A gas containing H2S and CO2 is bubbled through methanol at 4 degrees C. From th

Question

A gas containing H2S and CO2 is bubbled through methanol at 4 degrees C. From the bottom of the tank to the top, 80% of the H2S is absorbed by the methanol. The volume fraction of H2S in the entering gas is .6. The tank is 20 ft tall, and the gas is introduced at the bottom of the tank. The volumetric flow rate of the entering gas is 15 L/min. The gas enters at room temperature, but by the time it has bubbled through the tank, the temperature can be assumed to be the same temp as the methanol.

a.) What is the volumetric flow rate of the exiting gas?

b.) What pressure is the tank under?

Explanation / Answer

a) Amount of H2S present in the gas enetring = 0.6* 15 = 9 L/min

Taking room temperature to be 25 C = 298 K

If "M' is number of moles of gas entering,

If pressure of the tank is P

no. of moles enetring = (PV?RT) = (P*15 / 8.314*298) = 0.0060543 P

no.of moles of H2S absorbed = M*0.6*0.8 = 0.48 M

No.of moles in the leaving gas = M-0.48 M = 0.52 M

volume of this 0.52 M moles of gas at exit = (nRT/P) = (0.52 M * 8.314 * 277/P)

volumetric flow rate of the exiting gas = 0.52 * 0.0060543 P* 8.314 * 277/P = 7.25 L/min

volumetric flow rate of the exiting gas =7.25 L/min

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