(a) Consider the equilibrium of H₂O(I) and H₂O(g):

H₂O(I) = H₂O(g)

Calculate the partial pressure of water vapour at 1 bar and 25°C assuming that the vapour can be treated as an ideal gas  

(b) Find DG for the isothermal compression of 0.015 kg CCl₄(g) from a pressure of 0.2 atm to 2 atm at 306 K (assume ideal gas behaviour).

 

vapour pressure = 726.26 mmHg

Delta G = 5.71 ×10^-2 kJ

Step-by-step explanation

1. partial pressure of water vapour = laboratory pressure (P_lab) - saturated vapour pressure P_svp)

P _lab = 1 bar = 750.062mm Hg

Psvp= to a standard value of 23.8 mmHg at 25 °C

Now

Vapour pressure = 750.062 mmHg - 23.8 mmHg

= 726.262 mmHg

2 to calculate delta G, we need to follow several steps

Step 1. Calculate delta S

ΔG = delta H - Tdelta S

For isothermal process delta H = 0

Thus

ΔG = -T ΔS

Where

ΔS = nR In (P1/p2)

Where

N = number of moles = mass / molar mass

= 0.015 kg /153.82 g/ mol

= 15g / 153.82 g/ mol = 0.098 mol

R is gas constant = 8.324 J/ mol K

P1 is the initial pressure = 0.2

P2 is final pressure = 2atm

Thus

ΔS = 0.098 mol × 8.314 J/ mol K (In ( 0.2/2))

= -1.867 J/K

Step 2 calculate delta G

Now calculating delta G

ΔG = -T ΔS

We are given T = 306 K

Thus

Delta G = -(306 K × -1.867 J/K)

= 571.25 J

which can be converted to kJ to give

571.25 /1000 = 5.71 ×10^-1kJ