Question 19, case study question, second afternoon of 2010
Thread Content
19. Water is used to absorb ammonia from an air-ammonia mixture; when the total pressure of the mixture is 101 Kpa, the partial pressure of ammonia is 1300 Pa. It is known that this system obeys Henry’s law, and the following equilibrium relationship holds: when 2 g of ammonia is present in 100 g of water, the partial pressure of ammonia is 1600 Pa. When a 95% recovery rate for ammonia is required, what is the minimum water consumption (kg/h) for processing 1000 kg of air per hour? (A) 236 (B) 262 (C) 448 (D) 388Solution:
xa = (2/17) / (100/18 + 2/17) = 0.02074
E = Pa * xa = 1.6 / 0.02074 = 77.15 (kPa)
m = E / P = 77.15 / 101 = 0.764
Y = mX = 0.764X
y1 = 1.3 / 101 = 0.01287
Y1 = 0.01287 / (1 – 0.01287) = 0.01304 (L/V)min
mφA = 0.764 × 0.95 = 0.7258 Lmin
V = 0.7256 × 1000 × (1 – 0.01304) × 18 / 29 ≈ 448 (kg/h)
Answer: C
I can’t understand the parts in red font at the end of the answer provided by a forum user; could that user please explain them? Also, by directly using Lmin=V*(Y1-Y2)/(Y1/m-X2), should that formula work as well?
I can’t understand the parts in red font; could a fellow user help explain them?
(L/V)min = (Y1 – Y2) / (Y1/m – X2). Since Y1 – Y2 = 0.95Y1 and X2 = 0, then (L/V)min = 0.95m. V = 1000/29 mol, and L = 450.37 kg
In this problem, the air mentioned is to be understood as a mixture; the molar mass of this mixture is calculated first, followed by determining the amount in moles, from which the molar amount of air can be determined. It’s exactly 448