Chapter 10, Problem 72E

Interpretation Introduction

Interpretation:

The volume of dry gas when the volume of wet gas collected over water is $95.0\text{mL}$ at $20\text{°C}$ and $758\text{mm}\text{Hg}$, is to be calculated.

Concept introduction:

Ideal gas is defined as the gas in which the collisions between the molecules and the atoms are perfectly elastic and there are no intermolecular attractive forces found between them. $\text{22}\text{.4}\text{L}$ is occupied by one mole of an ideal gas.

Answer to Problem 72E

The volume of dry gas when the volume of wet gas collected over water is $95.0\text{mL}$ at $20\text{°C}$ and $758\text{mm}\text{Hg}$, is $86.472\text{mL}$.

Explanation of Solution

The relation between the initial and final pressure, volume and temperature of gas is shown below.

$\frac{P_1{V}_1}{T_1}=\frac{P_2{V}_2}{T_2}$ …(1)

Where,

• $P_1$ is the initial pressure.

• $V_1$ is the initial volume.

• $T_1$ is the initial temperature.

• $P_2$ is the final pressure.

• $V_2$ is the final volume.

• $T_2$ is the final temperature.

The above equation can be rearranged for the value of final volume $V_2$ as shown below.

$V_2=\frac{P_1{V}_1}{T_1}\times \frac{T_2}{P_2}$ …(2)

The volume of wet gas is $95.0\text{mL}$ at $20\text{°C}$ and $758\text{mm}\text{Hg}$.

The vapor pressure of water is $17.5\text{mm}\text{Hg}$.

According to the Dalton’s law of partial pressure, the total pressure is calculated as shown below.

${\text{P}}_{\text{Total}}={\text{P}}_{\text{1}}+{\text{P}}_{\text{2}}$

The above formula can be rearranged for the value of initial pressure ${\text{P}}_{\text{1}}$ as shown below.

${\text{P}}_{\text{1}}={\text{P}}_{\text{Total}}-{\text{P}}_{\text{2}}$

Substitute the value of vapor pressure and total pressure in the above expression.

$\begin{array}{rll}{\text{P}}_{\text{1}}& =& \left(\text{758}-\text{17}\text{.5}\right)\text{mm}\text{Hg}\\ & =& \text{740}\text{.5}\text{mm}\text{Hg}\end{array}$

The initial temperature is $20\text{°C}$ and the final temperature is at STP, that is, $0\text{°C}$.

Conversion of temperature from Celsius to Kelvin is done as shown below.

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

Therefore, $\text{20}{}^{\text{o}}\text{C}$ and $\text{0}{}^{\text{o}}\text{C}$ is calculated to Kelvin as shown below.

$\begin{array}{rll}{\text{T}}_{\text{initial}}\left(\text{K}\right)& =& \left(20+273\right)\text{K}\\ & =& \text{293}\text{K}\end{array}$

$\begin{array}{rll}{\text{T}}_{\text{final}}\left(\text{K}\right)& =& \left(\text{0}+273\right)\text{K}\\ & =& \text{273}\text{K}\end{array}$

So, the initial and final temperature are $\text{293}\text{K}$ and $\text{273}\text{K}$ respectively.

The value of initial volume is $95.0\text{mL}$, the value of initial and final pressure is $\text{740}\text{.5}\text{mm}\text{Hg}$ and $758\text{mm}\text{Hg}$ respectively.

Substitute the values of initial and final temperature, pressure and volume into the equation (2).

$\begin{array}{rll}{V}_2& =& \frac{P_1{V}_1}{T_1}\times \frac{T_2}{P_2}\\ V_2& =& \frac{740.5\text{mm}\text{Hg}\times 95\text{mL}}{293\text{K}}\times \frac{273\text{K}}{758\text{mm}\text{Hg}}\\ & =& \frac{19204867.5}{222094}\\ & =& 86.472\text{mL}\end{array}$

Therefore, the final volume is $86.472\text{mL}$.

Conclusion

The volume of dry gas is $86.472\text{mL}$.