Calculating S & G for Phase Change of Water

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SUMMARY

This discussion focuses on calculating the change in entropy (ΔS) and change in Gibbs free energy (ΔG) for the phase change of 1 mole of liquid water at 100°C and 1 bar pressure into vapor at the same temperature and a pressure of 0.1 bar. The molar enthalpy for vaporization of water at 100°C and 1 bar is established as 40.6 kJ/mol. The calculation involves two thermodynamic processes: vaporization and isothermal expansion, with ΔS calculated using the formula ΔS = (Q1 + Q2) / T, where Q1 is the heat of vaporization and Q2 is the heat from isothermal expansion.

PREREQUISITES
  • Understanding of thermodynamic processes, specifically vaporization and isothermal expansion.
  • Familiarity with the concepts of entropy (ΔS) and Gibbs free energy (ΔG).
  • Knowledge of the ideal gas law and its application in thermodynamics.
  • Basic proficiency in using the formula ΔS = ΔH / T for phase changes.
NEXT STEPS
  • Study the derivation and application of the Gibbs free energy equation.
  • Learn about the ideal gas law and its implications for phase changes.
  • Explore the concept of molar enthalpy and its role in thermodynamic calculations.
  • Investigate the relationship between pressure changes and entropy in thermodynamic systems.
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Students and professionals in thermodynamics, chemical engineering, and physical chemistry who are involved in phase change calculations and thermodynamic process analysis.

doomed
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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!
 
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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 transferred 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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doomed,

1.) Post only one thread per topic.

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