Q1: Acetone is to be removed from air in an isothermal dilute absorber using pure water as solvent. The incoming air contains 5 mole% of acetone (y_in = 0.05). The design equation to be used for obtaining the number of trays (N) of the absorber is, N+2 = 6 log (y_in/y_out).For 98% recovery of acetone, the number of trays required is/are

ANS:C - 9

To determine the number of trays required for 98% recovery of acetone from air in the isothermal dilute absorber using pure water as a solvent, we can use the given design equation: 𝑁+2=6log⁡(𝑦in𝑦out)N+2=6log(yout​yin​​) Given:

• 𝑦in=0.05yin​=0.05 (mole fraction of acetone in incoming air)
• 𝑦out=0.02yout​=0.02 (desired mole fraction of acetone in outgoing air for 98% recovery)

Substituting these values into the equation: 𝑁+2=6log⁡(0.050.02)N+2=6log(0.020.05​) 𝑁+2=6log⁡(2.5)N+2=6log(2.5) 𝑁+2=6×0.3979N+2=6×0.3979 𝑁+2=2.3874N+2=2.3874 𝑁≈2.3874−2N≈2.3874−2 𝑁≈0.3874N≈0.3874 Since the number of trays (N) must be a whole number, we round up to the nearest integer: 𝑁≈1N≈1 Therefore, the number of trays required for 98% recovery of acetone is 1.