Chapter 1, Problem 1A.8BE

(i)

Interpretation Introduction

Interpretation: The volume of the given gas mixture at $300\text{K}$ has to be calculated.

Concept introduction: The partial pressure of any gas in a mixture of gases is calculated using the total pressure of the mixture and the mole fraction of the gas in the mixture.  This is represented by the formula given below as,

  $p_{\text{total}}=\frac{p_{\text{a}}}{x_{\text{a}}}$

Answer to Problem 1A.8BE

The volume of the given gas mixture at $300\text{K}$ has been calculated as $\underset{\_}{3.14\times 10^{-3}{m}^3}$.

Explanation of Solution

The mass of neon given is equal to $225\text{mg}$ .  The number of moles of neon can be calculated using the formula given below as,

  $n=\frac{\text{Given}\text{mass}}{\text{Molar}\text{mass}}$

Where,

• Ø  $n$ is the number of moles.

The molar mass of neon is $\text{20}\text{.1797}\text{g}/\text{mol}$.  Substitute the values in the above equation for neon as given below.

    $\begin{array}{c}n=\frac{\text{Given}\text{mass}}{\text{Molar}\text{mass}}\\ n_{\text{Ne}}=\frac{225\times {10}^{-3}\text{g}}{\text{20}\text{.1797}\text{g}/\text{mol}}\\ =11.149\times {10}^{-3}\text{mol}\end{array}$

The molar mass of methane and argon are $\text{16}\text{g}/\text{mol}$ and $\text{39}\text{.948}\text{g}/\text{mol}$ respectively.  The given mass of methane and argon are $320\text{mg}$ and $175\text{mg}$ respectively.  Similarly, the number of moles of methane and argon are calculated as given below.

    $\begin{array}{c}n=\frac{\text{Given}\text{mass}}{\text{Molar}\text{mass}}\\ n_{{\text{CH}}_4}=\frac{320\times {10}^{-3}\text{g}}{\text{16}\text{g}/\text{mol}}\\ =20\times {10}^{-3}\text{mol}\end{array}$

    $\begin{array}{c}n=\frac{\text{Given}\text{mass}}{\text{Molar}\text{mass}}\\ n_{\text{Ar}}=\frac{175\times {10}^{-3}\text{g}}{\text{39}\text{.948}\text{g}/\text{mol}}\\ =4.38\times {10}^{-3}\text{mol}\end{array}$

The total number of moles of the gas mixture is calculated as given below.

  $n_{\text{total}}=n_{\text{Ne}}+n_{{\text{CH}}_4}+n_{\text{Ar}}$

Substitute the values in the above equation as follows.

    $\begin{array}{c}{n}_{\text{total}}=n_{\text{Ne}}+n_{{\text{CH}}_4}+n_{\text{Ar}}\\ =11.149\times {10}^{-3}\text{mol}+20\times {10}^{-3}\text{mol}+4.38\times {10}^{-3}\text{mol}\\ =35.529\times {10}^{-3}\text{mol}\end{array}$

The mole fraction of neon is calculated using the formula given below as,

    $x_{\text{Ne}}=\frac{n_{\text{Ne}}}{n_{\text{Total}}}$

Substitute the values in the above equation as follows.

    $\begin{array}{c}{x}_{\text{Ne}}=\frac{n_{\text{Ne}}}{n_{\text{Total}}}\\ =\frac{11.149\times {10}^{-3}\text{mol}}{35.529\times {10}^{-3}\text{mol}}\\ =0.313\end{array}$

Total pressure is calculated using the formula given below as,

    $p_{\text{total}}=\frac{p_{\text{Ne}}}{x_{\text{Ne}}}$

The partial pressure of neon is given as $8.87\text{kPa}$.  Substitute the values in the above equation as follows.

    $\begin{array}{c}{p}_{\text{total}}=\frac{p_{\text{Ne}}}{x_{\text{Ne}}}\\ =\frac{8.87\text{kPa}}{0.313}\\ =28.3\text{kPa}\end{array}$

The perfect gas equation is given by the formula as,

    $pV=nRT$

Where,

• Ø  $p$ is the total pressure.
• Ø  $V$ is the volume.
• Ø  $n$ is the number of moles.
• Ø  $R$ is the gas constant.
• Ø  $T$ is the temperature.

Rearrange the above equation for volume as follows.

    $\begin{array}{c}pV=nRT\\ V=\frac{nRT}{p}\end{array}$

Substitute the values in the above equation as follows.

