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

The new pressure is $102 k P a$

Explanation:

Step 1:

given data in the question  

T = $298 K$

RH = $60 \%$

P = $104 \mathrm{kPa}=104 \times 10^{3} \mathrm{Pa}=1.04 \times 10^{5} \mathrm{pa}$

RH(Relative humidity)=Vapour pressure of water vapour/Saturated vapour pressure    

=$\frac{60}{100}=0.6$

Saturated vapour pressure = $3.2 \times 10^{3} P a$

Vapour pressure of water vapour (VP)= $0.6 \times 3.2 \times 10^{3}=1.92 \times 10^{3} P a$

Step 2:

If the water vapor is removed from the air completely, then the net pressure is

= $1.04 \times 10^{5}-1.92 \times 10^{3}$      

=$10^{3}\left(1.04 \times 10^{2}-1.92\right)$

=$10^{3}(104-1.92)$  

=$10^{3}(102.08)$

=$102.08 \times 10^{3} P a$                                                                                                                                $=102.08 \mathrm{kPa} \approx 102 \mathrm{kPa}$