Question

The condition of air in a closed room is described as follows. Temperature = 25°C, relative humidity = 60%, pressure = 104 kPa. If all the water vapour is removed from the room without changing the temperature, what will be the new pressure? The saturation vapour pressure at 25°C − 3.2 kPa.

Answer 1

$$\begin{gathered} \begin{array}{l}\text{Here,}\\ \text{T = 298K}\\ \text{RH = 60\%}\\ \text{P = 1}\text{.04}\times {\text{10}}^5\text{pa}\\ RH=\frac{\text{Vapour pressure of water vapour}}{\text{Saturated vapour pressure}}\\ =0.6\\ \text{Saturated vapour pressure = 3}\text{.2}\times {\text{10}}^3\text{Pa}\\ \Rightarrow \text{Vapour pressure of water vapour (VP)}=\text{0}\text{.6}\times \text{3}\text{.2}\times {\text{10}}^3\text{=1}\text{.92}\times {\text{10}}^3\text{Pa}\\ \text{If the water vapour is completely removed from the air, then net pressure =1}\text{.04}\times {\text{10}}^5-1.92\times {10}^3\\ =1.02\times {10}^5\text{Pa}\end{array} \\ \text{=102 kPa} \end{gathered}$$