[trinhn812]

Consider the process:                A(l) ---> A(g)
                                                        75°C   155°C
which is carried out at constant pressure. DeltaS for this process is known to be 75.0 J/Kmol. For A(l) and A(g), the Cp values are 75 J/Kmol and 29 J/Kmol, respectively, and are not dependent on temperature. Calculate the
delta H(vaporization) for A(l) at 125°C (its boiling point).





Well, I ∆H = -29.8 kJ/mol;
Here's what I did:
I made delta G = 0
You said ∆S = 75.0 J/molK, or 0.0750 kJ/molK
∆G = ∆H - T∆S
∆H = ∆G - T∆S
∆H = 0 - (398K)(0.0750 kJ/molK)
∆H = - 29.85 kJ/mol = -29.8 kJ/mol

I don't know what the Cp and change in temperature mean.

[Yggdrasil]

Your answer is incorrect.  If you consider the entire process as a whole, you do not have a fixed temperature, so your starting equation doesn't make any sense.

It may be easier, however, to split the process into three separate transformations:

1)  A(l, 75^oC)  -->  A(l, 125^oC)
2)  A(l, 125^oC) --> A(g, 125^oC)
3)  A(g, 125^oC --> A(g, 155^oC)

Each of these steps will have their own value of ΔS and ΔH.  Now, try thinking about the problem this way: You are given the ΔS_total = ΔS₁ + ΔS₂ + ΔS₃, and you want to calculate ΔH₂.