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What is the vapor pressure of ethanol in mmHg at 52.5°C if its vapor pressure is 400.0 mmHg at 63.5°C? Given: ∆Hvap = 39.3 kJ/mol, R = 8.314 J/K•mol

A. 314.5 mmHg
B. 369.8 mmHg
C. 429.7 mmHg
D. 497.2 mmHg

Answer :

Final answer:

The vapor pressure of ethanol at 52.5 °C is calculated using the Clausius-Clapeyron equation, given its vapor pressure at 63.5 °C and its enthalpy of vaporization. By substituting the given values into the equation, one can solve for the vapor pressure at the desired temperature.Therefore, the vapor pressure of ethanol at 52.5°C is c)429.7 mmHg.

Explanation:

Using the Clausius-Clapeyron equation, we can calculate the vapor pressure of ethanol at 52.5°C:

  • ln(P2/P1) = (-∆Hvap/R) x (1/T2 - 1/T1)
  • ln(400.0/P1) = (-39,300 J/mol / 8.314 J/K·mol) x (1/(325.65 K) - 1/(336.65 K))
  • P1 = 429.7 mmHg

Therefore, the vapor pressure of ethanol at 52.5°C is 429.7 mmHg.Remember, understanding the steps and method is crucial for solving similar problems or when values change. This problem is a good demonstration of applying thermodynamics to real-life situations, like understanding the behavior of ethanol under different temperatures - a concept critical in various industrial and laboratory applications.The vapor pressure of ethanol at 52.5 °C is calculated using the Clausius-Clapeyron equation, given its vapor pressure at 63.5 °C and its enthalpy of vaporization. By substituting the given values into the equation, one can solve for the vapor pressure at the desired temperature.Therefore, the vapor pressure of ethanol at 52.5°C is c)429.7 mmHg.

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