Question

A sample of oxygen gas was collected via water displacement. Since the oxygen was collected via water displacement, the sample is saturated with water vapor. If the total pressure of the mixture at 26.4 °C is 805 torr, what is the partial pressure of oxygen? The vapor pressure of water at 26.4 °C is 25.81 mm Hg.

Answer 1

Answer :  The partial pressure of oxygen is, 799.19 torr

Solution :

According to the Dalton's law, the total pressure of the gas is equal to the sum of the partial pressure of the mixture of gasses.

$P_T=p_{H_2O}+p_{O_2}$

where,

$P_T$ = total partial pressure = 805 torr

$P_{O_2}$ = partial pressure of oxygen gas = ?

$P_{H_2O}$ = partial pressure of water = 25.81 mm Hg = 25.81 torr

Now put all the given values is expression, we get the partial pressure of the oxygen gas.

$805\text{ torr}=25.81\text{ torr}+p_{O_2}$

$p_{O_2}=779.19\text{ torr}$

Therefore, the partial pressure of oxygen gas is, 799.19 torr