The vapor pressure of liquid antimony is 400 mm Hg at 1.84*103 K. Assuming that its molar heat of vaporization is constant at 115 kJ/mol, the vapor pressure of liquid Sb is 394.98 mm Hg at a temperature of 1.81*103 K. Find the vapor pressure of liquid Sb?

Vapor pressure of liquid Sb = 8.19 x 10⁴ mm Hg

Explanation:

The vapor pressure can be calculated by using Clausius-Clapeyron equation.

ln (p₁/p₂) = (-ΔHvap/R) (1/T₁ - 1/T₂)

Where

p₁ is the vapor pressure at T₁ (Initial Temperature)

p₂ is the vapor pressure at T₂ (final Temperature)

ΔHvap is molar heat of vaporization of the substance

R is the real gas constant = 8.314 x 10⁻³ kJ/mol. K

Data Given:

p₁ = ?

p₂ = 394.98 mm Hg

T₁ = 1.84*10³ K

T₂ = 1.81*10³ K

ΔHvap = 115 kJ/mol

Put the values in the Clausius-Clapeyron equation

ln (p₁/p₂) = (-ΔHvap/R) (1/T₁ - 1/T₂)

ln (p₁/394.98 mm Hg) = (-115 kJ/mol / 8.314 x 10⁻³ kJ/mol. K) (1/1.84*10³ K - 1/1.81*10³ K)

ln (p₁ / 394.98 mm Hg) = ( - 13.8321 x 10³) (-0.5519)

ln (p₁ / 394.98 mm Hg) = 7633.936

ln cancel out by E, e is raise to a power x

So,

p₁/394.98 mm Hg = e^7633.936

p₁ / 394.98 mm Hg = 20.75 x 10³

p₁ = 20.75 x 10³ x 394.98 mm Hg

p₁ = 8.19 x 10⁴ mm Hg

Vapor pressure of liquid Sb = 8.19 x 10⁴ mm Hg