A glass contains some water at room temperature $20^{\circ} C$ . Refrigerated water is added to it slowly. When the temperature of the glass reaches $10^{\circ} C$ , small droplets condense on the outer surface. Calculate the relative humidity in the room. The boiling point of water at a pressure of 17.5 mm of mercury is $20^{\circ} C$ and at 8.9 mm of mercury it is $10^{\circ} C$ .

60 %

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75 %

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51 %

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35 %

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Solution

The correct option is C $51 %$
Given
SVP at the dew point = 8.9 mm SVP at room temp = 17.5 mm
Dew point = $10^{\circ} C$ as at this temp. the condensation starts
Room temp = $20^{\circ} C$
$R H = \frac{S V P a t d e w p o i n t}{S V P a t r o o m t e m p} = \frac{8.9}{17.5} = 0.508 = 51 %$

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The correct option is C 51%Given SVP at the dew point = 8.9 mm SVP at room temp = 17.5 mm Dew point =10∘C as at this temp. the condensation starts Room temp =20∘C RH=SVP at dew pointSVP at room temp=8.917.5=0.508=51%

The correct option is C $51 %$
Given
SVP at the dew point = 8.9 mm SVP at room temp = 17.5 mm
Dew point = $10^{\circ} C$ as at this temp. the condensation starts
Room temp = $20^{\circ} C$
$R H = \frac{S V P a t d e w p o i n t}{S V P a t r o o m t e m p} = \frac{8.9}{17.5} = 0.508 = 51 %$