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The vapor pressure of ethanol is 102 mmHg at 41.3°C. What is its vapor pressure at 79.6°C? The molar heat of vaporization of ethanol is 39.3 kJ/mol.

Answer :

The vapor pressure of ethanol at 41.3 °C is 102 mmHg, and we need to determine its vapor pressure at 79.6 °C. Given the molar heat of vaporization of ethanol (39.3 kJ/mol), we find that the vapor pressure of ethanol at 79.6 °C is approximately 398.3 mmHg.

The Clausius-Clapeyron equation relates the vapor pressure of a substance at different temperatures to its molar heat of vaporization. It can be written as:

ln(P₂/P₁) = -(ΔH_vap/R) * (1/T₂ - 1/T₁)

Where P₁ and P₂ are the initial and final vapor pressures, ΔH_vap is the molar heat of vaporization, R is the ideal gas constant, T₁ is the initial temperature, and T₂ is the final temperature.

To solve for the final vapor pressure, we rearrange the equation:

P₂ = P₁ * exp(-(ΔH_vap/R) * (1/T₂ - 1/T₁))

Substituting the given values, P₁ = 102 mmHg, T₁ = 41.3 °C (314.45 K), T₂ = 79.6 °C (352.75 K), and ΔH_vap = 39.3 kJ/mol, we can calculate the vapor pressure at 79.6 °C:

P₂ = 102 mmHg * exp(-(39.3 kJ/mol)/(8.314 J/(mol·K)) * (1/352.75 K - 1/314.45 K))

After evaluating the expression, we find that the vapor pressure of ethanol at 79.6 °C is approximately 398.3 mmHg.

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