Chapter 13, Problem 13.38P

Interpretation Introduction

Interpretation:

Total pressure and partial pressure of ${\text{N}}_{\text{2}}$ and ${\text{O}}_{\text{2}}$ has to be calculated.

Concept Introduction:

The combination of all the gas laws namely Boyle’s law, Charles law and Avogadro’s leads to the ideal gas equation that can be used to relate all the four properties such as given below:

  $PV=nRT$        (1)

Here,

$\text{R}$ denotes gas constant.

$V$ denotes the volume.

$n$ denotes the temperature.

$P$ denotes the pressure

The net pressure for gaseous mixture is equal to algebraic sum of partial pressures of its constituent gases. This is referred as Dalton’s law.

For a mixture of two gases, 1 and 2, $P_{\text{total}}$ is calculated as follows:

  $P_{\text{total}}=P_1+P_2$

In terms of mole fraction the partial pressure is calculated as follows:

  $P_{\text{i}}=x_{\text{i}}{P}_{\text{total}}$

Here,

$P_{\text{total}}$ denotes total pressure exerted by whole mixture of gases.

$x_{\text{i}}$ denotes the mole fraction.

$P_{\text{i}}$ represents the partial pressure.

Answer to Problem 13.38P

Total pressure and partial pressure of ${\text{N}}_{\text{2}}$ and ${\text{O}}_{\text{2}}$ is $2.04815\text{ atm}$, $1.2002\text{ atm}$ and $0.83093atm$.

Explanation of Solution

The conversion factor to convert $\text{mL}$ to $\text{L}$ is as follows:

  $1\text{ mL}={10}^{-3}\text{ L}$

Thus $\text{650 mL}$ is converted to $\text{L}$ as follows:

  $\begin{array}{c}\text{Volume}=\left(\text{650 mL}\right)\left(\frac{{10}^{-3}\text{ L}}{\text{1 mL}}\right)\\ =0.65\text{ L}\end{array}$

Thus $\text{500 mL}$ is converted to $\text{L}$ as follows:

  $\begin{array}{c}\text{Volume}=\left(\text{500 mL}\right)\left(\frac{{10}^{-3}\text{ L}}{\text{1 mL}}\right)\\ =0.5\text{ L}\end{array}$

Assume the temperature is constant throughout and is $25°\text{C}$.

The formula to convert degree Celsius to kelvin is as follows:

  $T\left(\text{K}\right)=T\left(°\text{C}\right)+273\text{ K}$        (1)

Substitute $\text{25 }°\text{C}$ in equation (1).

  $\begin{array}{c}T\left(\text{K}\right)=25°\text{C}+273\text{ K}\\ =298\text{ K}\end{array}$

The conversion factor to convert $\text{atm}$ to $\text{torr}$ is as follows:

  $1\text{ atm}=\text{760 torr}$

Thus, $\text{825 torr}$ is converted to $\text{atm}$ as follows:

  $\begin{array}{c}\text{Pressure}=\left(\text{825 torr}\right)\left(\frac{1\text{ atm}}{760\text{ torr}}\right)\\ =1.085\text{ atm}\end{array}$

Thus, $\text{732 torr}$ is converted to $\text{atm}$ as follows:

  $\begin{array}{c}\text{Pressure}=\left(\text{732 torr}\right)\left(\frac{1\text{ atm}}{760\text{ torr}}\right)\\ =0.96315\text{ atm}\end{array}$

The expression to calculate $n$ as per the ideal gas equation is as follows:

  $n=\frac{PV}{RT}$        (2)

Substitute $0.65\text{ L}$ for $V$, $1.085\text{ atm}$ for $P$, and $0.08206\text{ L}\cdot \text{atm/mol}\cdot \text{K}$ for $R$ and $298\text{ K}$ for $T$ in equation (2)to evaluate moles for ${\text{N}}_{\text{2}}$.

