STP: Standard Temperature and Pressure (0 o
C and 1 atmosphere)
1 mole of any gas at STP has a volume of 22.4 L Or 22.4L/mol
But not many laboratories around the world operate at a temperature of 0 o
C. A more conventional standard has been introduced and is gradually being accepted.
SATP: Standard Ambient Temperature and Pressure (25 o
C and 1 atmosphere) 1 mole of any gas
at SATP has a volume of 24.8 L
The relationship of the amount (n), volume (v) and the molar volume (V) is expressed as:
n = v/V or v = nV
Example: What volume is occupied by 0.024 mol of carbon dioxide gas at SATP?
n
CO2
= 0.024 mol, V
SATP
= 24.8 L/mol
v
CO2
=0.024 mol x 24.8 L/1 mol
v
CO2 =
0.60 L

Chemical reactions, which consumed and produced gases were studied carefully by many chemists at the beginning of the nineteenth century.

In 1809, the French chemist Joseph-Louis Gay-Lussac summarized the results of many experiments into what we now call Gay-Lussac's law of combining volumes:

When measured under the same conditions of temperature and pressure, the volumes of gases, which react together, are in the ratio of small whole numbers (mole ratio).

The measurements of volume made in 1809 were sufficiently accurate to show that the volume relationships were in fact integers.

For example, one volume of hydrogen reacts with one volume of chlorine to produce one volume of hydrogen chloride;

Two volumes of hydrogen react with one volume of oxygen to produce two volumes of water vapor; and three volumes of hydrogen react with one volume of nitrogen to produce three volumes of ammonia.

The law of combining volumes was further interpreted by the Italian chemist Amedeo
Avogadro in 1811, using what was then known as the Avogadro hypothesis .

We would now properly refer to it as Avogadro's law :

Equal volumes of gases under the same conditions of temperature and pressure contain equal numbers of molecules.

Most of the common gaseous elements actually exist as diatomic molecules: hydrogen, nitrogen, oxygen, fluorine, and chlorine.
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
Avogadro's interpretation cleared up many of these discrepancies; it enabled explanation of the empirical results of Gay-Lussac in terms of simple molecular reactions:
H
2(g)
+ Cl
2(g)
--> 2 HCl
(g)
2 H
2(g)
+ O
2(g)
--> 2 H
2
O
(g)
3 H
2(g)
+ N
2(g)
--> 2 NH
3(g)

The Avogadro law is equivalent to the statement that volume is directly proportional to number of atoms or molecules.

Since the fundamental unit of amount of substance, the mole, is equal to Avogadro's number of atoms or molecules, the amount of substance n, in moles.
Example: Calculate the volume of oxygen required for the complete combustion of 120 mL of
butane from a lighter
First: Write a balanced chemical equation:
2C
4
H
10 (g)
+13 O
2(g)
→ 8CO
2(g)
+ 10H
2
O(g)
M/R: 2 13 8 10
V : 120 mL v
v
O2
= 120 mL x 13/2
= 780 mL

An ideal gas is a hypothetical gas composed of particles that have zero size, travel in straight lines, and have no attraction to each other (zero intermolecular force)
Ideal gas Law: It describes the relationship among pressure, temperature, volume, and the
amount of matter, the four variables described in the gas laws;
According to Boyle’s Law; v α 1/p
According to Charles’s Law; v α T
According to Gay Lussac’s Law; p α T
The more particles in a can, the more the volume, v α n
Combining the mathematical relationship for volume,
v α 1/p x T x n that is, v = (constant, R) x 1/p x T x n
v = nRT Or PV = nRT
P
This is the ideal gas law. R is the gas constant with a constant value of 8.31 kpa
• L
Mol • K
Partial Pressures of Gases

When Dalton was conducting his studies, which led him to the atomic-molecular theory of matter, he also included studies of the behavior of gases.

These led him to propose, in 1803, what is now called Dalton's law of partial pressures:
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
For a mixture of gases in a container, the total pressure exerted is the sum of the pressures that each gas would exert if it were alone.

This law can be expressed in equation form as: p = p
1
+ p
2
+ p
3
+ ...
where p is the total or measured pressure and p
1
, p
2
, ... are the partial pressures of the individual gases.

