Answer :
Final answer:
The vapor pressure of ethanol at 65.0°C is approximately 199.6 mmHg. This is derived using the Clausius-Clapeyron equation.
Explanation:
The Clausius-Clapeyron equation establishes a connection between vapor pressure and temperature. It links the change in vapor pressure with temperature to the enthalpy of vaporization. The formula is represented as:
ln(P2/P1) = (-ΔHvap/R) * (1/T2 - 1/T1)
P1 and P2 are the vapor pressures at temperatures T1 and T2 respectively.
ΔHvap stands for the enthalpy of vaporization.
R is the universal gas constant, which is 8.314 J/(mol·K).
Using the provided information:
P1 is 100 mmHg at T1, which equals 34.9°C or 308.05 K.
ΔHvap is 39.3 kJ/mol, which can also be stated as 39300 J/mol.
T2 is 65.0°C, converting to 338.15 K.
Substituting these values into our equation, we have:
ln(P2/100) = (-39300 J/mol / 8.314 J/(mol·K)) * (1/338.15 K - 1/308.05 K)
From this, we can determine P2 to be about 199.6 mmHg.
Therefore, it's evident that the vapor pressure of ethanol at 65.0°C is roughly 199.6 mmHg. This relationship showcases how an increase in temperature will lead to a rise in vapor pressure.
Learn more about Vapor pressure
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