- 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

n_{total}=
n_{1}
+
n_{2}
+
n_{3}
+ ...

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 H_{2}O, and you want to know how much oxygen is present, we need to know that the oxygen is saturated with H_{2}O vapor. The total pressure is the sum of the pressure of oxygen plus that of H_{2}O.

p_{total} = p(H_{2}O) + p(O_{2})

**O _{2} gas can be collected over water by decomposing KClO_{3} according to the reaction.**

2 KClO_{3}
→
2 KCl
+
3 O_{2}

**If 0.044 liters of O _{2} saturated with H_{2}O vapor at 0.020 atmospheres and a temperature of 18.6°C are collected, what volume would the O_{2} gas occupy at STP if it was dry (i.e., without water vapor)? The vapor pressure of H_{2}O 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 O_{2} according to

p(O_{2}) = p_{total} - p(H_{2}O)

which gives

p(O_{2}) = 0.983 atm. - 0.020 atm. = 0.963 atm.

Under our experimental conditions we then calculate

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