Chapter 6, Problem 6.94QE

Interpretation Introduction

Interpretation:

The mole fraction of ${\text{CO}}_{\text{2}}$, ${\text{N}}_{\text{2}}$, $\text{Ar}$ and ${\text{O}}_2$ when total atmospheric pressure is $7\times {10}^2\text{ Pa}$ 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 6.94QE

The mole fraction of ${\text{CO}}_{\text{2}}$, ${\text{N}}_{\text{2}}$, $\text{Ar}$ and ${\text{O}}_2$ is $667.24\text{ Pa}$, $\text{18}\text{.9 Pa}$, $\text{11}\text{.2 Pa}$ and $0.91\text{ Pa}$ respectively.

Explanation of Solution

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

  $P_{{\text{CO}}_{\text{2}}}={\text{χ}}_{{\text{CO}}_{\text{2}}}{P}_{\text{T}}$        (1)

Here,

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

${\text{χ}}_{{\text{CO}}_{\text{2}}}$ denotes the mole fraction of ${\text{CO}}_{\text{2}}$.

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

Substitute 0.9532 for ${\text{χ}}_{{\text{CO}}_{\text{2}}}$ and $7\times {10}^2\text{ Pa}$ for $P_{\text{T}}$ in equation (1).

  $\begin{array}{c}{P}_{{\text{CO}}_{\text{2}}}=\left(0.9532\right)\left(7\times {10}^2\text{ Pa}\right)\\ =667.24\text{ Pa}\end{array}$

The formula to calculate partial pressure of ${\text{N}}_{\text{2}}$ is as follows:

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

Here,

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

${\text{χ}}_{{\text{N}}_{\text{2}}}$ 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 0.027 for ${\text{χ}}_{{\text{N}}_{\text{2}}}$ and $7\times {10}^2\text{ Pa}$ for $P_{\text{T}}$ in equation (2).

  $\begin{array}{c}{P}_{{\text{N}}_{\text{2}}}=\left(0.027\right)\left(7\times {10}^2\text{ Pa}\right)\\ =0.189\times {10}^2\text{ Pa}\\ =\text{18}\text{.9 Pa}\end{array}$

The formula to calculate partial pressure of $\text{Ar}$ is as follows:

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

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 0.016 for ${\text{χ}}_{\text{Ar}}$ and $7\times {10}^2\text{ Pa}$ for $P_{\text{T}}$ in equation (3).

  $\begin{array}{c}{P}_{\text{Ar}}=\left(0.016\right)\left(7\times {10}^2\text{ Pa}\right)\\ =0.112\times {10}^2\text{ Pa}\\ =\text{11}\text{.2 Pa}\end{array}$

The formula to calculate the partial pressure of ${\text{O}}_{\text{2}}$ is as follows:

  $P_{{\text{O}}_{\text{2}}}={\text{χ}}_{{\text{O}}_{\text{2}}}{P}_{\text{T}}$        (4)

Here,

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

${\text{χ}}_{{\text{O}}_{\text{2}}}$ denotes the mole fraction of ${\text{O}}_{\text{2}}$.

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

Substitute 0.0013 for ${\text{χ}}_{{\text{O}}_{\text{2}}}$ and $7\times {10}^2\text{ Pa}$ for $P_{\text{T}}$ in equation (4).

  $\begin{array}{c}{P}_{{\text{O}}_{\text{2}}}=\left(0.0013\right)\left(7\times {10}^2\text{ Pa}\right)\\ =0.91\text{ Pa}\end{array}$