The temperature and relative humidity in a room are 300 K and 20 % respectively. The volume of the room is 50 ma. The saturation vapour pressure at 300 K is 3-3 kPa. Calculate the mass of the water vapour present in the room.

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Solution

Given, T = 300K,

Relative humidity = 20 %,

V= 50 $m^{3}$

SVP at 300 K = 3.3 KPa

VP = Relative humidity \times SVP

= $(0.2 \times 3.3 \times 10^{3})$

PV = $\frac{m R T}{M}$

$\Rightarrow 0.2 \times 3.3 \times 10^{3} \times 50$

= $\frac{m \times 8.3 \times 300}{18}$

$\Rightarrow m = \frac{0.2 \times 3.3 \times 50 \times 18 \times 10^{3}}{8.3 \times 300}$

= 238.55 g $\approx$ 238 g

Mass of water present in the room = 238 g

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The temperature and relative humidity in a room are 300 K and 20 % respectively. The volume of the room is 50 ma. The saturation vapour pressure at 300 K is 3-3 kPa. Calculate the mass of the water vapour present in the room.

Given, T = 300K, Relative humidity = 20 %, V= 50 m3 SVP at 300 K = 3.3 KPa VP = Relative humidity \times SVP = (0.2×3.3×103) PV = mRTM ⇒0.2×3.3×103×50 = m×8.3×30018 ⇒m=0.2×3.3×50×18×1038.3×300 = 238.55 g ≈ 238 g Mass of water present in the room = 238 g