Question

Determine the equilibrium vapor pressure of germanium metal at 508.0 C. The standard enthalpy of vaporization of germanium is 376.6 kJ/mol and the vapor pressure at 100 C is 1.6E-32 Pa. Be sure to express your final answer in Pa.

Answer 1

P₂ = 5.550213 Pa

Step-by-step explanation:

Given data:

Heat of vaporization = 376.6 KJ/mol ( 376600 j/mol)

R = 8.3143 j mol⁻¹ K⁻¹

Temperature = T1 = 100 °C = 100 + 273 = 373 K

Temperature = T2 = 508.0 °C = 508.0+273 = 781 K

Pressure at 100°C= P1 = 1.6 × 10⁻³² Pa

Pressure at 508.0 °C = ?

Solution:

ln P₁/P₂ = ΔH /R ( 1/T₂ - 1/T₁)

ln (1.6 × 10⁻³² Pa) - ln (P₂) = 376600 j. mol⁻¹ / 8.3143 j mol⁻¹ K⁻¹ ( 1/T₂ - 1/T₁)

-73.212719 - ln (P₂) = 45295.45 (0.001280 - 0.002680)

-73.212719 - ln (P₂) = 45295.45 (-0.00414)

- ln (P₂) = -63.41363 + 73.212719

- ln (P₂) = 9.799089

P₂ = e⁻⁹°⁷⁹⁹⁰⁸⁹

P₂ = 5.550213 Pa