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Kinetic Theory — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysicsKinetic Theory5 Marks
Question
A bucket full of water is placed in a room at $15^\circ C$ with initial relative humidity $40\%$. The volume of the room is $50m^3.$
  1. How much water will evaporate?
  2. lf the room temperature is increased by $5^\circ C, $ how much more water will evaporate? The saturation vapour pressure of water at $15^\circ C$ and $20^\circ C$ are $1.6$kPa and $2.4$kPa respectively.
✓

Answer

  1. Rel. humidity $=\frac{\text{VP}}{\text{SVP at }15^\circ\text{C}}\Rightarrow0.4=\frac{\text{VP}}{1.6\times10^3}$
$\Rightarrow\text{VP}=0.4\times1.6\times10^3$
The evaporation occurs as along as the atmosphere does not become saturated.
Net pressure change $= 1.6 \times 10^3 - 0.4 \times 1.6 \times 10^3 = (1.6 - 0.4 \times 1.6)10^3 = 0.96 \times 10^3$
Net mass of water evaporated $=\text{m}\Rightarrow0.96\times10^3\times50=\frac{\text{m}}{18}\times8.3\times288$
$\Rightarrow\text{m}=\frac{0.96\times50\times18\times10^3}{8.3\times288}=361.45\approx361\text{g}$
  1. At $20^\circ C$ SVP $= 2.4$KPa, At $15^\circ C$ SVP $= 1.6$KPa
Net pressure charge $= (2.4 - 1.6) \times 10^3Pa = 0.8 \times 10^3Pa$
Mass of water evaporated $=\text{m}'=0.8\times10^3\ 50=\frac{\text{m}'}{18}\times8.3\times293$
$\Rightarrow\text{m}'=\frac{0.8\times50\times18\times10^3}{8.3\times293}=296.06\approx296\text{grams}$

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