[kido]

Hi! Just having trouble with a question about entropy. Here it is:

Consider a system of 2 moles of liquid water initially at 27 °C and 1 atm. Calculate the
change in entropy, ΔS, when the 2 moles of H2O(l) undergoes a simultaneous increase
in temperature, T, and pressure, P, to 37 °C and 40 atm, respectively. You will need to
make some assumptions to do this calculation. State clearly these assumptions.
The constant pressure heat capacity for H2O(l), Cp = 75 J K-1 mol-1.

Okay so I began by splitting the process up into two steps. The first step is reversible heating at constant pressure, so the equation can be given by dS = (Cp,m/T)dT, and
change in S = n x Cp x ln(Tf/Ti) = 2 x 75 x ln(37+273/27+273) = 4.92J.
The next step should be the change in entropy due to change in pressure...now this is the bit i'm struggling with. I know for an ideal gas this could be given by nR x ln(Pi/Pf), but i'm not sure how to go about it for a liquid. Can anyone help??

Thanks!! 

[Frank84]

I think you can despise the pressure factor in liquids , because doesnt affect entropy at all. You need to apply at least 100 atm for making the entropy decrease 1 J/K in liquids.
Anyways, dont you have thermodynamics water tables or diagram?

[kido]

Nope, no tables or diagrams. This is the whole question.

So when you change the pressure of a liquid, the entropy change is negligible?