Chapter 12, Problem 12.141P

(a)

Interpretation Introduction

Interpretation:

The mass of ethanol present in the vapor is to be calculated.

Concept introduction:

Clausius-Clapeyron equation is used to find the vapor pressure at one temperature when vapor pressure at another temperature and heat of enthalpy is given. It determines the change in vapor pressure with a change in temperature. The expression of the Clausius-Clapeyron equation is as follows:

  $\ln \left(\frac{P_2}{P_1}\right)=\frac{-\Delta H_{\text{vap}}}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)$        (1)

Here,

$P_1\text{ and }{P}_2$ are the vapor pressure.

$T_1\text{ and }{T}_2$ are the temperature.

$R$ is the gas constant.

$\Delta H_{\text{vap}}$ is the heat of vaporization.

The conversion factor to convert $°\text{C}$ to Kelvin is as follows:

  $T\left(\text{K}\right)=T\left(°\text{C}\right)+273$        (2)

The ideal gas equation can be expressed as follows,

  $PV=nRT$        (3)

Here,

$P$ is the pressure.

$V$ is the volume.

$T$ is the temperature.

$n$ is the mole of the gas.

$R$ is the gas constant.

(b)

Interpretation Introduction

Interpretation:

Whether all the ethanol vaporizes when the container is removed and warmed to room temperature is to be determined.

Concept introduction:

Clausius-Clapeyron equation is used to find the vapor pressure at one temperature when vapor pressure at another temperature and heat of enthalpy is given. It determines the change in vapor pressure with a change in temperature. The expression of the Clausius-Clapeyron equation is as follows:

  $\ln \left(\frac{P_2}{P_1}\right)=\frac{-\Delta H_{\text{vap}}}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)$        (1)

Here,

$P_1\text{ and }{P}_2$ are the vapor pressure.

$T_1\text{ and }{T}_2$ are the temperature.

$R$ is the gas constant.

$\Delta H_{\text{vap}}$ is the heat of vaporization.

The conversion factor to convert $°\text{C}$ to Kelvin is as follows:

  $T\left(\text{K}\right)=T\left(°\text{C}\right)+273$        (2)

The ideal gas equation can be expressed as follows,

  $PV=nRT$        (3)

Here,

$P$ is the pressure.

$V$ is the volume.

$T$ is the temperature.

$n$ is the mole of the gas.

$R$ is the gas constant.

(c)

Interpretation Introduction

Interpretation:

The mass of liquid ethanol that would be present at $0.0°\text{C}$ is to be calculated.

Concept introduction:

Clausius-Clapeyron equation is used to find the vapor pressure at one temperature when vapor pressure at another temperature and heat of enthalpy is given. It determines the change in vapor pressure with a change in temperature. The expression of the Clausius-Clapeyron equation is as follows:

  $\ln \left(\frac{P_2}{P_1}\right)=\frac{-\Delta H_{\text{vap}}}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)$        (1)

Here,

$P_1\text{ and }{P}_2$ are the vapor pressure.

$T_1\text{ and }{T}_2$ are the temperature.

$R$ is the gas constant.

$\Delta H_{\text{vap}}$ is the heat of vaporization.

The conversion factor to convert $°\text{C}$ to Kelvin is as follows:

  $T\left(\text{K}\right)=T\left(°\text{C}\right)+273$        (2)

The ideal gas equation can be expressed as follows,

  $PV=nRT$        (3)

Here,

$P$ is the pressure.

$V$ is the volume.

$T$ is the temperature.

$n$ is the mole of the gas.

$R$ is the gas constant.