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Ethanol (C\(_2\)H\(_5\)OH) is a volatile liquid used widely as a laboratory solvent. Ethanol has a standard boiling point of 78.5 °C, and its enthalpy of vaporization is 42.3 kJ mol\(^{-1}\).

What is its vapor pressure (in bar) at 36.3 °C?

Answer :

Final answer:

The vapour pressure of ethanol at 36.3 °C can be calculated using the Clausius-Clapeyron equation. This equation uses the standard boiling point and the enthalpy of vaporization of ethanol. The units must be handled carefully while solving the equation.

Explanation:

To calculate the vapour pressure of ethanol at 36.3°C from the standard boiling point, and the enthalpy of vaporization of ethanol, we can use the Clausius-Clapeyron equation. This equation is:

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

In this equation, P2 is the vapour pressure at the target temperature (36.3°C or 309.45 K), P1 is the standard vapour pressure at the boiling point (which is 1 bar at the standard boiling point 78.5°C or 351.65 K), ΔHvap is the enthalpy of vaporization (42.3 kJ mol-1), R is the ideal gas constant (8.314 J mol-1 K-1), T2 is the target temperature in Kelvin and T1 is the standard boiling temperature in Kelvin.

Filling the values into the equation and solving for P2 gives us the vapor pressure of ethanol at 36.3°C. Please be advised to handle the units carefully while solving the equation.

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