    $\begin{array}{c}V=\frac{nRT}{p}\\ =\frac{\left(35.529\times {10}^{-3}\text{mol}\right)\left(8.314\text{Pa}{\text{m}}^{\text{3}}{\text{K}}^{-1}{\text{mol}}^{-1}\right)\left(300\text{K}\right)}{28.3\times {10}^3\text{Pa}}\\ =\underset{\_}{3.14\times 10^{-3}{m}^3}\end{array}$

Thus, the volume of the gas mixture is $\underset{\_}{3.14\times 10^{-3}{m}^3}$.

(ii)

Interpretation Introduction

Interpretation: The total pressure of the given gas mixture at $300\text{K}$ has to be calculated.

Concept introduction: The partial pressure of any gas in a mixture of gases is calculated using the total pressure of the mixture and the mole fraction of the gas in the mixture.  This is represented by the formula given below as,

$p_{\text{total}}=\frac{p_{\text{a}}}{x_{\text{a}}}$

Answer to Problem 1A.8BE

The total pressure of the given gas mixture at $300\text{K}$ has been calculated as $\underset{\_}{28.3kPa}$.

Explanation of Solution

The mass of neon given is equal to $225\text{mg}$.  The number of moles of neon can be calculated using the formula given below as,

  $n=\frac{\text{Given}\text{mass}}{\text{Molar}\text{mass}}$

Where,

• Ø  $n$ is the number of moles.

The molar mass of neon is $\text{20}\text{.1797}\text{g}/\text{mol}$.  Substitute the values in the above equation for neon as given below.

    $\begin{array}{c}n=\frac{\text{Given}\text{mass}}{\text{Molar}\text{mass}}\\ n_{\text{Ne}}=\frac{225\times {10}^{-3}\text{g}}{\text{20}\text{.1797}\text{g}/\text{mol}}\\ =11.149\times {10}^{-3}\text{mol}\end{array}$

The molar mass of methane and argon are $\text{16}\text{g}/\text{mol}$ and $\text{39}\text{.948}\text{g}/\text{mol}$ respectively.  The given mass of methane and argon are $320\text{mg}$ and $175\text{mg}$ respectively.  Similarly, the number of moles of methane and argon are calculated as given below.

    $\begin{array}{c}n=\frac{\text{Given}\text{mass}}{\text{Molar}\text{mass}}\\ n_{{\text{CH}}_4}=\frac{320\times {10}^{-3}\text{g}}{\text{16}\text{g}/\text{mol}}\\ =20\times {10}^{-3}\text{mol}\end{array}$

    $\begin{array}{c}n=\frac{\text{Given}\text{mass}}{\text{Molar}\text{mass}}\\ n_{\text{Ar}}=\frac{175\times {10}^{-3}\text{g}}{\text{39}\text{.948}\text{g}/\text{mol}}\\ =4.38\times {10}^{-3}\text{mol}\end{array}$

The total number of moles of the gas mixture is calculated as given below.

  $n_{\text{total}}=n_{\text{Ne}}+n_{{\text{CH}}_4}+n_{\text{Ar}}$

Substitute the values in the above equation as follows.

    $\begin{array}{c}{n}_{\text{total}}=n_{\text{Ne}}+n_{{\text{CH}}_4}+n_{\text{Ar}}\\ =11.149\times {10}^{-3}\text{mol}+20\times {10}^{-3}\text{mol}+4.38\times {10}^{-3}\text{mol}\\ =35.529\times {10}^{-3}\text{mol}\end{array}$

The mole fraction of neon is calculated using the formula given below as,

    $x_{\text{Ne}}=\frac{n_{\text{Ne}}}{n_{\text{Total}}}$

Substitute the values in the above equation as follows.

    $\begin{array}{c}{x}_{\text{Ne}}=\frac{n_{\text{Ne}}}{n_{\text{Total}}}\\ =\frac{11.149\times {10}^{-3}\text{mol}}{35.529\times {10}^{-3}\text{mol}}\\ =0.313\end{array}$

Total pressure is calculated using the formula given below as,

    $p_{\text{total}}=\frac{p_{\text{Ne}}}{x_{\text{Ne}}}$

The partial pressure of neon is given as $8.87\text{kPa}$.  Substitute the values in the above equation as follows.

    $\begin{array}{c}{p}_{\text{total}}=\frac{p_{\text{Ne}}}{x_{\text{Ne}}}\\ =\frac{8.87\text{kPa}}{0.313}\\ =\underset{\_}{28.3kPa}\end{array}$

Thus, the total pressure of the gas mixture is $\underset{\_}{28.3kPa}$.