2}, can decompose to form water
and gaseous molecular oxygen. (Write a balanced reaction for this decomposition)
Relationship Between the IdealGas Equation and the
Gas Laws
Boyle's law, Charles's law and Avogadro's law all represent
special cases of the Ideal Gas law (equation)
If the quantity of gas and the temperature are held constant
then:
pV = nRT
V = nRT / p
V= (nRT) * (1/p)
V= constant * (1/p)
(Boyle's law)
If the quantity of gas and the pressure are held constant
then:
pV = nRT
V = (nR/p) * T
V = constant * T
(Charles's law)
If the temperature and pressure are held constant then:
pV = nRT
V = n * (RT/p)
V = constant * n
(Avogadro's law)
{ n_{1}, p_{1}, V_{1}, T_{1} }
Later, the gas will be in a different state, defined by new variables:
{ n_{2}, p_{2}, V_{2}, T_{2} }
For any change in state of the gas, (pV/nT) = R remains unchanged, so
This equation is sometimes called the General Gas Equation
Numerical Example:
A 1.0 liter sample of air at room temperature (25 °C)
and pressure (1 atm) is compressed to a volume of 3.3 ml at a pressure
of 1000. atm. What is the temperature of the air sample?
If the sample did not change temperature,
the increase in pressure by 1000 fold would decrease the volume 1000 fold
to make the volume 1.0 ml.
But the actual volume is 3.3 times that. So the absolute temperature
must have increased by a factor of 3.3.
Therefore, the temperature is 3.3 * (298) =
983K or 710°C
Molar Mass and Gas Densities
Density
Mass Density has the units of mass per unit volume.
Number Density has the units of molecules (moles) per unit volume and is directly derived from the Ideal Gas Equation of State:
pV = nRT
(n/V) = p/RT
(n/V) is the number density and has the units of moles/liter. If we know the molecular mass of the gas, we can convert this into grams/liter (mass/volume). The molar mass (M) is the number of grams in one mole of a substance. If we multiply both sides of the above equation by the molar mass:
where d is the number of grams per unit volume, or the mass per unit volume (which is the MASS DENSITY)
Numerical Example:
What is the mass density of carbon tetrachloride vapor at
714 torr and 125°C?
The molar mass of CCl_{4} is 12.0 + (4*35.5) =
154 g/mol.
125°C is (273+125) = 398K.
714 torr is 714/760=0.9395 atm
The density of the gas is (154)(0.9395)/(0.0821)(398) = 4.43 g/l
Gas Mixtures and Partial Pressures
How do we deal with gases composed of a mixtures of two
or more different substances?
John Dalton (17661844)  (gave us Dalton's atomic theory)
The total pressure of a mixture of gases
equals the sum of the pressures that each would exert if it were present
alone, i.e. TOTAL PRESSURE is the sum of the PARTIAL PRESSURES of the
components of the mixture
Dalton's Law of Partial Pressures: (Mathematics)
Let's denote P_{t} as the total pressure of the gas sample mixture and P_{1}, P_{2}, P_{3}, etc. as the partial pressures of the component gases in the mixture. Clearly, Dalton's law would suggest:P_{t}
= P_{1} + P_{2 }+ P_{3} + ...
If each of the component gases behaves independently of the others,
the ideal gas equation of state applies to each gas individually.
If n_{1} is the number
of moles of component 1 in the sample, then the partial pressure of that gas is simply:
Therefore, the total pressure P_{t} is:
Note: At a given temperature and volume, the
total pressure of a gas sample is determined by the total number of moles
of gas present, whether this represents a single substance, or a
mixture, as long as the mixture may be approximated as an ideal gas.
Numerical Example:
A gaseous mixture made from 10.0 g of oxygen and 15.00 g of
methane is placed in a 10.0 L vessel at 25.0°C. What is the partial pressure
of each gas, and what is the total pressure in the vessel?
