Answer :
Final answer:
The vapor pressure of ethanol at 58.05°C is approximately 1.02 × 10² mmHg.
Explanation:
To calculate the vapor pressure of ethanol at 58.05°C, we can use the Clausius-Clapeyron equation:
ln(P2/P1) = (ΔHvap/R) * (1/T1 - 1/T2)
Where:
- P1 is the initial vapor pressure (1.00 × 10² mmHg)
- P2 is the final vapor pressure (unknown)
- 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 (34.90°C + 273.15 K)
- T2 is the final temperature (58.05°C + 273.15 K)
Let's plug in the values and solve for P2:
ln(P2/1.00 × 10²) = (39.3 * 10³)/(8.314) * (1/(34.90 + 273.15) - 1/(58.05 + 273.15))
Simplifying the equation:
ln(P2/1.00 × 10²) = 0.0147
Using the natural logarithm properties:
P2/1.00 × 10² = e(0.0147)
P2 = 1.00 × 10² * e(0.0147)
Calculating the value:
P2 ≈ 1.02 × 10² mmHg
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