A sample of gas collected over water at 20^\circ\text{C} exerts a total pressure of 748.0\… — JAMB UTME

A sample of gas collected over water at $20^\circ\text{C}$ exerts a total pressure of $748.0\text{ mmHg}$ . If the saturated vapor pressure of water at $20^\circ\text{C}$ is $17.5\text{ mmHg}$ , what is the partial pressure of the dry gas?

$730.5\text{ mmHg}$ Answer
B
$765.5\text{ mmHg}$
C
$748.0\text{ mmHg}$
D
$713.0\text{ mmHg}$

Answer

730.5\text{ mmHg}

According to Dalton's Law of Partial Pressures, when a gas is collected over water, the total pressure (

P_{\text{total}}

) is equal to the sum of the partial pressure of the dry gas (

P_{\text{dry gas}}

) and the saturated vapor pressure of water (

P_{\text{water vapor}}

). Therefore,

P_{\text{dry gas}} = P_{\text{total}} - P_{\text{water vapor}} = 748.0\text{ mmHg} - 17.5\text{ mmHg} = 730.5\text{ mmHg}

Step-by-Step Solution

Identify the relationship given by Dalton's Law of Partial Pressures for a gas collected over water.

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

Gas collected over water is saturated with water vapor, so the total pressure is the sum of the dry gas partial pressure and the saturated vapor pressure of water (aqueous tension).

Rearrange the equation to solve for the partial pressure of the dry gas.

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

Isolating the unknown variable

P_{\text{dry gas}}

Substitute the given values into the equation.

P_{\text{dry gas}} = 748.0\text{ mmHg} - 17.5\text{ mmHg} = 730.5\text{ mmHg}

Subtracting

17.5\text{ mmHg}

from

748.0\text{ mmHg}

yields the partial pressure exerted by the dry gas alone.

Key Concept

Dalton's Law of Partial Pressures and Collection of Gas over Water (Aqueous Tension Adjustment)

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