Ethanol has a heat of vaporization of 38.56 kJ / mol and a normal boiling point of 78.4^∘ C . Wh

step 1 Everyone, my name is Ahmad Ali. In this question, he wants to calculate the vapor pressure and t

Ethanol has a heat of vaporization of 38.56 kJ / mol and a...

Ethanol has a heat of vaporization of 38.56 $\mathrm{kJ} / \mathrm{mol}$ and a normal boiling point of $78.4^{\circ} \mathrm{C}$ . What is the vapor pressure of ethanol at $15^{\circ} \mathrm{C} ?$

Key Concepts

Vapor Pressure

Vapor pressure is the pressure exerted by the vapor in equilibrium with its liquid (or solid) phase at a given temperature. It reflects how readily molecules escape from the surface of the liquid into the gas phase and is a critical parameter in studying phase behavior and volatility.

Heat of Vaporization

The heat of vaporization is the amount of energy required to convert one mole of a liquid into its vapor without a change in temperature. It is a measure of the strength of intermolecular forces within the liquid and is essential in calculations that involve phase transitions, especially when using thermodynamic relations.

Clausius-Clapeyron Equation

The Clausius-Clapeyron equation provides a quantitative relationship between the vapor pressure of a substance and its temperature. It is commonly used to estimate the vapor pressure at different temperatures by relating the heat of vaporization to changes in pressure with temperature, assuming the latent heat remains approximately constant.

Normal Boiling Point

The normal boiling point is defined as the temperature at which the vapor pressure of a liquid equals the external atmospheric pressure (typically 1 atm). In thermodynamic calculations, it provides a reference point that can be used together with the Clausius-Clapeyron equation to determine vapor pressures at other temperatures.

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