The temperature and relative humidity in a room are $300 K$ and $20 %$ respectively. The volume of the room is $50 m^{3}$ . The saturation vapour pressure at $300 K$ is $3.3 k P a$ . Calculate the mass of the water vapour present in the room.

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Solution

$T = 300 K,$
Relative humidity $= 20 %$ , $V = 50 m^{3}$
SVP at $300 K = 3.3 K P a$ ,
$V . P =$ Relative humidity $\times S V P = 0.2 \times 3.3 \times 10^{3}$

$P V = \frac{m}{M} R T \Rightarrow 0.2 \times 3.3 \times 10^{3} \times 50 = \frac{m}{18} \times 8.3 \times 300$

$\Rightarrow m = \frac{0.2 \times 3.3 \times 50 \times 18 \times 10^{3}}{8.3 \times 300} = 238.55$ grams $\approx 238 g$
Mass of water present in the room $= 238 g$

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The temperature and relative humidity in a room are 300K and 20% respectively. The volume of the room is 50m3. The saturation vapour pressure at 300K is 3.3kPa. Calculate the mass of the water vapour present in the room.

T=300K, Relative humidity =20%, V=50m3SVP at 300K=3.3KPa, V.P= Relative humidity ×SVP=0.2×3.3×103PV=mMRT⇒0.2×3.3×103×50=m18×8.3×300⇒m=0.2×3.3×50×18×1038.3×300=238.55 grams ≈238gMass of water present in the room =238g

The temperature and relative humidity in a room are $300 K$ and $20 %$ respectively. The volume of the room is $50 m^{3}$ . The saturation vapour pressure at $300 K$ is $3.3 k P a$ . Calculate the mass of the water vapour present in the room.