Chemistry Labs

Problem 5

A Dubai desalination plant uses multi-stage flash (MSF) desalination: seawater is heated at high pressure then flashed, releasing pure water vapour. Dubai seawater is at T0=25.00T_0 = 25.00 °C and contains 3.45 % NaCl by mass (assume complete ionisation); the boiling point elevation constant is Kb=0.5120K_b = 0.5120 K kg mol−1^{-1} and the latent heat of vaporisation of water is Evap=2260E_{\text{vap}} = 2260 kJ kg−1^{-1} (40.716 kJ mol−1^{-1}). (a) Calculate the boiling point of Dubai seawater at atmospheric pressure. (b) Calculate its boiling point at p=2.50p = 2.50 atm using the Clausius–Clapeyron equation. (c) A 100 L flash chamber holds 1.00 kg of seawater at 90.0 °C; a second 1.00 kg portion overheated to 110.0 °C is added, pressure is reduced, and the chamber equilibrates at Tf=97.0T_f = 97.0 °C. With cp=3.85c_p = 3.85 kJ kg−1^{-1} K−1^{-1}, calculate the amount nn (mol) of water that vaporised. (d) The plant produces 50 000 m³ of pure water per day, extracting overall 85 % of the water present in the seawater; calculate the mass of Dubai seawater needed per day.
Step 2 of 4: Boiling under pressure
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ln⁡2.501.00=−ΔvapHR(1T2−1373.78)⇒T2=401.9 K≈128.7 °C\ln\dfrac{2.50}{1.00} = -\dfrac{\Delta_{\text{vap}}H}{R}\left(\dfrac{1}{T_2}-\dfrac{1}{373.78}\right) \Rightarrow T_2 = 401.9\ \text{K} \approx 128.7\ \text{°C}
Analysis

Clausius–Clapeyron with ΔvapH=40716\Delta_{\text{vap}}H = 40716 J mol−1^{-1}: 1T2=1373.78−8.314ln⁡2.540716=2.4883×10−3\frac{1}{T_2} = \frac{1}{373.78} - \frac{8.314\ln 2.5}{40716} = 2.4883\times 10^{-3} K−1^{-1}, so T2=401.9T_2 = 401.9 K ≈128.7\approx 128.7 °C.