Previously in Chemistry one has learned about Avogadro’s hypothesis:
Equal volumes of any gas, at the same temperature and pressure, contain the same number of molecules.
(from Modern Chemistry, Davis, HRW)
In 1811, Avogadro found a way to explain Gay-Lussac’s simple ratios of combining volumes without violating Dalton’s idea of indivisible atoms. He did this by rejecting Dalton’s idea that reactant elements are always in monatomic form when they combine to form products. He reasoned that these molecules could contain more than one atom.
Avogadro also put forth an idea known today as Avogadro’s law: equal volumes of gases at the same temperature and pressure contain equal numbers of molecules.
It follows that at the same temperature and pressure, the volume of any given gas varies directly with the number of molecules.
Avogadro’s law also indicates that gas volume is directly proportional to the amount of gas, at a given temperature and pressure.
Note the equation for this relationship.
V = kn
Here, n is the amount of gas, in moles, and k is a constant.
Avogadro’s reasoning applies to the combining volumes for the reaction of hydrogen and oxygen to form water vapor.
Dalton had guessed that the formula of water was HO, because this formula seemed to be the most likely formula for such a common compound.
But Avogadro’s reasoning established that water must contain twice as many H atoms as O atoms, consistent with the formula H2O.
As shown below, the coefficients in a chemical reaction involving gases indicate the relative numbers of molecules, the relative numbers of moles, and the relative volumes.
The simplest hypothetical formula for oxygen indicated 2 oxygen atoms, which turns out to be correct. The simplest possible molecule of water indicated 2 hydrogen atoms and 1 oxygen atom per molecule, which is also correct.
Experiments eventually showed that all elements that are gases near room temperature, except the noble gases, normally exist as diatomic molecules.
As an equation
Avogadro’s Law – also known as Avogadro–Ampère law
when temperature and pressure are held constant:
volume of a gas is directly proportional to the # moles (or # particles) of gas
n1 / V1 = n2 / V2
What does this imply?
As # of moles of gas increases, the volume of the gas also increases.
As # of moles of gas is decreased, the volume also decreases.
Thus, # of molecules (or atoms) in a specific volume of ideal gas is independent of their size (or molar mass) of the gas.
These problems are from The Chem Team, Kinetic Molecular Theory and Gas Laws
Example #1: 5.00 L of a gas is known to contain 0.965 mol. If the amount of gas is increased to 1.80 mol, what new volume will result (at an unchanged temperature and pressure)?
I’ll use V1n2 = V2n1
(5.00 L) (1.80 mol) = (x) (0.965 mol)
x = 9.33 L (to three sig figs)
Example #2: A cylinder with a movable piston contains 2.00 g of helium, He, at room temperature. More helium was added to the cylinder and the volume was adjusted so that the gas pressure remained the same. How many grams of helium were added to the cylinder if the volume was changed from 2.00 L to 2.70 L? (The temperature was held constant.)
1) Convert grams of He to moles:
2.00 g / 4.00 g/mol = 0.500 mol
2) Use Avogadro’s Law:
V1 / n1 = V2 / n2
2.00 L / 0.500 mol = 2.70 L / x
x = 0.675 mol
3) Compute grams of He added:
0.675 mol – 0.500 mol = 0.175 mol
0.175 mol x 4.00 g/mol = 0.7 grams of He added
Example #3: A balloon contains a certain mass of neon gas. The temperature is kept constant, and the same mass of argon gas is added to the balloon. What happens?
(a) The balloon doubles in volume.
(b) The volume of the balloon expands by more than two times.
(c) The volume of the balloon expands by less than two times.
(d) The balloon stays the same size but the pressure increases.
(e) None of the above.
We can perform a calculation using Avogadro’s Law:
V1 / n1 = V2 / n2
Let’s assign V1 to be 1 L and V2 will be our unknown.
Let us assign 1 mole for the amount of neon gas and assign it to be n1.
The mass of argon now added is exactly equal to the neon, but argon has a higher gram-atomic weight (molar mass) than neon. Therefore less than 1 mole of Ar will be added. Let us use 1.5 mol for the total moles in the balloon (which will be n2) after the Ar is added. (I picked 1.5 because neon weighs about 20 g/mol and argon weighs about 40 g/mol.)
1 / 1 = x / 1.5
x = 1.5
answer choice (c).
Example #4: A flexible container at an initial volume of 5.120 L contains 8.500 mol of gas. More gas is then added to the container until it reaches a final volume of 18.10 L. Assuming the pressure and temperature of the gas remain constant, calculate the number of moles of gas added to the container.
V1 / n1 = V2 / n2
|5.120 L||18.10 L|
x = 30.05 mol <— total moles, not the moles added
30.05 – 8.500 = 21.55 mol (to four sig figs)