Question

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?

Answer 1

Answer:

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

Answer 2

Answer:

338 mm Hg

Explanation:

$ln(\frac{p2}{400mmHg} )=-115*\frac{1}{0.008314} *(\frac{1}{1.8*10^3} -\frac{1}{1.84*10^3}$)

ln(p2/400)=-0.16705

p2/400= e^-0.16705=0.84615

p2=0.84615*400=338

Note that T2 is lower than T1 and that the vapor pressure decreases as the temperature decreases.