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The normal boiling point for acetone is 56.5°C. At an elevation of 6500 ft, the atmospheric pressure is 606 torr. What would be the boiling point of acetone (ΔHvap = 32.0 kJ/mol) at this elevation?

What would be the vapor pressure of acetone at 25.0°C at this elevation?
Answered by DrBob222
Use the Clausius-Clapeyron equation.
Answered by Doc48
ln(VP₂/VP₁) = ΔHᵥ/R(T₂-T₁/T₁∙T₂) substitute given data (units compatible with R = 0.008314Kj/mol-K), VP₂ = 606Torr, VP₁ = 760Torr @ T₁ = 56.5C = 329.5K, ΔHᵥ = 32.0Kj/mole and solve for T₂.
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