We are interested in recovering acetone from 100 mol/s (initial temperature is 6

Question

We are interested in recovering acetone from 100 mol/s (initial temperature is 65 C) of a vapor mixture containing 66.9% (mole basis) acetone and rest nitrogen. For this purpose, the vapor mixture is cooled to 20 C where 95% of initial acetone leaves as liquid and the rest of acetone leaves as vapor along with all the nitrogen at 20 C. Calculate the required heat exchange rate.

At 65C, enthalpy of acetone is -214 kJ/mol and enthalpy of nitrogen is 1.2 kJ/mol. At 20 C, enthalpy of nitrogen is -0.15 kJ/mol. For acetone (liquid), enthalpy at 20 C is -250 kJ/mol.

Explanation / Answer

Initial temp = 65 °C

Vapour mixture contains:

   66.9% acetone   , 33.1 % nitrogen

At 20°C

     95 % of acetone leaves as vapour = ( 95/100 )* 66.9    = 63.555%, N is same that is   33.1%

So acetone used

= 66.9- 63.555 = 3.345

At 65C, enthalpy of acetone is -214 kJ/mol and enthalpy of nitrogen is 1.2 kJ/mol. At 20 C, enthalpy of nitrogen is -0.15 kJ/mol. For acetone (liquid), enthalpy at 20 C is -250 kJ/mol.

So enthalpy exchange (according to above data,)

=66.9*214 + 33*1.2 -(63.555* *250 + 33.1*0.15) = 1537.395 kJ/mol

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