High School

What is the vapor pressure of ethanol (in mmHg) at 55.7 °C if its vapor pressure is 400.0 mmHg at 63.5 °C?

Given:
- \(\Delta H_{vap} = 39.3 \text{ kJ/mol}\)
- \(R = 8.314 \text{ J/K} \cdot \text{mol}\)

Answer :

Final answer:

The question asks to calculate the vapor pressure of ethanol at 55.7 °C using the Clausius-Clapeyron equation, given its vapor pressure at 63.5 °C and ethanol's enthalpy of vaporization.

Explanation:

The question revolves around using the Clausius-Clapeyron equation to calculate the vapor pressure of ethanol at a different temperature given the vapor pressure at a known temperature and the enthalpy of vaporization. We can rearrange the Clausius-Clapeyron equation into:

ln(P1/P2) = (∆Hvap/R)((1/T2) - (1/T1))

Where P1 and P2 are the vapor pressures at temperatures T1 and T2 respectively, ∆Hvap is the enthalpy of vaporization, and R is the gas constant. Given the known parameters, one can solve for the vapor pressure at 55.7 °C (328.85 K) using the vapor pressure at 63.5 °C as a starting point.

To solve the problem, you would first convert the given ∆Hvap from kJ/mol to J/mol (since R is given in J/K•mol), calculate the temperatures in Kelvin, then solve for P2 using the equation above.

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