
#1
Oct905, 04:30 PM

P: 1

How do I calculate the values of change in entropy (S) and change in Gibbs free enery (G) for the conversion of n=1 mol of liquid water at 100 C and 1 bar pressure in to 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! 



#2
Oct905, 05:49 PM

P: 120

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 [tex]\Delta S=\frac{Q_1+Q_2}{T}[/tex] where Q1+Q2 represents the total transfered heat. Now you have [tex]Q_1=m\cdot \lambda_v[/tex] for the vaporization at 100 C and [tex]Q_2=\nu R T \ln \frac{V_2}{V_1}[/tex] for the isothermal expansion and so on.......([tex]p_1 V_1=p_2 V_2[/tex] is the answer at your last question) 


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