Asked by sjhfds
chloroform (chcl3) has a Hvap of 29.6 kj/mol and its boiling point occurs at 61.7c. the specific heat capacity of liquid chloroform is 0.622 J/gC while that of the gas is 0.714 J/gC. How much energy (in kJ) needs to be removed to condense 250 g of gaseous chloroform from 75.0C to a liquid at 40.0C? Molar mass of chloroform is 119.37 g/mol.
Answers
Answered by
DrBob222
I'm not going to go through this this will all of the temps but you can. Here is how you do the problem.
q1 = heat removed to move chloroform vapor from starting T to boiling point.
q1 = mass CHCl3 x specific heat vapor x (Tfinal-Tinitial)
q2 = heat removed to condense vapor at boiling point to liquid at boiling point.
q2 = mass x heat vap.
q3 = heat removed to move T from boiling point to final T in liquid phase.
q3 = mass CHCl3 x specific heat liquid CHCl3 x (Tfinal-Tinitial)
Total q = q1 + q2 + q3.
Note: Everything is in J/g except for dHvap. I would change that to J/g first thing out of the box.
q1 = heat removed to move chloroform vapor from starting T to boiling point.
q1 = mass CHCl3 x specific heat vapor x (Tfinal-Tinitial)
q2 = heat removed to condense vapor at boiling point to liquid at boiling point.
q2 = mass x heat vap.
q3 = heat removed to move T from boiling point to final T in liquid phase.
q3 = mass CHCl3 x specific heat liquid CHCl3 x (Tfinal-Tinitial)
Total q = q1 + q2 + q3.
Note: Everything is in J/g except for dHvap. I would change that to J/g first thing out of the box.
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