Question

How much energy must be removed from a 125 g sample of benzene (C6H6) vapor at 425.0 K to make it a solid at 200.0 K? Use the following steps to get to your answer. You MUST show your work to receive any credit. The following physical data for benzene may be useful. The quiz will be due at the beginning of class on 2/17/16. ΔHvap = 33.9 kJ/mol ΔHfus = 9.8 kJ/mol Cliq = 1.73 J/g°C Cgas = c Csol = 1.51 J/g°C Tmelting = 279.0 K Tboiling = 353.0 K

4) How much energy must be removed to bring the liquid from its ___________J boiling point to its melting point?

Answer 1

q released = msolid*s*DT + n*DHfus + mliquid*s*DT + n*DHvap + mgas*s*DT

msolid = 125 grams

s = 1.51 j/g.c

DT = (279-200) = 79

               = 125*1.51*(279-200) + (125/78)*9.8*10^3 + 125*1.73*(353-279) + (125/78)*33.9*10^3 + 125*1.255*(425-353)

   = 112240 joule

energy released when gas gets condensed from 425 k to 353 k

= mgas*s*DT = 125*1.255*(425-353) =11295 joule.

phase transition from gas to liquid = (125/78)*33.9*10^3 = 54326.923 joule.

liquid cooling 353 to 279 k    = 125*1.73*(353-279) = 16002.5 joule

phase transition from liquid to solid = (125/78)*9.8*10^3 = 15705.13 joule

cooling solid from 279 to 200 k = 125*1.51*(279-200) = 14911.25 j

total energy released = 14911.25+15705.13+16002.5+54326.923+11295

                 = 112240.8 joule.