Answer :
Final answer:
The vapor pressure of ethanol at 61.06°C is approximately 0.014 mmHg.
Explanation:
To calculate the vapor pressure of ethanol at 61.06°C, we can use the Clausius-Clapeyron equation:
ln(P2/P1) = (-ΔHvap/R) * (1/T2 - 1/T1)
Where:
- P1 is the vapor pressure at the initial temperature (34.90ºC)
- P2 is the vapor pressure at the final temperature (61.06°C)
- ΔHvap is the enthalpy of vaporization for ethanol (39.3 kJ/mol)
- R is the ideal gas constant (8.314 J/(mol·K))
- T1 is the initial temperature in Kelvin (34.90ºC + 273.15 = 308.05 K)
- T2 is the final temperature in Kelvin (61.06°C + 273.15 = 334.21 K)
Substituting the given values into the equation:
ln(P2/1.00 x 10^2) = (-39.3 * 10^3 J/mol / (8.314 J/(mol·K))) * (1/334.21 K - 1/308.05 K)
Simplifying the equation:
ln(P2/1.00 x 10^2) = -4.732
Using the natural logarithm properties, we can solve for P2:
P2/1.00 x 10^2 = e^(-4.732)
P2 = 1.00 x 10^2 * e^(-4.732)
Calculating the value:
P2 ≈ 0.014 mmHg
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