Answer :
Final Answer:
The vapor pressure of ethanol at 40.9°C is approximately 262.2 mmHg.
Explanation:
The relationship between vapor pressure and temperature is described by the Clausius-Clapeyron equation:
ln(P₂/P₁) = -ΔHvap / R x (1/T₂ - 1/T₁)
Where:
P₁ = vapor pressure at temperature T₁
P₂ = vapor pressure at temperature T₂
ΔHvap = molar heat of vaporization
R = ideal gas constant (8.314 J/mol·K)
T₁ = initial temperature in Kelvin
T₂ = final temperature in Kelvin
Given that P₁ = 450 mmHg at
T₁ = 68.5°C = 341.65 K, and
ΔHvap = 39.3 kJ/mol, we want to find P₂ at
T₂ = 40.9°C = 314.05 K.
Converting ΔHvap to joules/mol:
ΔHvap = 39.3 kJ/mol x 1000 J/kJ
= 39300 J/mol
Plugging the values into the Clausius-Clapeyron equation:
ln(P₂/450) = -39300 / (8.314 x (1/314.05 - 1/341.65))
Solving for P₂:
P₂ = 450 x e^(-39300 / (8.314 x (1/314.05 - 1/341.65)))
P₂ ≈ 262.2 mmHg
Therefore, the vapor pressure of ethanol at 40.9°C is approximately 262.2 mmHg.
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