Chapter 13, Problem 13.37P

Interpretation Introduction

Interpretation:

Partial pressure of ${\text{H}}_{\text{2}}$, ${\text{N}}_{\text{2}}$, $\text{Kr}$ and total pressure 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 13.37P

The partial pressure of ${\text{H}}_{\text{2}}$, ${\text{N}}_{\text{2}}$ and $\text{Kr}$ is $0.1999\text{ atm}$, $2.3999atm$ and $0.1999\text{ atm}$ respectively while the total pressure is $2.7997\text{ atm}$.

Explanation of Solution

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

  $n=\frac{PV}{RT}$        (2)

Substitute $\text{2 L}$ for $V$, $\text{1 atm}$ 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 ${\text{H}}_{\text{2}}$.

  $\begin{array}{c}n=\frac{\left(\text{1 atm}\right)\left(\text{2 L}\right)}{\left(0.08206\text{ L}\cdot \text{atm/mol}\cdot \text{K}\right)\left(298\text{ K}\right)}\\ =0.0817866\text{ mol}\end{array}$

Substitute $\text{8 L}$ for $V$, $\text{3 atm}$ 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 ${\text{N}}_{\text{2}}$.

  $\begin{array}{c}n=\frac{\left(\text{3 atm}\right)\left(\text{8 L}\right)}{\left(0.08206\text{ L}\cdot \text{atm/mol}\cdot \text{K}\right)\left(298\text{ K}\right)}\\ =0.981439\text{ mol}\end{array}$

Substitute $\text{4 L}$ for $V$, $\text{0}\text{.5 atm}$ 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 $\text{Kr}$.

  $\begin{array}{c}n=\frac{\left(\text{0}\text{.5 atm}\right)\left(\text{4 L}\right)}{\left(0.08206\text{ L}\cdot \text{atm/mol}\cdot \text{K}\right)\left(298\text{ K}\right)}\\ =0.0817866\text{ mol}\end{array}$

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

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

Substitute $0.0817866\text{ mol}$ for $n_{{\text{H}}_{\text{2}}}$, $0.981439\text{ mol}$ for $n_{{\text{N}}_{\text{2}}}$ and $0.0817866\text{ mol}$ for $n_{\text{Kr}}$ in equation (3).

  $\begin{array}{c}{n}_{\text{total}}=0.0817866\text{ mol}+0.981439\text{ mol}+0.0817866\text{ mol}\\ =\text{1}\text{.1450122 mol}\end{array}$

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 $0.0817866\text{ mol}$ for $n$ and $\text{1}\text{.1450122 mol}$ for $n_{\text{total}}$ in equation (4) to calculate mole fraction of ${\text{H}}_{\text{2}}$.

  $\begin{array}{c}x=\frac{0.0817866\text{ mol}}{\text{1}\text{.1450122 mol}}\\ =0.0714285\end{array}$

Similarly, substitute $0.981439\text{ mol}$ for $n$, $\text{1}\text{.1450122 mol}$ for $n_{\text{total}}$ in equation (4).

  $\begin{array}{c}{x}_{{\text{H}}_{\text{2}}}=\frac{0.981439\text{ mol}}{\text{1}\text{.1450122 mol}}\\ =0.8571428\end{array}$

Likewise, $0.0817866\text{ mol}$ for $n$, $\text{1}\text{.1450122 mol}$ for $n_{\text{total}}$ in equation (4) to calculate mole fraction of $\text{Kr}$.

  $\begin{array}{c}x=\frac{0.0817866\text{ mol}}{\text{1}\text{.1450122 mol}}\\ =0.0714285\end{array}$

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

  $P=\frac{nRT}{V}$        (5)

Substitute $298\text{ K}$ for $T$, $\text{10 L}$ for $V$, $\text{1}\text{.1450122 mol}$ for $n$, and $0.08206\text{ L}\cdot \text{atm/mol}\cdot \text{K}$ for $R$ in equation (5) to evaluate total pressure in vessel.

  $\begin{array}{c}P=\frac{\left(\text{1}\text{.1450122 mol}\right)\left(0.08206\text{ L}\cdot \text{atm/mol}\cdot \text{K}\right)\left(298\text{ K}\right)}{\left(\text{10 L}\right)}\\ =2.7999\text{ atm}\end{array}$

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

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

Here,

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

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

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

Substitute 0.0714285for $x_{{\text{H}}_{\text{2}}}$ and $2.7999\text{ atm}$ for $P_{\text{total}}$ in equation (6).

  $\begin{array}{c}{P}_{{\text{H}}_{\text{2}}}=\left(0.0714285\right)\left(2.7999\text{ atm}\right)\\ =0.1999\text{ atm}\end{array}$

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

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

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 0.8571428 for $x_{{\text{N}}_{\text{2}}}$ and $2.7999\text{ atm}$ for $P_{\text{total}}$ in equation (7).

  $\begin{array}{c}{P}_{N_2}=\left(0.8571428\right)\left(2.7999\text{ atm}\right)\\ =2.3999atm\end{array}$

Partial pressure of $\text{Kr}$ is calculated as follows:

  $P_{\text{Kr}}=x_{\text{Kr}}{P}_{\text{total}}$        (8)

Here,

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

$x_{\text{Kr}}$ denotes the mole fraction of $\text{Kr}$.

$P_{\text{Kr}}$ denotes the partial pressure exerted by $\text{Kr}$.

Substitute 0.0714285 for $x_{\text{Kr}}$ and $2.7999\text{ atm}$ for $P_{\text{total}}$ in equation (8).

  $\begin{array}{c}{P}_{\text{Kr}}=\left(0.0714285\right)\left(2.7999\text{ atm}\right)\\ =0.1999\text{ atm}\end{array}$

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

  $P_{\text{total}}=P_{{\text{H}}_{\text{2}}}+P_{N_2}+P_{\text{Kr}}$        (9)

Substitute $0.1999\text{ atm}$ for $P_{{\text{H}}_{\text{2}}}$, $2.3999atm$ for $P_{N_2}$ and $0.1999\text{ atm}$ for $P_{\text{Kr}}$ in equation (9).

  $\begin{array}{c}{P}_{\text{total}}=0.1999\text{ atm}+2.3999atm+0.1999\text{ atm}\\ =2.7997\text{ atm}\end{array}$