The condition of air in a closed room is described as follows. Temperature = $25^{0}$ 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 $25^{0}$ C = 3.2 kPa.

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Solution

T = $25^{0}$ C,

P = 104 kPa

$R_{H} = \frac{V P}{S V P}$

[SVP =3.2 KPa, R_H = 0.6]

$V P = 0.6 \times 3.2 \times 10^{3}$

= $1.92 \times 10^{3} = 2 \times 10^{3}$

When vapour presusre removed VP reduces to zero

Net pressure inside the room now

= $104 \times 10^{3} - 2 \times 10^{3}$

= $102 \times 10^{3}$ = 102 KPa

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The condition of air in a closed room is described as follows. Temperature = 250 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 250 C = 3.2 kPa.

T = 250 C, P = 104 kPa RH=VPSVP [SVP =3.2 KPa, R_H = 0.6] VP=0.6×3.2×103 = 1.92×103=2×103 When vapour presusre removed VP reduces to zero Net pressure inside the room now = 104×103−2×103 = 102×103 = 102 KPa