Gibbs Energy

Question

Estimate the change in the Gibbs energy and molar Gibbs energy of 100 cm^3 of water when the pressure acting on it is increased from 100 kPa to 500 kPa. Mass density of water = 0.997 g/cm^3

 

Please, please be descriptive, show all steps and explain. Thank you so much

Answer 1

Start with fundamental equation of chemical thermodynamics:

 

dG = VdP − SdT

 

We'll assume this is isothermal since no other information is given. Then dT = 0

 

dG = VdP

 

integration gives:

 

ΔG = ∫ V dP  (from p1 to p2)

 

for a liquid the volume is constant

 

ΔG = V(p2 - p1)

 

ΔG = 100 cm³ * (500 kPa - 100 kPa)

 

Convert volume cm³ to dm³

100 cm³ = 100 mL = 0.1 L = 0.1 dm³

 

ΔG = 0.1 dm³ * (400 kPa) = 40 kPa.dm³

 

Convert to energy units, 1 kPa.dm³ = 1 J

 

40 kPa.dm³ = 40 J

 

To find the molar Gibbs free energy, divide by the number of moles of water. You'll need to convert volume to grams using the given density and the atomic mass of water.