Answer to Question #98407 in Physical Chemistry for cindy

Question

The vapor pressure of liquid lead is 400. mm Hg at 1900 K. Assuming that its molar heat of vaporization is constant at 179 kJ/mol, the vapor pressure of liquid Pb is
mm Hg at a temperature of 1860 K.

Answer 1

Solution.

$\Delta H(vap.) = \frac{R \times T1 \times T2 \times ln(\frac{p2}{p1})}{T2-T1}$

$p2 = p1 \times \exp(\frac{\Delta H(vap.) \times (T2-T1)}{R \times T1 \times T2}$

$p2 = 400 \times \exp(\frac{149*10^3 \times (1860-1900)}{8.31 \times 1900 \times 1860}$

p2 = 313.46 mm Hg

Answer:

p2 = 313.46 mm Hg