Chapter 10, Problem 101AP

Interpretation Introduction

Interpretation:

The partial pressure and total pressure of the given mixtures are to be calculated.

Concept introduction:

The mole fraction of an individual gas for the combination of gases is the ratio of the moles of the individual gas with the total number of moles.

${\chi }_i=\frac{n_i}{n_{\text{total}}}$

Here, ${\chi }_i$ is the mole fraction, $n_i$ is the number of moles of the individual gas, and $n_{\text{total}}$ is the total number of moles.

The mole fraction of an individual gas for the combination of gases can be calculated from the ratio of the partial pressure of the individual gas with the total pressure of the combination.

${\chi }_i=\frac{P_i}{P_{\text{total}}}$

Here, ${\chi }_i$ is the mole fraction, $P_i$ is the partial pressure of the individual gas, and $P_{\text{total}}$ is the total pressure.

Answer to Problem 101AP

Solution:

(a)The pressure of flask (iii) is $2.67\text{ atm}$ and the pressure of flask (ii) is $4\text{ atm}$.

(b)The total pressure after opening the valve is $2.67\text{ atm}$ and partial pressure of gas A (red) is $1.33\text{ atm}$ and partial pressure of gas B (blue) is $1.33\text{ atm}$.

Explanation of Solution

Given information:

Volume: $\begin{array}{l}{V}_i=2\text{ L}\\ V_{ii}=1\text{ L}\\ V_{iii}=2\text{ L}\end{array}$

Pressure: $P_i=2\text{ atm}$

a)The pressure in flak (ii) and (iii)

The number of molecules in flask (i) is 9, whereas in flask (ii) the number of molecules is also 9.

The volume of flask (i) is $2\text{ L}$ and the volume of flask (ii) is $1\text{ L}$.

Now, the temperature and number of moles are constant and the volume of flask (ii) is half of the volume of flask (i), so the pressure of flask (ii) will be:

$P_{ii}=2P_i$

Substitute $2\text{ atm}$ for $P_i$ in the above equation

$\begin{array}{c}{P}_{ii}=2\times 2\text{ atm}\\ =4\text{ atm}\end{array}$

The number of molecules in flask (iii) is $12$.

The volume of flask (iii) is $2\text{ L}$.

So, the pressure of flask (ii) will be:

$P_{ii}=\frac{12}{9}{P}_i$

Substitute $2\text{ atm}$ for $P_i$ in the above equation

$\begin{array}{c}{P}_{ii}=\frac{12}{9}\times 2\text{ atm}\\ =2.67\text{ atm}\end{array}$

Hence, the pressure of flask (iii) is $2.67\text{ atm}$ and the pressure of flask (ii) is $4\text{ atm}$.

b) The total pressure and the partial pressure of each gas after the valves are opened.

Before opening the valves, flask (i) is considered.

So,

$\begin{array}{l}{n}_1=9\text{ moles}\\ V_1=2\text{ L}\\ P_1=2\text{ atm}\end{array}$

After opening the valves, the total of all flasks is considered.

So,

$\begin{array}{l}{n}_2=30\text{ moles}\\ V_2=5\text{ L}\end{array}$

The combined gas law forms the relationship between pressure, volume, temperature, and number of moles. This can be shown as

$\frac{P_1{V}_1}{T_1{n}_1}=\frac{P_2{V}_2}{T_2{n}_2}$

For constant temperature:

$\frac{P_1{V}_1}{n_1}=\frac{P_2{V}_2}{n_2}$

Rearrange the above equation for final pressure as follows:

$P_2=\frac{P_1{V}_1}{n_1}\times \frac{n_2}{V_2}$

Substitute $9\text{ moles}$ for $n_1$, $30\text{ moles}$ for $n_2$, $2\text{ L}$ for $V_1$, $5\text{ L}$ for $V_2$, and $2\text{ atm}$ for $P_1$ in the above equation

$\begin{array}{c}{P}_2=\frac{\left(2\text{ atm}\right)\left(2\text{ L}\right)}{9\text{ moles}}\times \frac{30\text{ moles}}{5\text{ L}}\\ =\frac{4\times 30}{9\times 5}\text{ atm}\\ =\frac{120}{45}\text{ atm}\\ =2.67\text{ atm}\end{array}$

The number of red color sphere is 15.

The number of blue color sphere is 15.

So, the total number of moles is as follows:

$n_{\text{total}}=n_{\text{red}}+n_{\text{blue}}$

Substitute $15$ for $n_{\text{red}}$ and $15$ for $n_{\text{blue}}$ in the above equation

$\begin{array}{c}{n}_{\text{total}}=15+15\\ =30\end{array}$

Calculate the mole fraction of red color sphere as follows:

${\chi }_{\text{red}}=\frac{n_{\text{red}}}{n_{\text{total}}}$

Substitute $30$ for $n_{\text{total}}$ and $15$ for $n_{\text{red}}$ in the above equation

$\begin{array}{c}{\chi }_{\text{red}}=\frac{15}{30}\\ =\frac{1}{2}\end{array}$

Calculate the mole fraction of the blue color sphere as follows:

${\chi }_{\text{blue}}=\frac{n_{\text{blue}}}{n_{\text{total}}}$

Substitute $30$ for $n_{\text{total}}$ and $15$ for $n_{\text{blue}}$ in the above equation

$\begin{array}{c}{\chi }_{\text{blue}}=\frac{15}{30}\\ =\frac{1}{2}\end{array}$

Calculate the partial pressure of the red gas as

$P_{\text{red}}={\chi }_{\text{red}}\times P_{\text{total}}$

Substitute $1/2$ for ${\chi }_{\text{red}}$ and $2.67\text{ atm}$ for $P_{\text{total}}$ in the above equation

$\begin{array}{c}{P}_{\text{red}}=\frac{1}{2}\times 2.67\text{ atm}\\ =1.33\text{ atm}\end{array}$

Calculate the partial pressure of the blue gas as follows:

$P_{\text{blue}}={\chi }_{\text{blue}}\times P_{\text{total}}$

Substitute $1/2$ for ${\chi }_{\text{red}}$ and $2.67\text{ atm}$ for $P_{\text{total}}$ in theabove equation

$\begin{array}{c}{P}_{\text{blue}}=\frac{1}{2}\times 2.67\text{ atm}\\ =1.33\text{ atm}\end{array}$

Hence, the total pressure after opening the valve is $2.67\text{ atm}$ and the partial pressure of gas A (red) is $1.33\text{ atm}$ and the partial pressure of gas B (blue) is $1.33\text{ atm}$.