Chapter 6, Problem 6.85QE

Interpretation Introduction

Interpretation:

The partial pressure of each gas in a flask that contains $0.22\text{ mol}$ neon, $0.33\text{ mol}$ for nitrogen and $0.22\text{ mol}$ oxygen has to be calculated.

Concept Introduction:

The net pressure of a mixture of gases is equal to the sum of the partial pressures of its constituent gases. This is known as Dalton’s law of partial pressure.

The total pressure for a mixture of two gases A and B is calculated as follows:

  $P_{\text{T}}=P_{\text{A}}+P_{\text{B}}$

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

  $P_{\text{A}}={\text{χ}}_{\text{A}}{P}_{\text{T}}$

Here,

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

${\text{χ}}_{\text{A}}$ denotes the mole fraction of component gas A.

$P_{\text{A}}$ denotes the pressure exerted by the gaseous component A.

$P_{\text{B}}$ denotes the pressure exerted by the gaseous component B.

Answer to Problem 6.85QE

The partial pressure of neon, nitrogen, and oxygen is $0.741\text{ atm}$, $1.114\text{ atm}$ and $0.741\text{ atm}$ respectively.

Explanation of Solution

The formula to calculate mole fraction is as follows:

  ${\text{χ}}_{\text{Ne}}=\frac{n_{\text{Ne}}}{n_{\text{Ne}}+n_{{\text{N}}_2}+n_{{\text{O}}_2}}$        (1)

Here,

$n_{\text{Ne}}$ denotes the number of moles of neon.

$n_{{\text{N}}_2}$ denotes the number of moles of ${\text{N}}_2$.

$n_{{\text{O}}_2}$ denotes the number of moles of ${\text{O}}_2$.

${\text{χ}}_{\text{Ne}}$ denotes the mole fraction of neon.

Substitute $0.22\text{ mol}$ for $n_{\text{Ne}}$, $0.33\text{ mol}$ for $n_{{\text{N}}_2}$ and $0.22\text{ mol}$ for $n_{{\text{O}}_2}$ to calculate the mole fraction of neon in equation (1).

  $\begin{array}{c}{\text{χ}}_{\text{Ne}}=\frac{0.22\text{ mol}}{0.22\text{ mol}+0.33\text{ mol}+0.22\text{ mol}}\\ =\frac{0.22\text{ mol}}{0.77\text{ mol}}\\ =0.285\end{array}$

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

  $P_{\text{Ne}}={\text{χ}}_{\text{Ne}}{P}_{\text{T}}$        (2)

Here,

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

${\text{χ}}_{\text{Ne}}$ denotes the mole fraction of neon.

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

Substitute 0.285 for ${\text{χ}}_{\text{Ne}}$ and $\text{2}\text{.6 atm}$ for $P_{\text{T}}$ in equation (2) to calculate the partial pressure of neon.

  $\begin{array}{c}{P}_{\text{Ne}}=\left(0.285\right)\left(\text{2}\text{.6 atm}\right)\\ =0.741\text{ atm}\end{array}$

The formula to calculate mole fraction is as follows:

  ${\text{χ}}_{{\text{N}}_2}=\frac{n_{{\text{N}}_2}}{n_{\text{Ne}}+n_{{\text{N}}_2}+n_{{\text{O}}_2}}$        (3)

Here,

$n_{\text{Ne}}$ denotes the number of moles of neon.

$n_{{\text{N}}_2}$ denotes the number of moles of ${\text{N}}_2$.

$n_{{\text{O}}_2}$ denotes the number of moles of ${\text{O}}_2$.

${\text{χ}}_{{\text{N}}_2}$ denotes the mole fraction of ${\text{N}}_2$.

Substitute $0.22\text{ mol}$ for $n_{\text{Ne}}$, $0.33\text{ mol}$ for $n_{{\text{N}}_2}$, $0.22\text{ mol}$ for $n_{{\text{O}}_2}$ to calculate the mole fraction of ${\text{N}}_2$ in equation (3).

  $\begin{array}{c}{\text{χ}}_{{\text{N}}_2}=\frac{0.33\text{ mol}}{0.22\text{ mol}+0.33\text{ mol}+0.22\text{ mol}}\\ =\frac{0.33\text{ mol}}{0.77\text{ mol}}\\ =0.4285\end{array}$

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

  $P_{{\text{N}}_2}={\text{χ}}_{{\text{N}}_2}{P}_{\text{T}}$        (4)

Here,

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

${\text{χ}}_{{\text{N}}_2}$ denotes the mole fraction of ${\text{N}}_2$.

$P_{{\text{N}}_2}$ denotes the partial pressure exerted by ${\text{N}}_2$.

Substitute 0.4285 for ${\text{χ}}_{\text{Ne}}$ and $\text{2}\text{.6 atm}$ for $P_{\text{T}}$ in equation (4) to calculate the partial pressure of

  $\begin{array}{c}{P}_{{\text{N}}_2}=\left(0.4285\right)\left(\text{2}\text{.6 atm}\right)\\ =1.114\text{ atm}\end{array}$

The formula to calculate mole fraction is as follows:

  ${\text{χ}}_{{\text{O}}_2}=\frac{n_{{\text{O}}_2}}{n_{\text{Ne}}+n_{{\text{N}}_2}+n_{{\text{O}}_2}}$        (5)

Here,

$n_{\text{Ne}}$ denotes the number of moles of neon.

$n_{{\text{N}}_2}$ denotes the number of moles of ${\text{N}}_2$.

$n_{{\text{O}}_2}$ denotes the number of moles of ${\text{O}}_2$.

${\text{χ}}_{{\text{O}}_2}$ denotes the mole fraction of ${\text{O}}_2$.

Substitute $0.22\text{ mol}$ for $n_{\text{Ne}}$, $0.33\text{ mol}$ for $n_{{\text{N}}_2}$ and $0.22\text{ mol}$ for $n_{{\text{O}}_2}$ to calculate the mole fraction of neon in equation (5).

  $\begin{array}{c}{\text{χ}}_{{\text{O}}_2}=\frac{0.22\text{ mol}}{0.22\text{ mol}+0.33\text{ mol}+0.22\text{ mol}}\\ =\frac{0.22\text{ mol}}{0.77\text{ mol}}\\ =0.285\end{array}$

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

  $P_{{\text{O}}_2}={\text{χ}}_{{\text{O}}_2}{P}_{\text{T}}$        (6)

Here,

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

${\text{χ}}_{{\text{O}}_2}$ denotes the mole fraction of ${\text{O}}_2$.

$P_{{\text{O}}_2}$ denotes the partial pressure exerted by ${\text{O}}_2$.

Substitute 0.285 for ${\text{χ}}_{\text{Ne}}$ and $\text{2}\text{.6 atm}$ for $P_{\text{T}}$ in equation (6) to calculate the partial pressure of ${\text{O}}_2$.

  $\begin{array}{c}{P}_{{\text{O}}_2}=\left(0.285\right)\left(\text{2}\text{.6 atm}\right)\\ =0.741\text{ atm}\end{array}$