Chemistry

The equilibrium partial pressure of water vapor in contact with a certain silica gel on which water is adsorbed is, at 25°C, as follows: Partial pressure of water, mmHg 0 2.14 4.74 7.13 9.05 10.9 12.6 14.3 16.7 Ib water/ 100 lb dry gel – kg/100 kg 0 5 10 15 20 25 30 35 40 (a) Plot the equilibrium data as P – partial pressure of water vapor against x – wt fraction water in the gel (6) Plot the equilibrium data as X – mol water/mass dry gel, Y = mol water vapor/mol dry air, for a total pressure of 1 std atm. (c) When 10 lb (4.54 kg) of silica gel containing 5 wt% adsorbed water is placed in a flowing airstream containing a partial pressure of water vapor – 12 mmHg, the total pressure is 1 std atm and the temperature 25°C. When equilibrium is reached, what mass of additional water will the gel have adsorbed? Air is not adsorbed. (d) One pound mass (0.454 kg) of silica gel containing 5 wt% adsorbed water is placed in a vessel in which there are 400 ft (11.33 m) moist air whose partial pressure of water is 15 mmHg. The total pressure and temperature are kept at 1 std atm and 25°C, respectively. At equilibrium, what will be the moisture content of the air and gel and the mass of water adsorbed by the gel? Ans: 0.0605 kg water. (e) Write the equation of the operating line for part (d) in terms of X and Y. Convert this into an equation in terms of p and x, and plot the operating curve on P, x coordinates. (1) If 1 lb (0.454 kg) of silica gel containing 18% adsorbed water is placed in a vessel containing 500 ft (14.16 m?) dry air and the temperature and pressure are maintained at 25°C and 1 std atm, respectively, what will be the final equilibrium moisture content of the air and the gel. (8) Repeat part ( for a total pressure of 2 std atm. Note that the equilibrium curve in terms of X and Y previously used is not applicable.