Chemistry 122 d

Question

A gas is collected over water at 25•C, and the volume inside the inverted container is equalized with the water level outside the container. The barometric pressure is 738.5 Torr, and the vapor pressure of the water is 23.8 Torr at 25•C. Calculate the “dry” pressure of the gas.

Answer 1

You need to use Dalton’s Law of Partial Pressure where Pt = Pa + Pa ...

The total pressure of a gas within a container equals the partial pressure of each individual gas that is found within that container added together.

When you collect a gas produced over water, as the gas moves through the water it combines with the water vapor to give you a mixture of gas that was produced (collected) plus the water vapor giving you the total pressure of both gases combined per Dalton’s Law of Partial Pressure.

In this problem you are given the Total Pressure of the gas that is being collected added to the H2O vapor, Pt = 738.5 Torr (unknown gas + H2O) and the vapor pressure of the water alone, Pa H2O = 23.8 Torr

Then using the above equation:

Pt = Pa (unknown gas) + Pa H20

738.5 Torr = Pa (unknown gas) + 23.8 Torr

Pa(dry unknown gas) = 738.5 - 23.8

Pa( dry unknown gas) = 714.7 Torr