Question

Find the Gibbs energy of vaporization of water (in J/mol) at a temperature of 373.15K and a pressure of 10 bar.

Answer 1

Given:

T = 373.15 K

P = 10 bar.

Solution:

We have to find delta G of vaporization.

Conversion of pressure to atm.

Pressure in atm = 10 bar x 1.0 atm / 1.01325 bar

= 0.9869 atm

Let’s show the equation.

H2O (l) --- > H2O (g)
Lets show the kp expression for this reaction.

Kp = p (H2O )

From this equation we can say kp of this reaction = pressure of H2O (g)

So kp = 0.9869 atm

We know the relation between kp and delta G

Delta G = - RTlnKp

Here Delta G is gibbs free energy , R is gas constant = 8.314 J / (K mol)

T = 373.15 K

Lets plug all the values in order to get Delta G

Delta G = -[ 8.314 x 373.15 x ln (0.9869)] J /mol

Delta G = + 40.91 J /mol

Answer : Delta G of this process = 40.91 J/mol