Chapter 6, Problem 6.86QE

Interpretation Introduction

Interpretation:

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 6.86QE

Partial pressure exerted by each gas in a flask that contains $\text{2}\text{.3 g}$ neon,  $\text{0}\text{.33 g}$ for nitrogen and $\text{1}\text{.1 g}$ oxygen are $2.5324\text{ atm}$, $0.005553\text{ atm}$ and $0.06107\text{ atm}$ respectively.

Explanation of Solution

The formula to convert mass in gram to moles is as follows:

  $\text{Number of moles}=\frac{\text{Given mass}}{\text{molar mass}}$        (1)

Substitute $\text{2}\text{.3 g}$ for mass and $\text{20 g/mol}$ for molar mass to calculate the number of moles of neon in equation (1).

  $\begin{array}{c}\text{Number of moles}=\frac{\text{2}\text{.3 g}}{\text{20}\text{.17 g/mol}}\\ =1.1403\text{ mol}\end{array}$

Substitute $\text{0}\text{.33 g}$ for mass and $\text{131}\text{.29 g/mol}$ for molar mass to calculate the number of moles of xenon in equation (1).

  $\begin{array}{c}\text{Number of moles}=\frac{\text{0}\text{.33 g}}{\text{131}\text{.29 g/mol}}\\ =0.0025\text{ mol}\end{array}$

Substitute $\text{1}\text{.1 g}$ for mass and $\text{39}\text{.948 g/mol}$ for molar mass to calculate the number of moles of argon in equation (1).

  $\begin{array}{c}\text{Number of moles}=\frac{\text{1}\text{.1 g}}{\text{39}\text{.948 g/mol}}\\ =0.0275\text{ mol}\end{array}$

The formula to calculate mole fraction is as follows:

  ${\text{χ}}_{\text{Ne}}=\frac{n_{\text{Ne}}}{n_{\text{Ne}}+n_{\text{Xe}}+n_{\text{Ar}}}$        (2)

Here,

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

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

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

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

Substitute $1.1403\text{ mol}$ for $n_{\text{Ne}}$, $0.0025\text{ mol}$ for $n_{\text{Xe}}$ and $0.0275\text{ mol}$ for $n_{\text{Ar}}$ to calculate the mole fraction of neon in equation (2).

  $\begin{array}{c}{\text{χ}}_{\text{Ne}}=\frac{1.1403\text{ mol}}{1.1403\text{ mol}+0.0025\text{ mol}+0.0275\text{ mol}}\\ =\frac{1.1403\text{ mol}}{\text{1}\text{.1703 mol}}\\ =0.974\end{array}$

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

  $P_{\text{Ne}}={\text{χ}}_{\text{Ne}}{P}_{\text{T}}$        (3)

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 0.974 for ${\text{χ}}_{\text{Ne}}$ and $\text{2}\text{.6 atm}$ for $P_{\text{T}}$ in equation (3) to calculate the partial pressure of neon.

  $\begin{array}{c}{P}_{\text{Ne}}=\left(0.974\right)\left(\text{2}\text{.6 atm}\right)\\ =2.5324\text{ atm}\end{array}$

The formula to calculate mole fraction is as follows:

  ${\text{χ}}_{\text{Xe}}=\frac{n_{\text{Xe}}}{n_{\text{Ne}}+n_{\text{Xe}}+n_{\text{Ar}}}$        (4)

Here,

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

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

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

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

Substitute $0.0025\text{ mol}$ for $n_{\text{Xe}}$, $1.1403\text{ mol}$ for $n_{\text{Ne}}$ and $0.0275\text{ mol}$ for $n_{\text{Ar}}$ to calculate the mole fraction of $\text{Xe}$ in equation (4).

  $\begin{array}{c}{\text{χ}}_{\text{Xe}}=\frac{0.0025\text{ mol}}{1.1403\text{ mol}+0.0025\text{ mol}+0.0275\text{ mol}}\\ =\frac{0.0025\text{ mol}}{\text{1}\text{.1703 mol}}\\ =0.002136\end{array}$

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

  $P_{\text{Xe}}={\text{χ}}_{\text{Xe}}{P}_{\text{T}}$        (5)

Here,

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

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

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

Substitute 0.002136 for ${\text{χ}}_{\text{Xe}}$ and $\text{2}\text{.6 atm}$ for $P_{\text{T}}$ in equation (5) to calculate the partial pressure of $\text{Xe}$.

  $\begin{array}{c}{P}_{\text{Xe}}=\left(0.002136\right)\left(\text{2}\text{.6 atm}\right)\\ =0.00555\text{ atm}\end{array}$

The formula to calculate mole fraction is as follows:

  ${\text{χ}}_{\text{Ar}}=\frac{n_{\text{Ar}}}{n_{\text{Ne}}+n_{\text{Xe}}+n_{\text{Ar}}}$        (6)

Here,

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

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

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

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

Substitute $0.0025\text{ mol}$ for $n_{\text{Xe}}$, $1.1403\text{ mol}$ for $n_{\text{Ne}}$ and $0.0275\text{ mol}$ for $n_{\text{Ar}}$ to calculate the mole fraction of $\text{Ar}$ in equation (6).

  $\begin{array}{c}{\text{χ}}_{\text{Ar}}=\frac{0.0275\text{ mol}}{1.1403\text{ mol}+0.0025\text{ mol}+0.0275\text{ mol}}\\ =\frac{0.0275\text{ mol}}{\text{1}\text{.1703 mol}}\\ =0.02349\end{array}$

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

  $P_{\text{Ar}}={\text{χ}}_{\text{Ar}}{P}_{\text{T}}$        (7)

Here,

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

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

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

Substitute 0.02349 for ${\text{χ}}_{\text{Ar}}$ and $\text{2}\text{.6 atm}$ for $P_{\text{T}}$ in equation (7) to calculate the partial pressure of $\text{Ar}$.

  $\begin{array}{c}{P}_{\text{Ar}}=\left(0.02349\right)\left(\text{2}\text{.6 atm}\right)\\ =0.06107\text{ atm}\end{array}$