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.

Answer to Problem 12.141P

The mass of ethanol present in the vapor is $8.8\times {10}^{-2}\text{g}$.

Explanation of Solution

The boiling point of ethanol is $78.5°\text{C}$ and pressure is $760\text{torr}$.

Substitute $-\text{11}°\text{C}$ in equation (2) to calculate the temperature $T_1$ in Kelvin as follows:

  $\begin{array}{c}{T}_1\left(\text{K}\right)=-\text{11}°\text{C}+273\\ =262\text{ K}\end{array}$

Substitute $78.5°\text{C}$ in equation (2) to calculate the temperature $T_2$ in Kelvin as follows:

  $\begin{array}{c}{T}_2\left(\text{K}\right)=78.5°\text{C}+273.15\\ =351.6\text{ K}\end{array}$

Substitute $8.314\text{ J/K}\cdot \text{mol}$ for $R$, $262\text{ K}$ for $T_1$, $351.6\text{ K}$ for $T_2$ and $40.5\text{ kJ/mol}$ for $\Delta H_{\text{vap}}$ in the equation (1).

  $\begin{array}{c}\ln \left(\frac{P_2}{P_1}\right)=\frac{-\left(40.5\text{ kJ/mol}\right)\left(\frac{1000\text{J}}{1\text{kJ}}\right)}{8.314\text{ J/K}\cdot \text{mol}}\left(\frac{1}{351.6\text{ K}}-\frac{1}{262\text{ K}}\right)\\ =4.7380851\end{array}$

Take the exponential of $4.7380851$ and substitute $760\text{torr}$ for $P_2$ as follows:

  $\begin{array}{c}\frac{760\text{torr}}{P_1}=e^{4.7380851}\\ \frac{760\text{torr}}{P_1}=114.2153\\ P_1=6.65410\text{torr}\\ \approx 6.65\text{torr}\end{array}$

Rearrange the equation (3) to calculate the number of moles of ethanol at $262\text{ K}$.

  $n=\frac{PV}{RT}$        (4)

Substitute the value $6.65410\text{torr}$ for $P$, $262\text{ K}$ for $T$, $4.7\text{L}$ for $V$ and $8.314\text{ J/K}\cdot \text{mol}$ for $R$ in the equation (4).

  $\begin{array}{c}n=\frac{\left(6.65410\text{torr}\right)\left(\frac{1\text{atm}}{760\text{torr}}\right)\left(4.7\text{L}\right)}{\left(8.314\text{ J/K}\cdot \text{mol}\right)\left(262\text{ K}\right)}\\ =0.00191306\text{mol}\end{array}$

The formula to calculate the mass of ethanol is as follows:

  $\text{Mass of ethanol}=\left(\text{moles of}\text{ethanol}\right)\left(\text{molar mass of ethanol}\right)$        (5)

Substitute $0.00191306\text{mol}$ for moles of ethanol and $46.07\text{g/mol}$ for molar mass of ethanol in the equation (5).

  $\begin{array}{c}\text{Mass of ethanol}=\left(0.00191306\text{mol}\right)\left(46.07\text{g/mol}\right)\\ =0.08813467\text{g}\\ \approx 8.8\times {10}^{-2}\text{g}\text{.}\end{array}$

Conclusion

The vapor contains $8.8\times {10}^{-2}\text{g}$ of ethanol.

(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.

Answer to Problem 12.141P

All the ethanol vaporizes when the container is removed and warmed to room temperature.

Explanation of Solution

The boiling point of ethanol is $78.5°\text{C}$ and pressure is $760\text{torr}$.

Substitute $20°\text{C}$ in equation (2) to calculate the temperature $T_1$ in Kelvin as follows:

  $\begin{array}{c}{T}_1\left(\text{K}\right)=20°\text{C}+273\\ =293\text{ K}\end{array}$

Substitute $78.5°\text{C}$ in equation (2) to calculate the temperature $T_2$ in Kelvin as follows:

  $\begin{array}{c}{T}_2\left(\text{K}\right)=78.5°\text{C}+273.15\\ =351.6\text{ K}\end{array}$

Substitute $8.314\text{ J/K}\cdot \text{mol}$ for $R$, $293\text{ K}$ for $T_1$, $351.6\text{ K}$ for $T_2$ and $40.5\text{ kJ/mol}$ for $\Delta H_{\text{vap}}$ in the equation (1).

  $\begin{array}{c}\ln \left(\frac{P_2}{P_1}\right)=\frac{-\left(40.5\text{ kJ/mol}\right)\left(\frac{1000\text{J}}{1\text{kJ}}\right)}{8.314\text{ J/K}\cdot \text{mol}}\left(\frac{1}{351.6\text{ K}}-\frac{1}{293\text{ K}}\right)\\ =2.7709337\end{array}$

Take the exponential of $2.7709337$ and substitute $760\text{torr}$ for $P_2$ as follows:

  $\begin{array}{c}\frac{760\text{torr}}{P_1}=e^{2.7709337}\\ \frac{760\text{torr}}{P_1}=15.97354\\ P_1=47.5787\text{torr}\\ \approx 47.6\text{torr}\end{array}$

Rearrange the equation (3) to calculate the number of moles of ethanol at $293\text{ K}$.

