A glass contains some water at room temperature $20^{0}$ C. Refrigerated water is added to it slowly. When the temperature of the glass reaches $10^{0}$ 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 $20^{0}$ C and at 8.9 mm of mercury it is $10^{0}$ C.

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Solution

Given:

SVF at the dew point = 8.9 mm

SVP at room temperature = 17.5 mm

Dew point = $10^{0}$ C as at this temperature the condensation starts

Room temperature = $20^{0}$ C

$R_{H} = \frac{SVP at dew point}{SVP at room temp.}$

= $\frac{8.9}{17.5}$ = 51 %

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Given: SVF at the dew point = 8.9 mm SVP at room temperature = 17.5 mm Dew point = 100 C as at this temperature the condensation starts Room temperature = 200 C RH=SVP at dew pointSVP at room temp. = 8.917.5 = 51 %

Given:

SVF at the dew point = 8.9 mm

SVP at room temperature = 17.5 mm

Dew point = $10^{0}$ C as at this temperature the condensation starts

Room temperature = $20^{0}$ C

$R_{H} = \frac{SVP at dew point}{SVP at room temp.}$

= $\frac{8.9}{17.5}$ = 51 %