Ideal Gases


The Ideal Gas Equation of State

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 Ideal-Gas 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

m3 . 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

0 °C (273.15K) and 1 atm pressure are referred to as conditions of:
Standard Temperature and Pressure (
STP)


--- The MOLAR VOLUME of any ideal gas is 22.4 liters at STP ---



Another Numerical Example:
Hydrogen peroxide, H2O2, can decompose to form water and gaseous molecular oxygen. (Write a balanced reaction for this decomposition)


2 H2O2 --> 2 H2O + O2


A sample of hydrogen peroxide is heated and the O2 gas produced is collected in a 750 ml flask. The pressure of the gas in the flask is 2.80 atmospheres and the temperature is recorded at 53.6 °C.

How many moles of O2 gas were produced?
How many moles of peroxide decomposed?

pV = nRT
n = pV/RT
n = (2.80 atm * 0.750 L) / ((0.0821 L.atm/mol.K) * (53.6 + 273.15)K)

Note conversion to ABSOLUTE TEMPERATURE!
n = 0.0783 mol O2 were produced
n=0.157 mol peroxide decomposed

Relationship Between the Ideal-Gas 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)

Check out this Boyle's Law Simulation

If the quantity of gas and the pressure are held constant then:
pV = nRT
V = (nR/p) * T
V = constant * T
(Charles's law)

Check out this Charle's [sic] law Simulation

If the temperature and pressure are held constant then:
pV = nRT
V = n * (RT/p)

V = constant * n
(Avogadro's law)


The 'General"case...
Suppose everything changes at once. One thing we are very sure of is that the gas constant, R, is in fact a constant. If we label the properties of the state of the gas initially by the subscript 1, then the state of the gas initially is defined by:

{ n1, p1, V1, T1 }

Later, the gas will be in a different state, defined by new variables:

{ n2, p2, V2, T2 }

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 CCl4 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 (1766-1844) - (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 Pt as the total pressure of the gas sample mixture and P1, P2, P3, etc. as the partial pressures of the component gases in the mixture. Clearly, Dalton's law would suggest:

Pt = P1 + P2 + P3 + ...

If each of the component gases behaves independently of the others, the ideal gas equation of state applies to each gas individually. If n1 is the number of moles of component 1 in the sample, then the partial pressure of that gas is simply:


Similarly for the partial pressure of component 2:


... and so on for all components.

Therefore, the total pressure Pt is:


All components will share the SAME TEMPERATURE, T, and VOLUME V.



Since the ideal gas equation works for any gas regardless of its identity, it should work for the gas sample as a whole, with the mole number simply the sum of the moles of each component in the mixture:


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 O2)(1 mol/32.0 g) = 0.3125 mol O2
(15.00 g CH4)(1 mol/16.0 g) = 0.9375 mol CH4

V=10.0 l
T=(273.15+25.0)K=298.15K

pO2 = (0.3125)(0.08206)(298.15)/(10) = 0.765 atm
pCH4 = (0.9375)(0.08206)(298.15)/(10) = 2.29 atm

ptotal = 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 value (n1/nt) = X1 is termed the mole fraction of component 1 in the gas mixture.

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 CO2, 20.0 mol percent O2 and 78.0 mol percent N2. 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 O2 are needed?


PO2 = 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:

3NO2(g) + H2O(l) --> 2HNO3(aq) + NO(g)

How many moles of nitric acid can be prepared using 450 L of NO2 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 NO2

92.9 mol NO2 (2 HNO3 / 3 NO2) = 62 mol HNO3


Collecting Gases Over Water

When heated, potassium chlorate gives off oxygen through the reaction:

2 KClO3(s) --> 2 KCl(s) + 3 O2(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 semi-volatile 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.

Ptotal = PO2 + PH2O

The pressure of the water vapor is determined by the equilibrium between the liquid phase and the gaseous phase of water. Thus the pressure of the water vapor is a function of temperature only, as long as there is liquid present:
The Vapor Pressure of Pure Water

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 KClO3 is partially decomposed, producing O2 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 O2 are collected?

Pt = 765 torr = PO2 + PH2O = PO2 + 23.76 torr
PO2 = 765 - 23.76 = 741.24 torr
PO2 = 741.2 torr (1 atm/760 torr) = 0.9753 atm
PV = nRT
(0.9753 atm)(0.250 L) = n(0.08206 L atm/mol K)(273.15 + 25.0)K
n = 9.97 x 10-3 mol O2

b) How many grams of KClO3 were decomposed?

9.96 x 10-3 mol O2 (2KC lO3/3 O2) = 6.64 x 10-3 mol KClO3
6.64 x 10-3 mol KClO3 (122.6 g/mol) = 0.815 g KClO3

c) If the O2 were dry, what volume would it occupy at the same T and P?

PO2 = (Pt)(XO2) = 765 torr (1.0) = 765 torr (1 atm/760 torr) = 1.007 atm
(1.00658 atm)(V) = (9.97 x 10-3 mol)(0.08206 L atm/mol K)(273.15 + 25)K
V = 0.242 L

Alternatively

If the number of moles, n, and the temperature, T, are held constant then we can use Boyle's Law:
P1V1 = P2V2
V2 = (0.250 L)*((741.2 torr)/(765 torr))
V2 = 0.242 L


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