For air, an appropriate form of Dalton's law would be: p(air) = p(N
2
) + p(O
2
) + p(CO
2
) + ...

At temperatures near ordinary room temperature, the partial pressure of each of the components of air is directly proportional to the number of moles of that component in any volume of air.

When the total pressure of air is 100 kPa, the partial pressures of each of its components
(in kPa) are numerically equal to the mole per cent of that component (Table below).

Thus the partial pressures of the major components of dry air at 100 kPa are nitrogen, 78 kPa; oxygen, 21 kPa; argon, 0.9 kPa; and carbon dioxide, 0.03 kPa.

The same substance may be found in different physical states under different conditions.

Water, for example, can exist as a solid phase (ice), a liquid phase (water), and a gas phase (steam or water vapor) at different temperatures.

The processes by which a substance is converted from one phase to another are called by specific names.

The conversion from solid to liquid is melting and the reverse conversion from liquid to solid is freezing.

The conversion from liquid to gas is called boiling or vaporization and the reverse conversion from gas to liquid is called condensation.

The conversion from solid to gas, when it occurs directly without going through a liquid state as in the case of iodine and carbon dioxide, is called sublimation; the reverse conversion from gas to solid shares the name of condensation.

All solids and liquids, that is, all substances in condensed phases, exhibit a vapor pressure.

This is a pressure of the substance in the gas phase, which is established at a particular temperature.

The vapor pressure of a substance depends upon the temperature. A table of vapor pressures at 25 o
C for a few selected substances is normally given.
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
Gases are often collected over any liquid, which has a vapor pressure, such as water.

The total pressure measured, p, is the sum of the partial pressure of the collected gas and the partial pressure of the liquid, which is its vapor pressure.

If the liquid is water, as is most often the case, the pressure of the gas collected must be:
P
(gas)
= P total
- P
(H2O)

The measured or total pressure is obtained using a barometer and manometer.

The vapor pressure of water at the appropriate temperature is obtained from a table such as that given below.
Table: Vapor Pressures and Densities of Pure Substances at
25 o
C
Substance Vapor Pressure
(kPa)
Density
(kg/m
3
)
0.99702 H
2
O(l)
CH
3
OH(l) 16.8511
C
2
H
5
OH(l) 7.8279
C
6
H
6
(l)
Hg(l)
3.1691
12.6893
0.2460
0.791
0.785
0.899
13.5340
I
2
(s) 0.1889 4.93
Example: A volume of 546.3 cm
3
of hydrogen is collected over water at 30 o
C when the
atmospheric pressure is 100.45 kPa. The actual pressure of the hydrogen and the
amount of hydrogen present are calculated as follows.
The vapor pressure of water at 30 o
C is 4.246 kPa, so the actual pressure of the hydrogen is:
100.45 kpa – 4.246 kpa = 96.20 kPa.
The amount of hydrogen present is: PV = nRT, n = PV/RT
(96.20 kPa)(0.5463 dm
3
)/(8.314 kPa dm
3
/mol K)(303K) = 0.0209 mol H
2
This extends stoichiometric calculations to gases, using gas volume, molar volume, and ideal gas law
Example: If 275 g of propane burns in a gas barbecue, what volume of oxygen measured at STP is required for the reaction?
4

First: write a balanced chemical equation
C
3
H
8(g)
+ 5 O
2 (g)
→ 3CO
2 (g)
+ 4H
2
O
(g)
M/R: 1 5 3 4 m: 275 g v?
M: 44.11 g/mol
n : 275g x1mol n
O2
=6.23 mol x 5 mol
44.11g 1 mol
6.23 mol 31.2 mol
v
O2
=nV
31.2 mol x 22.4 L/mol
698 L
Summary:
For all stoichiometry involving Gravimetric, Solution and Gas
Write a balanced chemical equation and list the measurements and the conversion factors and the one to be calculated
Convert the measurement to an amount in moles using the appropriate conversion factor
Calculate the amount of the other substance by using the mole ratio from the balanced equation
Convert the calculated amount to the final quantity requested by using the appropriate conversion factor.
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