Introduction into:
Dalton’s Law of Partial
Pressures
So far, we’ve assumed that all the gases we’ve been
working with are pure. However, this isn’t true most
of the time. Most gases we use are actually
mixtures of several different elements or
compounds. To deal with this, we will examine the
idea of partial pressure.
Dalton’s Law of Partial Pressures
• To work with mixtures, we usually treat each
gas as if it were alone, rather than in a mixture.
This is a reasonable assumption because gases
don’t interact with each other much anyway. As
a result, the total pressure of the mixture of
gases would be the same as the sums of all the
partial pressures.
• Dalton’s Law of Partial Pressures:
o Ptot = P1 + P2 + P3 + … Pn
o Where n is the total number of gases in the mixture.
Partial Pressure
• Partial pressure: The pressure that each of the
components in a mixture of gases would exert if the
other gases in the mixture were removed.
• Let’s back up at little:
o For any pure gas (let's use helium), PV = nRT holds
true. Therefore, P is directly proportional to n if V
and T remain constant. As n goes up, so would P. Or
the reverse.
o Suppose you were to double the moles of helium
gas present. What would happen?
o Answer: the gas pressure doubles.
Partial Pressure Continued
• However, suppose the new quantity of gas added was
a DIFFERENT gas. Suppose that, instead of helium,
you added neon.
• What would happen to the pressure?
• Answer: the pressure doubles, same as before.
• Dalton's Law immediately follows from this example
since each gas is causing 50% of the pressure.
Summing their two pressures gives the total
pressure.
• Written as an equation, it looks like this:
o PHe + PNe = Ptotal
Example Problems
• Example 1: A container contains three gases. If the partial
pressure of carbon dioxide is 0.50 atm, the partial pressure of
nitrogen is 0.30 atm, and the partial pressure of methane is 0.40
atm, what’s the total pressure inside the container?
• Ptot = 0.50 atm + 0.30 atm + 0.40 atm = 1.20 atm
• Example 2: A container has a total internal pressure of 1.50
atm. This container contains two gases, nitrogen (which has a
partial pressure of 0.80 atm) and oxygen (with unknown partial
pressure). Given this information, what’s the partial pressure of
oxygen in the container?
• Ptot = Pnitrogen + Poxygen
o 1.50 atm = 0.80 atm + X
o X = 0.70 atm
Graham’s Law
• Have you ever noticed when air freshener is sprayed into
the air, it can take seconds before you notice the smell
filling the room? You begin noticing the smell because the
molecules diffuse through the surrounding air.
• To begin to think about this law, please consider samples
of two different gases at the same Kelvin temperature.
• Since temperature is proportional to the kinetic energy of
the gas molecules, the kinetic energy (KE) of the two gas
samples is also the same.
• In equation form, we can write: KE1 = KE2
Graham’s Law Continued
• Since KE = ½mv2,
o (m = mass and v = velocity) we can write the following
equation:
• m1v12 = m2v22
**Note
that the value of one-half cancels.
• The equation above can be rearranged algebraically into the
following:
o v2 = √(m1)
v1 = √(m2)
• Graham's Law is often stated as follows:
• Rate of diffusion of gas1 = √(Mm2)
Rate of diffusion of gas2= √(Mm1)
o where MM means the molar mass of the substance in question
Graham’s Law
• Diffusion is the process in which particles in a system
move from an area of high concentration to an area of
low concentration.
• Diffusion continues until a uniform concentration of
particles is reached throughout the system.
• The rate at which gas molecules diffuse, at constant
temperature, decreases as the molar mass of the gas
increases (convince yourself of this statement based
on the equation given on the previous slide).
Graham’s Law Practice Problems
• How much faster does hydrogen gas escape through a
porous container than sulfur dioxide?
• 2.278 x 10-4 mol of an unidentified gaseous substance
effuses through a tiny hole in 95.70 s. Under
identical conditions, 1.738 x 10-4 mol of argon gas
takes 81.60 s to effuse. What is the molar mass of
the unidentified substance?