Ethanol has a heat of vaporization of 38.56 kJ/mol and a normal boiling point of 78.4 °C. What is the vapor pressure of ethanol at 15 °C?

검증된 단계별 안내

Identify the known values: heat of vaporization (\(\Delta H_{vap}\)) = 38.56 \(\text{ kJ/mol}\), T_1 = 78.4 \(\degree\) C = 351.4 \(\text{ K}\), P_1 = 1 \(\text{ atm}\), T_2 = 15 \(\degree\) C = 288.15 \(\text{ K}\).

Use the Clausius-Clapeyron equation: \(\ln\left(\frac{P_2}{P_1}\right) = -\frac{\Delta H_{vap}}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right)\), where \(R = 8.314 \text{ J/mol K}\).

Convert \(\Delta H_{vap}\) from kJ/mol to J/mol by multiplying by 1000.

Substitute the known values into the Clausius-Clapeyron equation to solve for \(\ln\left(\frac{P_2}{1 \text{ atm}}\right)\).

Exponentiate both sides to solve for \(P_2\), the vapor pressure at 15 \(\degree\) C.

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주요 개념

질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.

Heat of Vaporization

The heat of vaporization is the amount of energy required to convert a unit mass of a liquid into vapor without a change in temperature. For ethanol, this value is 38.56 kJ/mol, indicating the energy needed to overcome intermolecular forces during the phase transition from liquid to gas.

Boiling Point

The boiling point of a substance is the temperature at which its vapor pressure equals the external pressure surrounding the liquid. For ethanol, the normal boiling point is 78.4 °C, meaning at this temperature, ethanol will transition from liquid to gas under standard atmospheric pressure.

Vapor Pressure

Vapor pressure is the pressure exerted by a vapor in equilibrium with its liquid at a given temperature. It reflects the tendency of molecules to escape from the liquid phase into the gas phase. The vapor pressure of ethanol at 15 °C can be estimated using the Clausius-Clapeyron equation, which relates vapor pressure to temperature and heat of vaporization.