Question

The temperature and the relative humidity are 300 K and 20% in a room of volume 50 m3. The floor is washed with water, 500 g of water sticking on the floor. Assuming no communication with the surrounding, find the relative humidity when the floor dries. The changes in temperature and pressure may be neglected. Saturation vapour pressure at 300 K = 3.3 kPa.

Answer 1

The relative humidity when the floor dries is 62%.

• Given,
• T=300K
• RH=20%
• V=50m^3
• m=500g
• SVP=3300Pa
• RH=20%=$\frac{20}{100}$
• ⇒$\dfrac{VP}{SVP}=0.2$
• ∴VP=0.2×3300=660Pa
• ⇒$P_1=660Pa$
• for the evaporated water, the partial pressure is given by,
• $P_2=\dfrac{m}{M}\dfrac{RT}{V}$

• here, M=18 g for water
• $P_2=\dfrac{500}{18}\dfrac{8.314*300}{50}$
• ⇒$P_2=1385Pa$
• Total pressure,
• $P=P_1+P_2$
• =660+1385=2045Pa
• Now,

• $RH=\dfrac{P}{SPV} *100\\\\=\dfrac{2045}{3300} *100%\\\\\\\\=61.9%$

Answer 2

Explanation:

Given data in the question  

Step 1 :

Mass for water$(M)=18 g$  

$V=50 \mathrm{m}^{3}$

$m=500 g$

$T=300 K$

Saturated vapour pressure$=3300 P a$

RH$=20 \%$

(Saturated vapour pressure)/(vapour pressure)  $=\frac{20}{100}=0.2$  

vapour pressure=$P_{1}=0.2 \times 3300=660 P a$

Step 2:

Partial pressure is given by $P_{2}$ for evaporated water

$P_{2} V=\frac{m}{M} R T$

$P_{2}=\frac{500}{18 \times 50} \times 8.31 \times 300$

$P_{2}=\frac{500}{900} \times 2493$

$P_{2}=\frac{5}{9} \times 2493$

$P_{2}=0.55 \times 2493$

$P_{2}=1385 \mathrm{Pa}$

Step 3:

net pressure, $P=P_{1}+P_{2}=1385+660=2045 P a$

RH ( Relative humidity )=P/(Saturated vapour pressure)×100  

$=\frac{2045}{3300} \times 100 \%$

$=61.9 \approx 62 \%$