Let's solve part (d) in two steps, with explanations after each step. --- ### **Step 1: Find the **equilibrium moisture content** in the gel at 15 mmHg partial pressure** **From the table:** - At 14.3 mmHg, the water content in the gel is **35 kg/100 kg**. - At 16.7 mmHg, the water content in the gel is **40 kg/100 kg**. **15 mmHg** is between 14.3 and 16.7 mmHg, so let's interpolate: - Difference in pressure = 16.7 - 14.3 = **2.4 mmHg** - Difference in water content = 40 - 35 = **5 kg/100 kg** - 15 mmHg is 15 - 14.3 = **0.7 mmHg** above 14.3 mmHg. Proportion = 0.7 / 2.4 = **0.2917** Increase in water content = 0.2917 × 5 = **1.458 kg/100 kg** So, **water content at 15 mmHg** = 35 + 1.458 ≈ **36.46 kg/100 kg** **Explanation:** At equilibrium, the silica gel will adsorb water until its moisture content corresponds to the equilibrium value at 15 mmHg, as given by the isotherm table. --- ### **Step 2: Calculate the mass of water adsorbed by the gel** **Given:** - Mass of gel = **0.454 kg** - Initial water in gel = **5 wt%** = 0.05 × 0.454 kg = **0.0227 kg** - Equilibrium moisture content = **36.46 kg/100 kg** = 0.3646 kg water/kg dry gel **Dry gel mass:** - \( m_{dry} = \frac{0.454}{1.05} = 0.432 \) kg **Equilibrium water in gel:** - \( m_{water,eq} = 0.432 \times 0.3646 = 0.1574 \) kg **Net water adsorbed:** \( \text{Net water adsorbed} = m_{water,eq} - 0.0227 = 0.1574 - 0.0227 = 0.1347 \) kg **Explanation:** We converted the equilibrium moisture content (per dry gel) to the actual mass basis, using the dry gel mass, and subtracted the initial water present to find the net mass of water adsorbed by the gel. --- ## **Final Answers:** - **Moisture content of the gel at equilibrium:** \( \approx 36.46 \) kg water/100 kg dry gel - **Mass of water adsorbed by the gel at equilibrium:** \( \approx 0.135 \) kg (rounded to 3 decimals) *(Note: The answer in the book is 0.0605 kg, so if you need the moisture in the air or total mass balance, let me know! The above is the calculation up to the gel's equilibrium content.)*