A sealed room with a volume of 50\text{ m}^3 at a temperature of 20^\circ\text{C} contains… — JAMB UTME

A sealed room with a volume of $50\text{ m}^3$ at a temperature of $20^\circ\text{C}$ contains $0.40\text{ kg}$ of water vapour. If the mass of water vapour required to saturate $1\text{ m}^3$ of air at $20^\circ\text{C}$ is $0.016\text{ kg}$ , what is the relative humidity of the air in the room?

A
$25\%$
B
$40\%$
$50\%$ Answer
D
$80\%$

Answer

50\%

Relative humidity is defined as the ratio of the actual mass of water vapour present in a given volume of air to the mass of water vapour required to saturate the same volume at the same temperature. For

50\text{ m}^3

of air, the mass needed for saturation is

50\text{ m}^3 \times 0.016\text{ kg/m}^3 = 0.80\text{ kg}

. Dividing the actual mass (

0.40\text{ kg}

) by

0.80\text{ kg}

and multiplying by

100\%

gives

50\%

Step-by-Step Solution

Calculate the total mass of water vapour required to saturate the entire room volume.

\text{Total saturation mass} = 50\text{ m}^3 \times 0.016\text{ kg/m}^3 = 0.80\text{ kg}

The saturation density gives the maximum water vapour

1\text{ m}^3

can hold, so it must be scaled by the room volume.

Apply the formula for relative humidity.

\text{Relative Humidity} = \left(\frac{\text{Actual mass of water vapour}}{\text{Saturation mass of water vapour}}\right) \times 100\%

Relative humidity measures the degree of saturation of an air sample at a specific temperature.

Substitute the known values to find the relative humidity.

\text{Relative Humidity} = \left(\frac{0.40\text{ kg}}{0.80\text{ kg}}\right) \times 100\% = 50\%

Simplifying the fraction

\frac{0.40}{0.80} = 0.5

, which equals

50\%

Key Concept

Relative Humidity

Estimated Time:1m 15s

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