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 3-3 kPa- Calculate the mass of the water vapour present in the room-

Here,T=300KSVP=3300 #160;Pa #160;at #160;300 #160;KRH=20 ...

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Standard XII

Physics

Anomalous Expansion of Water

The temperatu...

Question

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

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Solution

$\begin{array}{l} Here, \\ T = 300 K \\ S V P = 3300 Pa at 300 K \\ R H = 20 % \\ \Rightarrow \frac{P}{S V P} = 0.2 \\ \Rightarrow P = 0.2 \times S V P = 0.2 \times 3300 = 660 \\ V {=50 m}^{3} \\ M = 18 g \\ Now, \\ PV = nRT \\ ⇒ PV = \frac{m}{M} R T \\ \Rightarrow 660 \times 50 = \frac{m}{18} \times 8.3 \times 300 \\ \Rightarrow m = 238.55 g≈238 g \end{array}$

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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 m3. The saturation vapour pressure at 300 K 3.3 kPa. Calculate the mass of the water vapour present in the room.

Here,T=300KSVP=3300 Pa at 300 KRH=20%⇒PSVP=0.2⇒P=0.2×SVP=0.2×3300=660V=50 m3M=18 g Now,PV = nRT⇒ PV = mMRT⇒660×50=m18×8.3×300⇒m=238.55 g≈238 g

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