  $\begin{array}{c}n=\frac{\left(\text{1}\text{.085 atm}\right)\left(0.65\text{ L}\right)}{\left(0.08206\text{ L}\cdot \text{atm/mol}\cdot \text{K}\right)\left(298\text{ K}\right)}\\ =0.02884\text{ mol}\end{array}$

Substitute $0.5\text{ L}$ for $V$, $0.96315\text{ atm}$ for $P$, and $0.08206\text{ L}\cdot \text{atm/mol}\cdot \text{K}$ for $R$ and $298\text{ K}$ for $T$ in equation (2) to evaluate moles for ${\text{O}}_{\text{2}}$.

  $\begin{array}{c}n=\frac{\left(0.96315\text{ atm}\right)\left(\text{0}\text{.5 L}\right)}{\left(0.08206\text{ L}\cdot \text{atm/mol}\cdot \text{K}\right)\left(298\text{ K}\right)}\\ =0.01969\text{ mol}\end{array}$

The formula of total moles in mixture is given as follows:

  $n_{\text{total}}=n_{{\text{N}}_{\text{2}}}+n_{{\text{O}}_{\text{2}}}$        (3)

Substitute $0.02884\text{ mol}$ for $n_{{\text{N}}_{\text{2}}}$, $0.01969\text{ mol}$ for $n_{{\text{O}}_{\text{2}}}$ in equation (3).

  $\begin{array}{c}{n}_{\text{total}}=0.02884\text{ mol}+0.01969\text{ mol}\\ =0.04853\text{ mol}\end{array}$

The formula to calculate mole fraction is as follows:

  $x=\frac{n}{n_{\text{total}}}$        (4)

Here,

$n_{\text{total}}$ denotes total moles in mixture.

Substitute $0.02884\text{ mol}$ for $n$, $0.04853\text{ mol}$ for $n_{\text{total}}$ in equation (4) to calculate mole fraction of ${\text{N}}_{\text{2}}$.

  $\begin{array}{c}x=\frac{0.02884\text{ mol}}{0.04853\text{ mol}}\\ =0.5942\end{array}$

Similarly, substitute $0.01969\text{ mol}$ for $n$, $0.04853\text{ mol}$ for $n_{\text{total}}$ in equation (4) to calculate mole fraction of ${\text{O}}_{\text{2}}$.

  $\begin{array}{c}x=\frac{0.01969\text{ mol}}{0.04853\text{ mol}}\\ =0.40572\end{array}$

For given mixture of gases $P_{\text{total}}$ as per Dalton’s law is calculated as follows:

  $P_{\text{total}}=P_{N_2}+P_{{\text{O}}_{\text{2}}}$        (5)

Substitute $1.085\text{ atm}$ for $P_{N_2}$ and $0.96315\text{ atm}$ for $P_{{\text{O}}_{\text{2}}}$ in equation (5).

  $\begin{array}{c}{P}_{\text{total}}=1.085\text{ atm}+0.96315\text{ atm}\\ =2.04815\text{ atm}\end{array}$

The partial pressure is calculated as follows:

  $P_{{\text{N}}_{\text{2}}}=x_{{\text{N}}_{\text{2}}}{P}_{\text{total}}$        (6)

Here,

$P_{\text{total}}$ denotes total pressure exerted by mixture.

$x_{{\text{N}}_{\text{2}}}$ denotes the mole fraction of ${\text{N}}_{\text{2}}$.

$P_{{\text{N}}_{\text{2}}}$ denotes the partial pressure exerted by ${\text{N}}_{\text{2}}$.

Substitute 0.5962 for $x_{{\text{N}}_{\text{2}}}$ and $2.04815\text{ atm}$ for $P_{\text{total}}$ in equation (6).

  $\begin{array}{c}{P}_{{\text{N}}_{\text{2}}}=\left(0.5962\right)\left(2.04815\text{ atm}\right)\\ =1.2002\text{ atm}\end{array}$

In terms of mole fraction the partial pressure is calculated as follows:

  $P_{{\text{O}}_{\text{2}}}=x_{{\text{O}}_{\text{2}}}{P}_{\text{total}}$        (7)

Substitute 0.40572 for $x_{{\text{O}}_{\text{2}}}$ and $2.04815\text{ atm}$ for $P_{\text{total}}$ in equation (7).

  $\begin{array}{c}{P}_{{\text{O}}_{\text{2}}}=\left(0.4057\right)\left(2.04815\text{ atm}\right)\\ =0.83093atm\end{array}$