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Question
The condition of air in a closed room is described as follows. Temperature = 25°C, relative humidity = 60%, pressure = 104 kPa. If all the water vapour is removed from the room without changing the temperature, what will be the new pressure? The saturation vapour pressure at 25°C − 3.2 kPa.
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Solution
Here ,
T = 298 K
RH = 60%
`"P" = 1.04 xx 10^5 "Pa"`
RH = `"Vapour pressure of water vapour" /"saturated Vapour pressure"`
=0.6
`"Saturated vapour pressure"= 3.2 xx 10^3 "Pa"`
⇒`"vapour pressure of water vapour" ("VP") = 0.6 xx 3.2 xx 10^3 = 1.92 xx 10^3 "Pa"`
If the water vapour is completely removed from the air , then net pressure = `1.04 xx 10^5 - 1.92 xx 10^3`
=`1.02 xx 10^5 "Pa"`
=102 kPa
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