Chapter 16, Problem 16A.2BE

(a)

Interpretation Introduction

Interpretation:

The diffusion constant of nitrogen at $20°\text{C}$ and $100.0\text{ Pa}$ is to be calculated.  The flow of gas due to diffusion at $100.0\text{ Pa}$ has to be calculated.

Concept introduction:

Diffusion of gas is a process in which the gas follows from higher pressure area to lower pressure area.  The process of diffusion requires a potential gradient.  The diffusion coefficient of a gas is given by the expression as shown below.

    $D=\frac{1}{3}\left(\frac{kT}{\sigma p}\right){\left(\frac{8RT}{\text{π}M}\right)}^{1/2}$

Answer to Problem 16A.2BE

The diffusion constant of nitrogen at $20°\text{C}$ and $100.0\text{ Pa}$ is $\underset{\_}{1.476m^2{s}^{-1}}$.  The flow of gas due to diffusion at $100.0\text{ Pa}$ is $\underset{\_}{-72.708mol{m}^{-2}{s}^{-1}}$.

Explanation of Solution

The diffusion coefficient of a gas is given by the expression as shown below.

    $D=\frac{1}{3}\left(\frac{kT}{\sigma p}\right){\left(\frac{8RT}{\text{π}M}\right)}^{1/2}$        (1)

Where,

• $R$ is the gas constant $\left(8.314{\text{J K}}^{-1}{\text{ mol}}^{-1}\right)$.
• $T$ is the temperature.
• $k$ is the Boltzmann constant $\left(1.381\times {10}^{-23}{\text{J K}}^{-1}\right)$.
• $\sigma$ is the collision cross section area.
• $M$ is the molar mass.
• $p$ is the pressure.

The temperature of nitrogen gas is $20°\text{C}$.

The conversion of temperature in Kelvin is shown below.

    $\begin{array}{c}T=\left(273+20.0°\text{C}\right)\text{K}\\ =\text{293 K}\end{array}$

The pressure of nitrogen gas at $20°\text{C}$ is $100.0\text{ Pa}$.

The conversion of pressure in ${\text{kg m}}^{-1}{\text{ s}}^{-2}$ is shown below.

    $\begin{array}{c}p=\left(100.0\text{ Pa}\right)\left(\frac{{\text{1 kg m}}^{-1}{\text{ s}}^{-2}}{\text{1 Pa}}\right)\\ =\text{100}{\text{.0 kg m}}^{-1}{\text{ s}}^{-2}\end{array}$

The value of $\sigma$ for nitrogen has is $0.43{\text{ nm}}^2$.

The conversion of $\sigma$ in ${\text{m}}^2$ is shown below.

    $\begin{array}{c}\sigma =\left(0.43{\text{ nm}}^2\right)\left(\frac{1{\text{ m}}^2}{{10}^{20}\text{ nm}}\right)\\ =0.43\times {10}^{-20}{\text{ m}}^2\end{array}$

The molar mass of nitrogen is $28.02\times {10}^{-3}{\text{ kg mol}}^{-1}$.

Substitute the value of $R$, $T$, $k$, $\sigma$, $M$, and $p$ in the equation (1).

    $\begin{array}{c}D=\left(\begin{array}{c}\frac{1}{3}\left(\frac{\left(1.381\times {10}^{-23}{\text{J K}}^{-1}\right)\left(\text{293 K}\right)}{\left(0.43\times {10}^{-20}{\text{ m}}^2\right)\left(\text{100}{\text{.0 kg m}}^{-1}{\text{ s}}^{-2}\right)}\right)\\ {\left(\frac{8\left(8.314{\text{J K}}^{-1}{\text{ mol}}^{-1}\right)\left(\frac{1{\text{kg m}}^2{\text{ s}}^{-2}}{1\text{J}}\right)\left(\text{293 K}\right)}{\left(3.14\right)\left(28.02\times {10}^{-3}{\text{ kg mol}}^{-1}\right)}\right)}^{1/2}\end{array}\right)\\ =\left(3.1367\times {10}^{-3}\right){\left(221498.0201\right)}^{1/2}{\text{ m}}^2{\text{ s}}^{-1}\\ =\left(3.1367\times {10}^{-3}\right)\left(470.6358\right){\text{ m}}^2{\text{ s}}^{-1}\\ =\underset{\_}{1.476m^2{s}^{-1}}\end{array}$

Therefore, the diffusion constant of nitrogen at $20°\text{C}$ and $100.0\text{ Pa}$ is $\underset{\_}{1.476m^2{s}^{-1}}$. 

