A mixture containing 0.40\text{ mol} of neon and 0.10\text{ mol} of argon is collected ove… — JAMB UTME

A mixture containing $0.40\text{ mol}$ of neon and $0.10\text{ mol}$ of argon is collected over water at $27^\circ\text{C}$ . If the partial pressure of neon in the dry gas mixture is $196.0\text{ kPa}$ and the saturated vapor pressure of water at $27^\circ\text{C}$ is $4.0\text{ kPa}$ , what is the total pressure of the wet gas mixture?

$249.0\text{ kPa}$ Answer
B
$245.0\text{ kPa}$
C
$241.0\text{ kPa}$
D
$200.0\text{ kPa}$

Answer

The total pressure of the wet gas mixture is

249.0\text{ kPa}

The total dry gas pressure is calculated from the partial pressure of neon and its mole fraction (

0.80

), yielding

245.0\text{ kPa}

. Adding the saturated vapor pressure of water (

4.0\text{ kPa}

) gives the total pressure of the wet gas as

249.0\text{ kPa}

Step-by-Step Solution

Calculate total moles of dry gas and the mole fraction of neon.

Total dry moles

= 0.40\text{ mol} + 0.10\text{ mol} = 0.50\text{ mol}

. Mole fraction of neon

\chi_{\text{Ne}} = 0.40 / 0.50 = 0.80

Dalton's Law states that partial pressure is proportional to mole fraction in a dry gas mixture.

Calculate the total pressure of the dry gas mixture.

P_{\text{dry total}} = P_{\text{Ne}} / \chi_{\text{Ne}} = 196.0\text{ kPa} / 0.80 = 245.0\text{ kPa}

Rearranging

P_{\text{Ne}} = \chi_{\text{Ne}} \times P_{\text{dry total}}

yields the total dry pressure.

Calculate the total pressure of the wet gas mixture collected over water.

P_{\text{total}} = P_{\text{dry total}} + P_{\text{H}_2\text{O}} = 245.0\text{ kPa} + 4.0\text{ kPa} = 249.0\text{ kPa}

Gas collected over water is saturated with water vapor, so total pressure is the sum of dry gas pressure and saturated vapor pressure of water.

Key Concept

Dalton's Law of Partial Pressures & Gas Collection Over Water

Estimated Time:1m 30s

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