The temperature and the relative humidity are 300 K and 20% in a room of volume 50 m³. The floor is washed with water, 500 g of water sticking on the floor. Assuming no communication with the surrounding, find the relative humidity when the floor dries. The changes in temperature and pressure may be neglected. Saturation vapour pressure at 300 K = 3.3 kPa.

Open in App

Solution

$\begin{array}{l} Here, \\ M = 18 g for water \\ m = 500 g \\ V = 50 m^{3} \\ T = 300 K \\ S V P = 3300 P a \\ R H = 20 % \\ \frac{V P}{S V P} = 0.2 \\ \Rightarrow V P = P_{1} = 0.2 \times 3300 = 660 Pa \\ {Partial pressure P}_{2} for evaporated water is given by \\ P_{2} V = \frac{m}{M} R T \\ \Rightarrow P_{2} = \frac{500}{18 \times 50} \times 8.31 \times 300 \\ \Rightarrow P_{2} = 1385 P a \\ Total pressure, P = P_{1} + P_{2} = 1385 + 660 = 2045 Pa \\ RH= \frac{P}{S V P} \times 100= \frac{2045}{3300} \times 100 % = 61.9 \approx 62 % \end{array}$

Suggest Corrections

Similar questions

Q. The temperature and relative humidity in a room are 300 K and 20% respectively. The volume of the room is 50 m³. The saturation vapour pressure at 300 K 3.3 kPa. Calculate the mass of the water vapour present in the room.

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.

The condition of air in a closed room is described as follows. Temperature = $25^{0}$ 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 $25^{0}$ C = 3.2 kPa.

Q. 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.

Q. The pressure, dry bulb temperature and relative humidity of air in a room are 1 bar,

30^{\circ} C

and 70%, respectively. If the saturated steam pressure at

30^{\circ} C

is 4.25 kPa, the specific humidity of the room air in kg of water vapour/kg dry air is

Join BYJU'S Learning Program

Here,M=18g for waterm=500gV=50 m3T=300KSVP=3300PaRH=20%VPSVP=0.2⇒VP=P1=0.2×3300=660PaPartial pressure P2 for evaporated water is given byP2V=mMRT⇒P2=50018×50×8.31×300⇒P2=1385PaTotal pressure, P=P1+P2=1385+660=2045PaRH=PSVP×100=20453300×100%=61.9≈62%