How does pressure change affect entropy and Gibbs free energy in water vaporization?

[doomed]

phase change?

How do I calculate the values of change in entropy (S) and change in Gibbs free energy (G) for the conversion of n=1 mol of liquid water at 100 C and 1 bar pressure into vapor at the same temperatue and a pressure of 0.1 bar. Assume ideal behavior for the vapor. the molar enthalpy for vaporization of water at 100 C and 1 bar is 40.6 kJ/mol.

I know that delta S = delta H/T, but how the change in pressure play into this problem for delta S and delta G?

HELP me PLEASE!

 

[clive]

I'll give you some hints for the entropy:

You have 2 thermodynamic processes there:
1) vaporization
2) isothermal expansion form 1 bar to 0.1 bar.

Because the temperature is the same, the variation in entropy can easily be evaluated by
$$\Delta S=\frac{Q_1+Q_2}{T}$$
where Q1+Q2 represents the total transferred heat.

Now you have
$$Q_1=m\cdot \lambda_v$$
for the vaporization at 100 C
and
$$Q_2=\nu R T \ln \frac{V_2}{V_1}$$
for the isothermal expansion

and so on...($$p_1 V_1=p_2 V_2$$ is the answer at your last question)

 

[quantumdude]

doomed,

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