Dalton’s law of partial pressure states that in a mixture of two or more non-reacting gases, the sum of the partial pressures of each gas is equal to the overall pressure of the gas mixture. A gas’s partial pressure is the pressure it would exert on the walls if it were the only gas in a container [1-4].
English physicist and chemist John Dalton published the law in 1802.
For Dalton’s law to hold good, the following assumptions are made: [1-6]
- The gas molecules occupy an insignificant volume compared to the container’s volume.
- The gases are ideal, meaning there is no attraction among the molecules.
- The pressure is not excessively high. Otherwise, the volume occupied by the molecules becomes significant compared to the free space between them, resulting in a deviation from Dalton’s law.
Based on these assumptions, one can calculate the total pressure of the gas mixture in terms of the partial pressure of the gases. Suppose a mixture contains n gases whose partial pressures are p1, p2, p3, …, pn. Then, the total pressure p of the mixture is given by
p = p1 + p2 + p3 + … + p4
p = Σpi
Partial Pressure from Mole Fraction
Dalton’s law can also be expressed in terms of mole fraction. The mole fraction X is the fraction of the gas present in the mixture. It is evaluated by taking the ratio of the number of moles (ni) of a given gas and the total number of moles (Σni) of all the gases.
Xi = ni /(Σni)
Then, the partial pressure of the i-th gas is
pi = Xi p
Partial Pressure from Ideal Gas Equation
The partial pressure can be calculated from the ideal gas equation. Suppose ni moles of the i-th gas are present in the mixture. The partial pressure is given by
pi = niRT/V
T: Absolute temperature
V: Volume of the gas
Problem 1: A mixture of oxygen and hydrogen exerts a total pressure of 2.3 atm on the walls of the container. If the partial pressure of hydrogen is 1.2 atm, find the mole fraction of oxygen in the mixture.
Given, phydrogen = 1.2 atm p = 2.3 atm
Therefore, the partial pressure of oxygen is
poxygen = p – phydrogen = 2.3 atm – 1.2 atm = 1.1 atm
The mole fraction of oxygen can be found using the following formula.
poxtgen = Xoxygen p
=> Xoxygen = poxygen/p
=> Xoxygen = 1.1 atm/2.3 atm = 0.48
Problem 2: 14 L of oxygen gas at 2.4 atm and 18 L of nitrogen at 2.2 atm are added to a 12 L container at 300 K. Find the partial pressure of oxygen and nitrogen and then find the total pressure.
Given Voxygen = 14 L poxygen1 = 2.4 atm Vnitrogen = 18 L pnitrogen1 = 2.2 atm T = 300 K
V = 12 L R = 0.082 L-atm · mol-1 · K-1
The mole fraction of oxygen is
noxygen = (poxygen Voxygen)/(RT) = (2.4 atm · 14 L)/(0.082 L-atm · mol-1 · K-1 · 300 K) = 1.36 mol
nnitrogen = (pnitrogen Vnitrogen)/(RT) = (2.2 atm · 18 L)/(0.082 L-atm · mol-1 · K-1 · 300 K) = 1.61 mol
Therefore, the total number of moles is
n = noxygen + nnitrogen = 1.36 + 1.61 = 2.97 mol
The total pressure is
P = nRT/V = 2.97 mol · 0.082 L-atm · mol-1 · K-1 · 300 K/12 L = 6.09 atm
Therefore, the partial pressures are
poxygen = (noxygen/n) · P = (1.36 mol/2.97 mol) · 6.09 atm = 2.78 atm
pnitrogen = (nnitrogen/n) · P = (1.61 mol/2.97 mol) · 6.09 atm = 3.3 atm
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