Question

The vapor pressure of liquid antimony is 400 mm Hg at 1.84x10' K. Assuming that its molar heat of vaporization is constant at 115 kJ/mol, the vapor pressure of liquid Sb is m m Hg at a temperature of 1.81x10' K.

Answer 1

We have

P₁ = 400 mm Hg

T₁ = 1.84 $\times$ 10³ K = 1840 K

$\Delta$Hvap = 115 kJ/mol = 115 $\times$ 10³ J/mol = 115000 J/mol

P₂ = ?

T₂ = 1.81 $\times$ 10³ K = 1810 K

R = ideal gas constant= 8.314 J/mol.K

We know Clausius-Clapeyron equation

ln(P₂/P₁) = - ($\Delta$Hvap/R).(1/T₂ - 1/T₁)

So by putting above values in above formula we get

ln(P₂ / 400 mm Hg) = -(115000 J/mol / 8.314 J/mol.K ) .(1/1810 K - 1/1840 K)

l(P₂ / 400 mm Hg = -0.125

P₂ / 400 mm Hg = e^-0.125 = 0.88

P₂ = 0.88 $\times$ 400 mm Hg = 352 mm Hg

So

Vapor Pressure of Sb is 352 mm Hg at temperature of  1.81 $\times$ 10³ K