Partial Pressures

Law of Partial Pressures:
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.

$p_{\mbox{total}} = p_1+p_2+p_3 + \cdots$

where $p_1$ is the partial pressure of the first gas, $p_2$ is the partial pressure of the second gas, $p_3$ is the partial pressure of third gas, etc... If each gas behaves ideally, then the partial pressure of each gas can be calculated from the ideal gas law.

$p_1 =n_1 R T/V$,        $p_2=n_2 R T/V$,             $p_3 =n_3 R T/V$,     ...

Remember all the different gases are together in a single container of volume V and at a temperature T. We can substitute these expressions into the sum above and get

$p_{\mbox{total}} = n_1 R T/V + n_2 R T/V + n_3 R T/V + \cdots$

which we can rewrite as

$p_{\mbox{total}} = \left(n_1 + n_2 + n_3\right) R T/V + \cdots$

or as

$p_{\mbox{total}} = n_{\mbox{total}} R T/V + \cdots$

where

ntotal= n1 + n2 + n3 + ...

Thus for a mixture of ideal gases, it is the total number of moles of particles that is important, not the identity or composition of the individual gas particles. Thus we define...

Mole Fraction:
The fraction of moles of that component in the total moles of gas mixture.

For example, the mole fraction of the first gas is given by...

$x_1 = \displaystyle \frac{n_1}{n_{\mbox{total}}} =\displaystyle \frac{n_1}{n_1+n_2+n_3+\cdots} $

A common use of partial pressures is when a gas is collected over a liquid which has an appreciable vapor pressure. For example, if oxygen is collected by displacement of H2O, and you want to know how much oxygen is present, we need to know that the oxygen is saturated with H2O vapor. The total pressure is the sum of the pressure of oxygen plus that of H2O.

ptotal = p(H2O) + p(O2)

O2 gas can be collected over water by decomposing KClO3 according to the reaction.

2 KClO3 2 KCl + 3 O2

If 0.044 liters of O2 saturated with H2O vapor at 0.020 atmospheres and a temperature of 18.6°C are collected, what volume would the O2 gas occupy at STP if it was dry (i.e., without water vapor)? The vapor pressure of H2O at 18.6°C is 0.020 atmospheres. The experiment was performed on a day when the atmospheric pressure of 0.983 atmospheres.

First we need to calculate the partial pressure of O2 according to

p(O2) = ptotal - p(H2O)

which gives

p(O2) = 0.983 atm. - 0.020 atm. = 0.963 atm.

Under our experimental conditions we then calculate

alternatetext

  • Partial Pressures:
  • Gas Stoichiometry:
  • Gas Stoichiometry - Analysis of Mixtures:
  • Gas Stoichiometry - Analysis of Mixtures with Collection Over Water: