Question

Part A - The vapor pressure of water at 25∘C and 1 atm is about 18 torr. What is the approximate vapor pressure of water at 80∘ ∘ C?

Answer 1

Given,

Vapour pressure of water at 25 ^oC (P₁) = 18 Torr

Vapour pressure of water at 80^oC (P₂) = ?

Temperature(T₁) = 25 ^oC + 273.15 = 298.15 K

Temperature(T₂) = 80 ^oC + 273.15 = 353.15 K

Now, We know, the Clausius-Claperyon equation,

ln[P₁ /P₂] = H_vap / R [ 1/T₂ - 1/T₁]

Here, H_vap of water = 40700 J/mol

R = 8.314 J/K.mol

Substituting the known values in the above formula,

ln[18 /P₂] = 40700 J/mol / 8.314 J/K.mol [ 1/353.15 K - 1/ 298.15 K]

ln[18 /P₂] = 4895.357 x -0.00052

[18 /P₂] = e^-2.55713

P₂ = 232.2 Torr

Thus, the vapour pressure of water at 80 ^oC is 230 Torr [ 2 s.f]