Question

# 1- The normal Boiling Point of liquid benzene is 80.1 oC, with 31.0 kJ/mol of ΔHvap...

1- The normal Boiling Point of liquid benzene is 80.1 oC, with 31.0 kJ/mol of ΔHvap at that temperature. The molar heat capacities of the liquid and the gaseous benzene are given as [Cp(l) = 33.44 + 0.334 T] and [Cp(g) = 10.28 + 0.252 T] respectively. i- Calculate ΔH, ΔS and ΔG when 1 mole of liquid benzene is converted to 1 mole of gaseous benzene at 1.00 atm and 80.1 oC. (10 pts) ii- Calculate the work done and ΔU for the process in (i). (5 pts) iii- The pressure is now raised to 2.00 atm. Calculate the new Boiling Point, assuming perfect gas behavior for the gas and molar volume of the liquid benzene (90.0 cm3/mol) does not change significantly in this pressure range. (10 pts) iv- Calculate ΔH, ΔS ΔG when 1 mole of liquid benzene is converted to 1 mole of gaseous benzene at 2.00 atm and its new Boiling Point. (10 pts) v- Calculate the work done and ΔU for the process in (iv).

Part 1:

In first place we have to trace a path of the phase change for the benzene, having to:

According to the data given the phase change to the given conditions of T and P occurs isothermally and isobarically, so the heat for this system is given only by a normal latent phase heat.In that sense if one mole of liquid benzene completely changes to the gas phase as planned, then the energy for this change is:

And to the other propiertys

#### Earn Coins

Coins can be redeemed for fabulous gifts.