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# Gas Law Relationships for AP Chemistry (page 2)

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By McGraw-Hill Professional
Updated on Feb 9, 2011

### Dalton's Law of Partial Pressures

Dalton's law says that in a mixture of gases (A + B + C …) the total pressure is simply the sum of the partial pressures (the pressures associated with each individual gas). Mathematically, Dalton's law looks like this:

PTotal = PA + PB + PC + …

Commonly Dalton's law is used in calculations involving the collection of a gas over water, as in the displacement of water by oxygen gas. In this situation there is a gas mixture: O2 and water vapor, H2O(g). The total pressure in this case is usually atmospheric pressure, and the partial pressure of the water vapor is determined by looking up the vapor pressure of water at the temperature of the water in a reference book. Simple subtraction generates the partial pressure of the oxygen.

If you know how many moles of each gas are in the mixture and the total pressure, you can calculate the partial pressure of each gas by multiplying the total pressure by the mole fraction of each gas:

PA = (PTotal)(XA)

where XA = mole fraction of gas A. The mole fraction of gas A would be equal to the moles of gas A divided by the total moles of gas in the mixture.

### Graham's Law of Diffusion and Effusion

Graham's law defines the relationship of the speed of gas diffusion (mixing of gases due to their kinetic energy) or effusion (movement of a gas through a tiny opening) and the gases' molecular mass. The lighter the gas, the faster is its rate of effusion. Normally this is set up as the comparison of the effusion rates of two gases, and the specific mathematical relationship is:

where r1 and r2 are the rates of effusion/diffusion of gases 1 and 2 respectively, and M2 and M1 are the molecular masses of gases 2 and 1 respectively. Note that this is an inverse relationship.

For example, suppose you wanted to calculate the ratio of effusion rates for hydrogen and nitrogen gases. Remember that both are diatomic, so the molecular mass of H2 is 2.016 g/mol and the molecular mass of N2 would be 28.02 g/mol. Substituting into the Graham's law equation:

Hydrogen gas would effuse through a pinhole 3.728 times as fast as nitrogen gas. The answer is reasonable, since the lower the molecular mass, the faster the gas is moving. Sometimes we measure the effusion rates of a known gas and an unknown gas, and use Graham's law to calculate the molecular mass of the unknown gas.

### Gas Stoichiometry

The gas law relationships can be used in reaction stoichiometry problems. For example, suppose you have a mixture of KClO3 and NaCl, and you want to determine how many grams of KClO3 are present. You take the mixture and heat it. The KCIO3 decomposes according to the equation:

2 KClO3(s) → 2 KCl(s) + 3 O2(g)

The oxygen gas that is formed is collected by displacement of water. It occupies a volume of 542 mL at 27°C. The atmospheric pressure is 755.0 torr. The vapor pressure of water at 27°C is 26.7 torr.

First, you need to determine the pressure of just the oxygen gas. It was collected over water, so the total pressure of 755.0 torr is the sum of the partial pressures of the oxygen and the water vapor:

The partial pressure of water vapor at 27°C is 26.7 torr, so the partial pressure of the oxygen can be calculated by:

At this point you have 542 mL of oxygen gas at 728.3 torr and 300. K (27°C + 273). From this data you can use the ideal gas equation to calculate the number of moles of oxygen gas produced:

PV = nRT
PV/RT = n

You will need to convert the pressure from torr to atm:

(728.3 torr) × (1 atm/760.0 torr) = 0.9583 atm

and express the volume in liters: 542 mL = 0.542 L

Now you can substitute the quantities into the ideal gas equation:

(0.9583 atm)(0.542 L)/(0.0821 L · atm/K · mol)(300. K) = n
0.02110 mol O2 = n

Now you can use the reaction stoichiometry to convert from moles O2 to moles KClO3 and then to grams KClO3:

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