Chapter 11, Problem 11.76MP

Interpretation Introduction

Interpretation:

The vapor pressure of ethanol at $\text{25}\text{.}°\text{C}$ is to be calculated.

Concept introduction:

Clausius-Clapeyron equation is used to determine the vapor pressure at another temperature if vapor pressure at one temperature and enthalpy of vaporization is known.

The expression of Clausius-Clapeyron equation is as follows:

$\ln P_{\text{vap}}=\left(\frac{-\Delta H_{\text{vap}}}{\text{R}}\right)\left(\frac{1}{T}\right)+C$

Where,

• $P_{\text{vap}}$ is the vapor pressure.
• $\Delta H_{\text{vap}}$ is the heat of vaporization.
• $T$ is the temperature.
• $C$ is the constant.
• $\text{R}$ is universal gas constant.

For the two pressures and temperature, the Clausius-Clapeyron equation is modified as follows:

$\ln \frac{P{}_1}{P_2}=\left(\frac{\Delta H_{\text{vap}}}{\text{R}}\right)\left(\frac{1}{T_2}-\frac{1}{T_1}\right)$

The conversion factor to convert $°\text{C}$ to Kelvin is,

$T\left(\text{K}\right)=T\left(°\text{C}\right)+273$

The conversion factor to convert $\text{kJ}$ to $\text{J}$ is as follows:

$1\text{ kJ}=\text{1000 J}$

The formula to relate enthalpy and entropy change is as follows:

$\Delta G=\Delta H-T\Delta S$

Where,

• $\Delta G$ is free energy change.
• $\Delta H$ is enthalpy change.
• $\Delta S$ is entropy change.

At equilibrium, free energy change $\left(\Delta G\right)$ is 0. The relation between enthalpy change $\left(\Delta H\right)$ and degree of randomness $\left(\Delta S\right)$ is as follows:

$\Delta S=\frac{\Delta H}{T}$