Question

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.

Answer 1

Step-by-step 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