(10.0 g O_{2})(1 mol/32.0 g) = 0.3125 mol O_{2}
(15.00 g CH_{4})(1 mol/16.0 g) = 0.9375 mol CH_{4}
V=10.0 l
T=(273.15+25.0)K=298.15K
p_{O2} = (0.3125)(0.08206)(298.15)/(10) = 0.765 atm
p_{CH4} = (0.9375)(0.08206)(298.15)/(10) = 2.29 atm
p_{total} = 2.294 + 0.7646 = 3.06 atm
Partial Pressures and Mole Fractions
Partial Pressures and Mole Fractions
The ratio of the partial pressure of one component of a gas to the total pressure is:
or
The mole fraction (X) is a dimensionless number. The above equation can be rearranged:
The partial pressure of a component of a gas mixture
is equal to the mole fraction of that component times the total pressure
of the mixture
Numerical Example
a) A synthetic atmosphere is created by blending 2.00 mol
percent CO_{2}, 20.0 mol percent O_{2} and 78.0 mol percent
N_{2}. If the total pressure is 750 torr, calculate the partial
pressure of the oxygen component.
Mole fraction of oxygen is (20/100) = 0.200
Therefore, partial pressure of oxygen = (0.200)(750 torr) = 150. torr
b) If 25.0 liters of this atmosphere, at 37.0°C, have
to be produced, how many moles of O_{2} are needed?
P_{O2} = 150 torr (1 atm/760 torr) = 0.197 atm
V = 25.0 L
T = (273.15+37.0)K=310.15K
R = 0.0821 L^{.}atm/mol^{.}K
P V = n R T
n = (PV)/(RT) = (0.197 atm * 25.0 L)/(0.0821 L^{.}atm/mol^{.}K
* 310K)
n = 0.194 mol
Volumes of Gases in Chemical Reactions
For a gas that is approximately ideal, the number of moles of the gas sample is related to its pressure (p), volume (V) and temperature (T) only, regardless of its chemical identity.
Example:
The synthesis of nitric acid involves the reaction of nitrogen dioxide
gas with water:
3NO_{2}(g) + H_{2}O(l) >
2HNO_{3}(aq) + NO(g)
How many moles of nitric acid can be prepared using 450 L of NO_{2} at a pressure of 5.0 atm and a temperature of 295 K?
(5.0 atm)(450 L) = n(0.0821 L^{.}atm/mol^{.}K)(295K)
= 92.9 mol NO_{2
}92.9 mol NO_{2} (2 HNO_{3} / 3 NO_{2}) = 62
mol HNO_{3}
Collecting Gases Over Water
When heated, potassium chlorate gives off oxygen through the
reaction:
2 KClO_{3}(s) > 2 KCl(s)
+ 3 O_{2}(g)
The oxygen evolved can easily be be collected and measured (b) in a beaker that is initially filled with water (a). {Water is a ubiquitous semivolatile liquid}
The volume of gas collected is measured by adjustment of the the beaker so that the water level in the beaker is the same as in the pan. When the levels are the same, the pressure inside the beaker is the same as the ambient pressure in the lab (approximately 1 atm of pressure; Accurate determinations simultaneously measure the barometric pressure in the laboratory) The total pressure inside the beaker is equal to the sum of the partial pressure of gas collected and the partial pressure of water vapor due to evaporation.
P_{total} = P_{O2} + P_{H2O}
 
Temperature (°C) 
Vapor Pressure (torr) 
0 
4.58 
25 
23.76 
35 
42.2 
65 
187.5 
100 
760.0 
Numerical Example:
A sample of KClO_{3} is partially decomposed,
producing O_{2} gas that is collected over water. The volume of
gas collected is 0.250 liter at 25 °C and 765 torr total pressure.
a) How many moles of O_{2} are collected?
b) How many grams of KClO_{3} were decomposed?
c) If the O_{2} were dry, what volume would
it occupy at the same T and P?
Alternatively