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Estimate the normal boiling temperature of propan-1-ol, C3H7OH, given that its vapour
pressure is 2701 Pa at 298 K and 9129 Pa at 318 K.

Respuesta :

mickymike92

Based on the data provided, the normal boiling point of propan-1-ol is 93.7° C

What is the normal boiling point of a liquid?

The normal boiling point of a liquid is the temperature at which the clvapor pressure of the liquid is equal to the atmospheric pressure.

The Clausius-Clapeyron Equation is used to determine the normal boiling point of liquids.

The Clausius-Clapeyron Equation is given as:

  • ln (P₁/P₂) = ΔHvap/R × (1/T₂ - 1/T₁)

where

  • ΔHvap is in J/mol
  • R = 8.314 J/mol.K

From the given data:

T1 = 298 K

T2 = 318 K

P1 = 2710 Pa

P2 = 9129

ΔHvap = ?

Substituting the values:

ln ln (P₁/P₂) = ΔHvap/R × (1/T₂ - 1/T₁)

ln (2710/9129) = ΔHvap/(8.314) × (1/318 - 1/298)

  • ΔHvap = 47843.57 J/mol

Then at P2 = 101325 Pa

Normal boiling point T₂ can be determined thus:

ln (2710/101325) = 47843.57/8.314 × ( 1/T₂ - 1/298)

T₂ = 366.7 K

T₂ = 93.7° C

Therefore, the normal boiling point of propan-1-ol is 93.7° C.

Learn more about normal boiling point at: https://brainly.com/question/40140

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