Answer:
P₂ = 5.550213 Pa
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
