Q1: At equilibrium the concentration of water in vapour phase (C^*)in kg/m³ of air space and the amount of water (m) adsorbed per kg of dry silica gel are related by, C^* = 0.0667m. To maintain dry conditions in a room of air space 100m³ containing 2.2 kg of water vapour initially, 10 kg of dry silica gel is kept in the room. The fraction of initial water remaining in the air space after a long time (during which the temperature is maintained constant) is

ANS:C - 0.4

To solve this problem, we need to use the concept of equilibrium between the water vapor in the air and the water adsorbed onto the silica gel. The equilibrium relationship given is 𝐶∗=0.0667𝑚C∗=0.0667m, where 𝐶∗C∗ is the concentration of water vapor in the air space in kg/m³ and 𝑚m is the amount of water adsorbed per kg of dry silica gel. Initially, the room contains 2.2 kg of water vapor in 100 m³ of air space, so the initial concentration of water vapor is 𝐶initial=2.2 kg100 m3=0.022 kg/m3Cinitial​=100 m32.2 kg​=0.022 kg/m3. Let's denote the amount of water adsorbed per kg of dry silica gel as 𝑚m. Initially, the total amount of water adsorbed onto the silica gel 𝑀initialMinitial​ is 𝑚×10 kgm×10 kg (since there are 10 kg of dry silica gel in the room). At equilibrium, the concentration of water vapor in the air space 𝐶∗C∗ will decrease as water is adsorbed onto the silica gel. According to the equilibrium relationship, 𝐶∗=0.0667𝑚C∗=0.0667m. At equilibrium, the total amount of water remaining in the air space will be equal to the initial amount minus the amount adsorbed onto the silica gel. Let's denote the fraction of initial water remaining in the air space as 𝑓f. So, at equilibrium: 𝐶∗=𝐶initial−𝑀initialVolume of air spaceC∗=Cinitial​−Volume of air spaceMinitial​​ Substituting the values we have: 0.0667𝑚=0.022−𝑚×101000.0667m=0.022−100m×10​ Solving for 𝑚m: 0.0667𝑚=0.022−0.1𝑚0.0667m=0.022−0.1m 0.1667𝑚=0.0220.1667m=0.022 𝑚=0.0220.1667≈0.132 kg/kg of silica gelm=0.16670.022​≈0.132 kg/kg of silica gel Now, at equilibrium, the amount of water remaining in the air space: 𝑀equilibrium=𝐶∗×Volume of air spaceMequilibrium​=C∗×Volume of air space =0.0667×100≈6.67 kg=0.0667×100≈6.67 kg Therefore, the fraction of initial water remaining in the air space after a long time is: 𝑓=𝑀equilibriumInitial amount of waterf=Initial amount of waterMequilibrium​​ =6.672.2=2.26.67​ =3.032=3.032 Since the fraction cannot exceed 1, it means all water has been adsorbed onto the silica gel. Thus, the correct answer is 1.01.0.