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Answer :
Vapor pressure of ethanol is 286 mmHg.
We can use the Clausius-Clapeyron equation to calculate the vapor pressure of ethanol at 55.32°C:
- ln(P2/P1) = (ΔHvap/R)(1/T1 - 1/T2)
where:
- P1 is the vapor pressure of ethanol at 34.90°C
- P2 is the vapor pressure of ethanol at 55.32°C
- ΔHvap is the heat of vaporization of ethanol (39.3 kJ/mol)
- R is the gas constant (8.314 J/mol K)
- T1 is the temperature of ethanol at 34.90°C (34.90 + 273.15 = 308.25 K)
- T2 is the temperature of ethanol at 55.32°C (55.32 + 273.15 = 328.67 K)
We can plug these values into the equation to solve for P2:
- ln(P2/1.00×10^2 mmHg) = (39.3 kJ/mol)/(8.314 J/mol K)(1/308.25 K - 1/328.67 K)
- P2/1.00×10^2 mmHg = 0.286
- P2 = 0.286 * 1.00×10^2 mmHg
- P2 = 286 mmHg
Therefore, the vapor pressure of ethanol at 55.32°C is 286 mmHg.
About Vapor pressure
In chemistry, vapor pressure is the pressure exerted by a vapor in thermodynamic equilibrium with its condensed phases (solid or liquid) at a given temperature in a closed system. The equilibrium vapor pressure is an indication of a liquid's thermodynamic tendency to evaporate. The pressure exhibited by vapor present above a liquid surface is known as vapor pressure. As the temperature of a liquid increases, the attractive interactions between liquid molecules become less significant in comparison to the entropy of those molecules in the gas phase, increasing the vapor pressure.
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