Chapter 11, Problem 11.96QE

(a)

Interpretation Introduction

Interpretation:

Pressure of the container has to be calculated if all the water present in it is vaporized.

Answer to Problem 11.96QE

The pressure of container that contains $\text{1}\text{.50}\text{g}$ of vaporized water is $2.07\text{atm}$.

Explanation of Solution

Mass of water present in the container is given as $\text{1}\text{.50}\text{g}$.  Molar mass of water is $\text{18}\text{.01}\text{g/mol}$.  The number of moles of water present in $\text{1}\text{.50}\text{g}$ is calculated as shown below;

    $\begin{array}{c}\text{Number}\text{of}\text{moles}=\frac{\text{Mass}\text{of}\text{water}}{\text{Molar}\text{mass}\text{of}\text{water}}\\ =\frac{\text{1}\text{.50}\text{g}}{18.01\text{g/mol}}\\ =0.0833\text{mol}\end{array}$

Therefore, the number of moles present in $\text{1}\text{.50}\text{g}$ of water is $0.0833\text{mol}$.

The ideal gas equation is given as shown below;

    $PV=nRT$        (1)

Rearranging the above equation in terms of pressure as follows;

    $P=\frac{nRT}{V}$

Substituting the values in above equation, the pressure of container can be calculated as shown below;

    $\begin{array}{c}P=\frac{nRT}{V}\\ =\frac{0.0833\text{mol}\times 0.0821\text{L}\cdot \text{atm}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}\times 303\text{K}}{1.00\text{L}}\\ =2.07\text{atm}\end{array}$

Therefore, the pressure of container that contains $\text{1}\text{.50}\text{g}$ of vaporized water is $2.07\text{atm}$.

(b)

Interpretation Introduction

Interpretation:

Vapor pressure of water at $\text{30}\text{°C}$ has to be given.

Explanation of Solution

From the table 6.3 present in the book, the vapor pressure of water at $\text{30 °C}$ is found to be $\text{31}\text{.84}\text{torr}$.

(c)

Interpretation Introduction

Interpretation:

Mass of water that evaporates has to be calculated and also the liquid is in equilibrium with vapor in the vessel or not has to be given.

Answer to Problem 11.96QE

The mass of vaporized water is $\text{0}\text{.303}\text{g}$ and the liquid-vapor equilibrium is present in the container.

Explanation of Solution

Vapor pressure of water is given as $\text{31}\text{.84}\text{torr}$ at temperature of $\text{30}\text{°C}$.

The vapor pressure of water can be converted into $\text{atm}$ units by using the conversion factor as shown below;

    $\begin{array}{c}\text{31}\text{.84}\text{torr}=\text{31}\text{.84}\text{torr}\times \frac{\text{1}\text{atm}}{\text{760}\text{torr}}\\ =0.0419atm\end{array}$

The ideal gas equation is given as shown below;

    $PV=nRT$        (1)

Rearranging the above equation in terms of number of moles as follows;

    $n=\frac{PV}{RT}$

Substituting the values in above equation, the number of moles of water can be calculated as shown below;

    $\begin{array}{c}n=\frac{PV}{RT}\\ =\frac{0.0419atm\times 1.00\text{L}}{0.0821\text{L}\cdot \text{atm}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}\times 303\text{K}}\\ =1.68\times {10}^{-3}\text{mol}\end{array}$

Therefore, the number of moles of water present in vapor state is $1.68\times {10}^{-3}\text{mol}$.

Molar mass of water is $\text{18}\text{.01}\text{g/mol}$.  From the number of moles and molar mass of water, the mass of water that is vaporized can be calculated as shown below;

    $\begin{array}{c}\text{Mass}\text{of}\text{water}=\text{Number}\text{of}\text{moles}\times \text{Molar}\text{mass}\\ =1.68\times {10}^{-3}\text{mol}\times 18.01\text{g/mol}\\ \text{=30}\text{.25}\times {\text{10}}^{\text{-3}}\text{g}\\ \text{=0}\text{.303}\text{g}\end{array}$

Therefore, the mass of water is $\text{0}\text{.303}\text{g}$.  As the total mass of water present in the container is $\text{1}\text{.50}\text{g}$ and the mass of water present in vapor state is $\text{0}\text{.303}\text{g}$, the liquid and vapor phase are in equilibrium.