  $n=\frac{PV}{RT}$        (6)

Substitute the value $47.5787\text{torr}$ for $P$, $293\text{ K}$ for $T$, $4.7\text{L}$ for $V$ and $8.314\text{ J/K}\cdot \text{mol}$ for $R$ in the equation (6).

  $\begin{array}{c}n=\frac{\left(47.5787\text{torr}\right)\left(4.7\text{L}\right)}{\left(8.314\text{ J/K}\cdot \text{mol}\right)\left(293\text{ K}\right)}\left(\frac{1\text{atm}}{760\text{torr}}\right)\\ =0.01223168\text{mol}\end{array}$

The formula to calculate the mass of ethanol is as follows:

  $\text{Mass of ethanol}=\left(\text{moles of}\text{ethanol}\right)\left(\text{molar mass of ethanol}\right)$        (5)

Substitute $0.01223168\text{mol}$ for moles of ethanol and $46.07\text{g/mol}$ for molar mass of ethanol in the equation (5).

  $\begin{array}{c}\text{Mass of ethanol}=\left(0.01223168\text{mol}\right)\left(46.07\text{g/mol}\right)\\ =0.5635135\text{g}\\ \approx 0.56\text{g}\text{.}\end{array}$

The mass of ethanol present in the vapor is less than the available liquid ethanol and therefore, all the ethanol vaporizes.

Conclusion

If the mass of substance present in the vapor phase is less than the liquid form then the substance will vaporize but if mass of substance present in the vapor phase is greater than the liquid form then the substance will not vaporize.

(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.

Answer to Problem 12.141P

The mass of liquid ethanol that would be present at $0.0°\text{C}$ is $0.15\text{g}$.

Explanation of Solution

The boiling point of ethanol is $78.5°\text{C}$ and pressure is $760\text{torr}$.

Substitute $78.5°\text{C}$ in equation (2) to calculate the temperature $T_1$ in Kelvin as follows:

  $\begin{array}{c}{T}_1\left(\text{K}\right)=78.5°\text{C}+273.15\\ =351.6\text{ K}\end{array}$

Substitute $0°\text{C}$ in equation (2) to calculate the temperature $T_2$ in Kelvin as follows:

  $\begin{array}{c}{T}_2\left(\text{K}\right)=0°\text{C}+273.2\\ =273.2\text{ K}\end{array}$

Substitute $8.314\text{ J/K}\cdot \text{mol}$ for $R$, $351.6\text{ K}$ for $T_1$, $273.2\text{ K}$ for $T_2$ and $40.5\text{ kJ/mol}$ for $\Delta H_{\text{vap}}$ in the equation (1).

  $\begin{array}{c}\ln \left(\frac{P_2}{P_1}\right)=\frac{-\left(40.5\text{ kJ/mol}\right)\left(\frac{1000\text{J}}{1\text{kJ}}\right)}{8.314\text{ J/K}\cdot \text{mol}}\left(\frac{1}{273.2\text{ K}}-\frac{1}{351.6\text{ K}}\right)\\ =–3.975864\end{array}$

Take the exponential of $–3.975864$ and substitute $760\text{torr}$ for $P_1$ as follows:

  $\begin{array}{c}\frac{P_2}{760\text{torr}}=e^{–3.975864}\\ \frac{P_2}{760\text{torr}}=0.0187631\\ P_2=14.259956\text{torr}\end{array}$

Rearrange the equation (3) to calculate the number of moles of ethanol at $273.2\text{ K}$.

  $n=\frac{PV}{RT}$        (7)

Substitute the value $14.259956\text{torr}$ for $P$, $273.2\text{ K}$ for $T$, $4.7\text{L}$ for $V$ and $8.314\text{ J/K}\cdot \text{mol}$ for $R$ in the equation (7).

  $\begin{array}{c}n=\frac{\left(14.259956\text{torr}\right)\left(\frac{1\text{atm}}{760\text{torr}}\right)\left(4.7\text{L}\right)}{\left(8.314\text{ J/K}\cdot \text{mol}\right)\left(273.2\text{ K}\right)}\\ =0.00393168\text{mol}\end{array}$

The formula to calculate the mass of ethanol is as follows:

  $\text{Mass of ethanol}=\left(\text{moles of}\text{ethanol}\right)\left(\text{molar mass of ethanol}\right)$        (5)

Substitute $0.00393168\text{mol}$ for moles of ethanol and $46.07\text{g/mol}$ for molar mass of ethanol in the equation (5).

  $\begin{array}{c}\text{Mass of ethanol}=\left(0.00393168\text{mol}\right)\left(46.07\text{g/mol}\right)\\ =0.18113\text{g}\end{array}$

The formula to calculate the mass of ethanol in liquid phase is as follows:

  $\text{Mass of ethanol in liquid}=\text{total mass}-\text{mass of ethanol in vapor}$        (8)

Substitute $0.33\text{g}$ for the total mass and $0.18113\text{g}$ for mass of ethanol in vapor in the equation (8).

  $\begin{array}{c}\text{Mass of ethanol in liquid}=\left(0.33\text{g}\right)-\left(0.18113\text{g}\right)\\ =0.14887\text{g}\\ \approx 0.15\text{g}\text{.}\end{array}$

Conclusion

At $0.0°\text{C}$, $0.15\text{g}$ of liquid ethanol would be present.