A sample of oxygen gas is collected over water at 27^\circ\text{C} and a total pressure of… — JAMB UTME

A sample of oxygen gas is collected over water at $27^\circ\text{C}$ and a total pressure of $755\text{ mmHg}$ . If the saturated vapor pressure of water at $27^\circ\text{C}$ is $25\text{ mmHg}$ , what is the partial pressure of the dry oxygen gas in $\text{mmHg}$ ?

Answer: 730 mmHg

Answer

The partial pressure of the dry oxygen gas is

730\text{ mmHg}

According to Dalton's Law of Partial Pressures, the total pressure exerted by a mixture of gases is equal to the sum of the partial pressures of the component gases. When a gas is collected over water, the gas absorbs water vapor, so

P_{\text{total}} = P_{\text{dry gas}} + P_{\text{water vapor}}

. To find the pressure of the dry oxygen, the water vapor pressure (

25\text{ mmHg}

) must be subtracted from the total barometric pressure (

755\text{ mmHg}

), giving

730\text{ mmHg}

Step-by-Step Solution

Identify Dalton's Law equation for a gas collected over water

P_{\text{total}} = P_{\text{dry gas}} + P_{\text{water}}

When a gas is collected over water, it becomes saturated with water vapor. The total observed pressure is the sum of the partial pressure of the dry gas and the vapor pressure of water (aqueous tension).

Subtract the aqueous tension from the total pressure

P_{\text{dry gas}} = 755\text{ mmHg} - 25\text{ mmHg} = 730\text{ mmHg}

Isolating

P_{\text{dry gas}}

gives the pressure exerted purely by the collected oxygen gas.

Key Concept

Dalton's Law of Partial Pressures and Collection of Gas over Water

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