Moist air is filled inside a room with total pressure of 100 kPa, and at $25^{\circ}$ C. If the moist air is saturated with mass of dry air as 50 kg, the mass of vapour in the room is (use saturation pressure of water at $25^{\circ} C$ as 3.17 kPa)

10.18 kg

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1.018 kg

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9.98 kg

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0.998 kg

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Solution

The correct option is B 1.018 kg
$ω = 0.622 \times \frac{P_{v}}{P_{t} - P_{v}} = 0.622 \times \frac{3.17}{100 - 3.17}$

$ω = 0.020363 k g / k g d . a .$

$∴ ω = \frac{mass of vapour}{mass of dry air} = 0.020363$

Mass of vapour $= 50 \times 0.020363 = 1.018 k g$

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Moist air is filled inside a room with total pressure of 100 kPa, and at 25∘C. If the moist air is saturated with mass of dry air as 50 kg, the mass of vapour in the room is (use saturation pressure of water at 25∘C as 3.17 kPa)

The correct option is B 1.018 kgω=0.622×PvPt−Pv=0.622×3.17100−3.17 ω=0.020363kg/kgd.a. ∴ω=mass of vapourmass of dry air=0.020363 Mass of vapour =50×0.020363=1.018kg

Moist air is filled inside a room with total pressure of 100 kPa, and at $25^{\circ}$ C. If the moist air is saturated with mass of dry air as 50 kg, the mass of vapour in the room is (use saturation pressure of water at $25^{\circ} C$ as 3.17 kPa)

The correct option is B 1.018 kg
$ω = 0.622 \times \frac{P_{v}}{P_{t} - P_{v}} = 0.622 \times \frac{3.17}{100 - 3.17}$

$ω = 0.020363 k g / k g d . a .$

$∴ ω = \frac{mass of vapour}{mass of dry air} = 0.020363$

Mass of vapour $= 50 \times 0.020363 = 1.018 k g$