The pressure gradient of the pipe is $1.20{\text{ bar m}}^{-1}$.

The flow of gas due to diffusion $\left(\frac{J_z}{N_{\text{A}}}\right)$ is given by the expression as shown below.

    $\frac{J_z}{N_{\text{A}}}=-\left(\frac{D}{RT}\right)\frac{\text{d}p}{\text{d}z}$

Where,

• $\frac{\text{d}p}{\text{d}z}$ is the pressure gradient.

Substitute the values of $\frac{\text{d}p}{\text{d}z}$, $D$, $R$, and $T$ in the above equation.

    $\begin{array}{c}\frac{J_z}{N_{\text{A}}}=-\left(\frac{1.476{\text{m}}^2{\text{ s}}^{-1}}{\left(8.314{\text{J K}}^{-1}{\text{ mol}}^{-1}\right)\left(\frac{1{\text{kg m}}^2{\text{ s}}^{-2}}{1\text{J}}\right)\left(\text{293 K}\right)}\right)\left(1.20{\text{ bar m}}^{-1}\right)\\ =-\left(6.059\times {10}^{-4}{\text{ mol s kg}}^{-1}\right)\left(1.20{\text{ bar m}}^{-1}\right)\left(\frac{{\text{10}}^5{\text{ kg m}}^{-1}{\text{ s}}^{-2}}{\text{1 bar}}\right)\\ =\underset{\_}{-72.708mol{m}^{-2}{s}^{-1}}\end{array}$

Therefore, the flow of gas due to diffusion at $100.0\text{ Pa}$ is $\underset{\_}{-72.708mol{m}^{-2}{s}^{-1}}$.

(b)

Interpretation Introduction

Interpretation:

The diffusion constant of nitrogen at $20°\text{C}$ and $100\text{ kPa}$ is to be calculated.  The flow of gas due to diffusion at $100\text{ kPa}$ has to be calculated.

Concept introduction:

As mentioned in the concept introduction in part (a).

Answer to Problem 16A.2BE

The diffusion constant of nitrogen at $20°\text{C}$ and $100\text{ kPa}$ is $\underset{\_}{1.476\times 10^{-3}{m}^2{s}^{-1}}$.  The flow of gas due to diffusion at $100\text{ kPa}$ is $\underset{\_}{-0.0727mol{m}^{-2}{s}^{-1}}$.

Explanation of Solution

The diffusion coefficient of a gas is given by the expression as shown below.

    $D=\frac{1}{3}\left(\frac{kT}{\sigma p}\right){\left(\frac{8RT}{\text{π}M}\right)}^{1/2}$        (1)

Where,

• $R$ is the gas constant $\left(8.314{\text{J K}}^{-1}{\text{ mol}}^{-1}\right)$.
• $T$ is the temperature.
• $k$ is the Boltzmann constant $\left(1.381\times {10}^{-23}{\text{J K}}^{-1}\right)$.
• $\sigma$ is the collision cross section area.
• $M$ is the molar mass.
• $p$ is the pressure.

The temperature of nitrogen gas is $20°\text{C}$.

The conversion of temperature in Kelvin is shown below.

    $\begin{array}{c}T=\left(273+20.0°\text{C}\right)\text{K}\\ =\text{293 K}\end{array}$

The pressure of nitrogen gas at $20°\text{C}$ is $100\text{ kPa}$.

The conversion of pressure in ${\text{kg m}}^{-1}{\text{ s}}^{-2}$ is shown below.

    $\begin{array}{c}p=\left(100\text{ kPa}\right)\left(\frac{{\text{10}}^3{\text{ kg m}}^{-1}{\text{ s}}^{-2}}{\text{1 kPa}}\right)\\ =\text{1}\text{.00}\times {10}^5{\text{ kg m}}^{-1}{\text{ s}}^{-2}\end{array}$

The value of $\sigma$ for nitrogen has is $0.43{\text{ nm}}^2$.

The conversion of $\sigma$ in ${\text{m}}^2$ is shown below.

    $\begin{array}{c}\sigma =\left(0.43{\text{ nm}}^2\right)\left(\frac{1{\text{ m}}^2}{{10}^{20}\text{ nm}}\right)\\ =0.43\times {10}^{-20}{\text{ m}}^2\end{array}$

The molar mass of nitrogen is $28.02\times {10}^{-3}{\text{ kg mol}}^{-1}$.

Substitute the value of $R$, $T$, $k$, $\sigma$, $M$, and $p$ in the equation (1).

