Chapter 11, Problem 11.95QE

A 1.50-g sample of methanol (CH₃OH) is placed in an evacuated 1.00-L container at 30 °C.

1. (a) Calculate the pressure in the container if all of the methanol is vaporized. (Assume the ideal gas law, PV = nRT.)
2. (b) The vapor pressure of methanol at 30 °C is 158 torr. What mass of methanol actually evaporates? Is liquid in equilibrium with vapor in the vessel?

Interpretation Introduction

Interpretation:

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

Answer to Problem 11.95QE

The pressure of container that contains $\text{1}\text{.50}\text{g}$ of vaporized methanol is $\text{1}\text{.16}\text{atm}$.

Explanation of Solution

Mass of methanol present in the container is given as $\text{1}\text{.50}\text{g}$.  Molar mass of methanol is $\text{32}\text{.04}\text{g/mol}$.  The number of moles of methanol present in $\text{1}\text{.50}\text{g}$ is calculated as shown below;

    $\begin{array}{c}\text{Number}\text{of}\text{moles}=\frac{\text{Mass}\text{of}\text{methanol}}{\text{Molar}\text{mass}\text{of}\text{methanol}}\\ =\frac{\text{1}\text{.50}\text{g}}{\text{32}\text{.04}\text{g/mol}}\\ =0.0468\text{mol}\end{array}$

Therefore, the number of moles present in $\text{1}\text{.50}\text{g}$ of methanol is $\text{0}\text{.0468}\text{mol}$.

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;

    $\begin{array}{c}P=\frac{nRT}{V}\\ =\frac{0.0468\text{mol}\times 0.0821\text{L}\cdot \text{atm}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}\times 303\text{K}}{1.00\text{L}}\\ =1.16\text{atm}\end{array}$

Therefore, the pressure of container that contains $\text{1}\text{.50}\text{g}$ of vaporized methanol is $\text{1}\text{.16}\text{atm}$.

Interpretation Introduction

Interpretation:

Mass of methanol 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 11.95QE

The mass of vaporized methanol is $\text{0}\text{.268}\text{g}$ and the liquid-vapor equilibrium is present in the container.

Explanation of Solution

Vapor pressure of methanol is given as $\text{158}\text{torr}$ at temperature of $\text{30}\text{°C}$.

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

    $\begin{array}{c}\text{158}\text{torr}=\text{158}\text{torr}\times \frac{\text{1}\text{atm}}{\text{760}\text{torr}}\\ =0.208atm\end{array}$

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 methanol can be calculated as shown below;

    $\begin{array}{c}n=\frac{PV}{RT}\\ =\frac{0.208atm\times 1.00\text{L}}{0.0821\text{L}\cdot \text{atm}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}\times 303\text{K}}\\ =8.36\times {10}^{-3}\text{mol}\end{array}$

Therefore, the number of moles of methanol present in vapor state is $8.36\times {10}^{-3}\text{mol}$.

Molar mass of methanol is $\text{32}\text{.04}\text{g/mol}$.  From the number of moles and molar mass of methanol, the mass of methanol that is vaporized can be calculated as shown below;

    $\begin{array}{c}\text{Mass}\text{of}\text{methanol}=\text{Number}\text{of}\text{moles}\times \text{Molar}\text{mass}\\ =8.36\times {10}^{-3}\text{mol}\times 32.04\text{g/mol}\\ \text{=268}\times {\text{10}}^{\text{-3}}\text{g}\\ \text{=0}\text{.268}\text{g}\end{array}$

Therefore, the mass of methanol is $\text{0}\text{.268}\text{g}$.  As the total mass of methanol present in the container is $\text{1}\text{.50}\text{g}$ and the mass of methanol present in vapor state is $\text{0}\text{.268}\text{g}$, the liquid and vapor phase are in equilibrium.