A bucket full of water is placed in a room at $15^{0}$ C with initial relative humidity 40 %. The volume of the room is 50 $m^{3}$ . (a) How much water will evaporate ? (b) If the room temperature is increased by $5^{0}$ C, how much more water will evaporate ? The saturation vapour pressure of water at $15^{0}$ C and $20^{0}$ C are 1.6 kPa and 2.4 kPa respectively.

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Solution

(a) Relative humidity

= $\frac{V P}{S V P at 15^{0} C}$

$\Rightarrow 0.4 = \frac{V P}{1.6 \times 10^{3}}$

$\Rightarrow V P = 0.4 \times 1.6 \times 10^{3}$

The evaporation occurs as long as the atmosphere does not saturated Net pressure change

= $1.6 \times 10^{3} - 0.4 \times 1.6 \times 10^{3}$

= $(1.6 - 0.4 \times 1.6) 10^{3}$

= $0.96 \times 10^{3}$

Mass of water evaporated = m

$\Rightarrow 0.96 \times 10^{3} \times 50 = \frac{m \times 8.3 \times 288}{18}$

$\Rightarrow m = \frac{0.96 \times 50 \times 18 \times 10^{3}}{8.33 \times 288}$

= 361.45 $\approx$ 361 g

(b) At $20^{0}$ C, SVp = 2.4 KPa,

At $15^{0}$ C, SVP = 1.6 KPa

Net pressure change

= $(2.4 - 1.6) \times 10^{8}$ Pa

= $0.8 \times 10^{3}$ Pa

Mass of water evaporated

= m = $\frac{m^{'} \times 8.3 \times 293}{18}$

$\Rightarrow m^{'} = \frac{0.8 \times 50 \times 18 \times 10^{3}}{8.3 \times 293}$

= 296.06 $\approx$ 296 grams.

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