    $\begin{array}{c}D=\left(\begin{array}{c}\frac{1}{3}\left(\frac{\left(1.381\times {10}^{-23}{\text{J K}}^{-1}\right)\left(\text{293 K}\right)}{\left(0.43\times {10}^{-20}{\text{ m}}^2\right)\left(\text{1}\text{.00}\times {10}^5{\text{ kg m}}^{-1}{\text{ s}}^{-2}\right)}\right)\\ {\left(\frac{8\left(8.314{\text{J K}}^{-1}{\text{ mol}}^{-1}\right)\left(\frac{1{\text{kg m}}^2{\text{ s}}^{-2}}{1\text{J}}\right)\left(\text{293 K}\right)}{\left(3.14\right)\left(28.02\times {10}^{-3}{\text{ kg mol}}^{-1}\right)}\right)}^{1/2}\end{array}\right)\\ =\left(3.1367\times {10}^{-6}\right){\left(221498.0201\right)}^{1/2}{\text{ m}}^2{\text{ s}}^{-1}\\ =\left(3.1367\times {10}^{-6}\right)\left(470.6358\right){\text{ m}}^2{\text{ s}}^{-1}\\ =\underset{\_}{1.476\times 10^{-3}{m}^2{s}^{-1}}\end{array}$

Therefore, the diffusion constant of nitrogen at $20°\text{C}$ and $100\text{ kPa}$ is $\underset{\_}{1.476\times 10^{-3}{m}^2{s}^{-1}}$.

The pressure gradient of the pipe is $1.20{\text{ bar m}}^{-1}$.

The flow of gas due to diffusion $\left(\frac{J_z}{N_{\text{A}}}\right)$ is given by the expression as shown below.

    $\frac{J_z}{N_{\text{A}}}=-\left(\frac{D}{RT}\right)\frac{\text{d}p}{\text{d}z}$

Where,

• $\frac{\text{d}p}{\text{d}z}$ is the pressure gradient.

Substitute the values of $\frac{\text{d}p}{\text{d}z}$, $D$, $R$, and $T$ in the above equation.

    $\begin{array}{c}\frac{J_z}{N_{\text{A}}}=-\left(\frac{1.476\times {10}^{-3}{\text{m}}^2{\text{ s}}^{-1}}{\left(8.314{\text{J K}}^{-1}{\text{ mol}}^{-1}\right)\left(\frac{1{\text{kg m}}^2{\text{ s}}^{-2}}{1\text{J}}\right)\left(\text{293 K}\right)}\right)\left(1.20{\text{ bar m}}^{-1}\right)\\ =-\left(6.059\times {10}^{-7}{\text{ mol s kg}}^{-1}\right)\left(1.20{\text{ bar m}}^{-1}\right)\left(\frac{{\text{10}}^5{\text{ kg m}}^{-1}{\text{ s}}^{-2}}{\text{1 bar}}\right)\\ =\underset{\_}{-0.0727mol{m}^{-2}{s}^{-1}}\end{array}$

Therefore, the flow of gas due to diffusion at $100\text{ kPa}$ is $\underset{\_}{-0.0727mol{m}^{-2}{s}^{-1}}$.

(c)

Interpretation Introduction

Interpretation:

The diffusion constant of nitrogen at $20°\text{C}$ and $20.0\text{ MPa}$ is to be calculated.  The flow of gas due to diffusion at $20.0\text{ MPa}$ has to be calculated.

Concept introduction:

As mentioned in the concept introduction in part (a).

Answer to Problem 16A.2BE

The diffusion constant of nitrogen at $20°\text{C}$ and $20.0\text{ MPa}$ is $\underset{\_}{7.38\times 10^{-6}{m}^2{s}^{-1}}$.  The flow of gas due to diffusion at $20.0\text{ MPa}$ is $\underset{\_}{-3.636\times 10^{-4}mol{m}^{-2}{s}^{-1}}$.

Explanation of Solution

The diffusion coefficient of a gas is given by the expression as shown below.

    $D=\frac{1}{3}\left(\frac{kT}{\sigma p}\right){\left(\frac{8RT}{\text{π}M}\right)}^{1/2}$        (1)

Where,

• $R$ is the gas constant $\left(8.314{\text{J K}}^{-1}{\text{ mol}}^{-1}\right)$.
• $T$ is the temperature.
• $k$ is the Boltzmann constant $\left(1.381\times {10}^{-23}{\text{J K}}^{-1}\right)$.
• $\sigma$ is the collision cross section area.
• $M$ is the molar mass.
• $p$ is the pressure.

The temperature of nitrogen gas is $20°\text{C}$.

The conversion of temperature in Kelvin is shown below.

