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Answer :
Final answer:
The vapor pressure of ethanol at 65 °C is approximately 130 torr.
Explanation:
To find the vapor pressure of ethanol at 65 °C, we can use the Clausius-Clapeyron equation:
[tex]ln(P2/P1) = (-ΔHvap/R) * (1/T2 - 1/T1)[/tex]
Where:
- P1 is the vapor pressure at the initial temperature (35 °C) = 100 torr
- P2 is the vapor pressure at the final temperature (65 °C)
- ΔHvap is the enthalpy of vaporization = 39.3 kJ/mol
- R is the ideal gas constant = 8.314 J/(mol·K)
- T1 is the initial temperature in Kelvin (35 °C + 273.15)
- T2 is the final temperature in Kelvin (65 °C + 273.15)
Let's calculate the vapor pressure at 65 °C:
[tex]ln(P2/100) = (-39.3 * 10^3 / 8.314) * (1/(65 + 273.15) - 1/(35 + 273.15))[/tex]
Solving for P2:
[tex]P2 = 100 * e^((-39.3 * 10^3 / 8.314) * (1/(65 + 273.15) - 1/(35 + 273.15)))[/tex]
Calculating the value:
P2 ≈ 130 torr
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