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Calculate the change in molar Gibbs Free Energy for liquid water when pressure is increased from 50 kPa to 900 kPa at a constant 373 K, where the density of water is 0.997 g/cm3. Now calculate the change in molar Gibbs Free Energy for water vapor (ideal) when pressure is increased from 50 kPa to 900 kPa at a constant 373 K. Suggest a physical reason that would explain the large differences in your answers.

Respuesta :

Answer:

Explanation:

Given ;

  • Initial pressure = 50 kPa = P1
  • final pressure = 900 kPa = P2
  • density of water is 0.997 g/cm3 = 997kg/m3
  • Gibb's free energy is given as ; Δ G = VdP
  • ΔG = V (P2 - P1)
  • = 0.01 (900 - 50) x 1000

= 8500KJ/mol

Mass of water = volume x density

= 0.01 x 0.997 = 9970g

moles of water = mass /molar mass

= 9970g / 18g/mol

= 553.88moles

hence molar gibbs = Δ G /n

= 8500/553.88 = 15.35J/mol

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