    $\begin{array}{c}T=\left(273+20.0°\text{C}\right)\text{K}\\ =\text{293 K}\end{array}$

The pressure of nitrogen gas at $20°\text{C}$ is $20.0\text{ MPa}$.

The conversion of pressure in ${\text{kg m}}^{-1}{\text{ s}}^{-2}$ is shown below.

    $\begin{array}{c}p=\left(20.0\text{ MPa}\right)\left(\frac{{\text{10}}^6{\text{ kg m}}^{-1}{\text{ s}}^{-2}}{\text{1 MPa}}\right)\\ =2.\text{00}\times {10}^7{\text{ kg m}}^{-1}{\text{ s}}^{-2}\end{array}$

The value of $\sigma$ for nitrogen has is $0.43{\text{ nm}}^2$.

The conversion of $\sigma$ in ${\text{m}}^2$ is shown below.

    $\begin{array}{c}\sigma =\left(0.43{\text{ nm}}^2\right)\left(\frac{1{\text{ m}}^2}{{10}^{20}\text{ nm}}\right)\\ =0.43\times {10}^{-20}{\text{ m}}^2\end{array}$

The molar mass of nitrogen is $28.02\times {10}^{-3}{\text{ kg mol}}^{-1}$.

Substitute the value of $R$, $T$, $k$, $\sigma$, $M$, and $p$ in the equation (1).

    $\begin{array}{c}D=\left(\begin{array}{c}\frac{1}{3}\left(\frac{\left(1.381\times {10}^{-23}{\text{J K}}^{-1}\right)\left(\text{293 K}\right)}{\left(0.43\times {10}^{-20}{\text{ m}}^2\right)\left(2.\text{00}\times {10}^7{\text{ kg m}}^{-1}{\text{ s}}^{-2}\right)}\right)\\ {\left(\frac{8\left(8.314{\text{J K}}^{-1}{\text{ mol}}^{-1}\right)\left(\frac{1{\text{kg m}}^2{\text{ s}}^{-2}}{1\text{J}}\right)\left(\text{293 K}\right)}{\left(3.14\right)\left(28.02\times {10}^{-3}{\text{ kg mol}}^{-1}\right)}\right)}^{1/2}\end{array}\right)\\ =\left(1.568\times {10}^{-8}\right){\left(221498.0201\right)}^{1/2}{\text{ m}}^2{\text{ s}}^{-1}\\ =\left(1.568\times {10}^{-8}\right)\left(470.6358\right){\text{ m}}^2{\text{ s}}^{-1}\\ =\underset{\_}{7.38\times 10^{-6}{m}^2{s}^{-1}}\end{array}$

Therefore, the diffusion constant of nitrogen at $20°\text{C}$ and $20.0\text{ MPa}$ is $\underset{\_}{7.38\times 10^{-6}{m}^2{s}^{-1}}$.

The pressure gradient of the pipe is $1.20{\text{ bar m}}^{-1}$.

The flow of gas due to diffusion $\left(\frac{J_z}{N_{\text{A}}}\right)$ is given by the expression as shown below.

    $\frac{J_z}{N_{\text{A}}}=-\left(\frac{D}{RT}\right)\frac{\text{d}p}{\text{d}z}$

Where,

• $\frac{\text{d}p}{\text{d}z}$ is the pressure gradient.

Substitute the values of $\frac{\text{d}p}{\text{d}z}$, $D$, $R$, and $T$ in the above equation.

    $\begin{array}{c}\frac{J_z}{N_{\text{A}}}=-\left(\frac{7.38\times {10}^{-6}{\text{m}}^2{\text{ s}}^{-1}}{\left(8.314{\text{J K}}^{-1}{\text{ mol}}^{-1}\right)\left(\frac{1{\text{kg m}}^2{\text{ s}}^{-2}}{1\text{J}}\right)\left(\text{293 K}\right)}\right)\left(1.20{\text{ bar m}}^{-1}\right)\\ =-\left(3.03\times {10}^{-9}{\text{ mol s kg}}^{-1}\right)\left(1.20{\text{ bar m}}^{-1}\right)\left(\frac{{\text{10}}^5{\text{ kg m}}^{-1}{\text{ s}}^{-2}}{\text{1 bar}}\right)\\ =\underset{\_}{-3.636\times 10^{-4}mol{m}^{-2}{s}^{-1}}\end{array}$

Therefore, the flow of gas due to diffusion at $20.0\text{ MPa}$ is $\underset{\_}{-3.636\times 10^{-4}mol{m}^{-2}{s}^{-1}}$.