Question

A bucket full of water is placed in a room at 15°C with initial relative humidity 40%. The volume of the room is 50 m3. (a) How much water will evaporate? (b) If the room temperature is increased by 5°C, how much more water will evaporate? The saturation vapour pressure of water at 15°C and 20°C are 1.6 kPa and 2.4 kPa respectively.

Answer 1

Explanation:

(a) given data in the question  

Relative humidity = 40 %

We know that  

Relative humidity =

$\frac{\text { vapour pressure }}{\text { Saturated vapour pressure at } 15^{\circ} \mathrm{C}}$

$0.4=\frac{v p}{1.6} \times 10^{3}$

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

$V P=0.64 \times 10^{3}$

$V P=640$

Evaporation takes place insofar as the atmosphere is not saturated.

Total pressure change  $=1.6 \times 10^{3}-640$

$=1600-640$

$=0.96 \times 10^{3}$

$=960$

Let the mass of evaporated water be m. Then  

$960 \times 50=\frac{m \times 8.3 \times 288}{18}$

$48000=\frac{m \times 2390.4}{18}$

$m=\frac{48000 \times 18}{2390.4}$

$\begin{aligned}&m=\frac{864000}{2390.4}\\& =361.45 \approx 361 g\end{aligned}$

(b) for  20°C,Saturated vapour pressure = 2.4 KPa

for 15°C,Saturated vapour pressure = 1.6 KPa  

Total pressure change $=(2.4-1.6) \times 10^{3} P a$

$=0.8 \times 10^{3} P a$

Mass of evaporated water is provided by

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

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

$=296.06 \approx 296 g$