mackenziegriggs
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Determine the vapor pressure (atm) of rubbing alcohol (isopropanol) at 20.0 °C. The normal boiling point of isopropanol is 82.3 °C and the heat of vaporization (ΔHvap) is 39.9 kJ/mol.

Respuesta :

scme1702
Here we apply the Clausius-Clapeyron equation:
ln(P₁/P₂) = ΔH/R x (1/T₂ - 1/T₁)

The normal vapor pressure is 4.24 kPa (P₁)
The boiling point at this pressure is 293 K (P₂)
The heat of vaporization is 39.9 kJ/mol (ΔH)
We need to find the vapor pressure (P₂) at the given temperature 355.3 K (T₂)

ln(4.24/P₂) = 39.9/0.008314 x (1/355.3 - 1/293)
P₂ = 101.2 kPa

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