The temperature and the relative humidity are 300 K and 20 % in a room of volume 50 m a. 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.

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Solution

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

$\Rightarrow 0.20 = \frac{V P}{3.3 \times 10^{3}}$

$\Rightarrow V P = 0.20 \times 3.3 \times 10^{3}$

= 660

$\Rightarrow$ Again PV = nRT

$\Rightarrow P = \frac{n R T}{V} = \frac{m}{M} \times \frac{R \times T}{V}$

= $\frac{500 \times 8.3 \times 300}{18 \times 50}$

= 1383.3

Net P = 1383.3 +660 = 2043.3

Now RH = $\frac{2043.3}{3300}$ = 0.619= 62 %

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RH=VPSVP ⇒0.20=VP3.3×103 ⇒VP=0.20×3.3×103 = 660 ⇒ Again PV = nRT ⇒P=nRTV=mM×R×TV =500×8.3×30018×50 = 1383.3 Net P = 1383.3 +660 = 2043.3 Now RH = 2043.33300 = 0.619= 62 %