question archive (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
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(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 (Plab) - saturated vapour pressure Psvp)
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