Chapter 1, Problem 1C.15P

Interpretation Introduction

Interpretation: The graph of $p/\rho$ versus $p$ is straight line graph with slope proportional to $B\text{'}$ has to be shown.  The values of $B\text{'}$ and $B$ have to be stated for the given data at $25°\text{C}$.

Concept introduction: The Virial equation of state is given by the formula as,

    $p{V}_{\text{m}}=RT\left(1+\frac{B}{V_{\text{m}}}+\frac{C}{V_{\text{m}}^2}+\mathrm{...}\right)$

Where,

• • $p$ is the pressure.
• • $V_{\text{m}}$ is the molar volume.
• • $R$ is the gas constant.
• • $T$ is the temperature.
• • $B,C$ are the Virial coefficients.

Answer to Problem 1C.15P

The graph of $p/\rho$ versus $p$ is straight line graph with slope proportional to $B\text{'}$ has been shown by the equation given below as,

    $\frac{p}{\rho }=\frac{RT}{M}+\frac{B\text{'}RTp}{M}$

The values of $B\text{'}$ and $B$ have been calculated as $\underset{\_}{9.07\times 10^{-7}P{a}^{-1}}$ and $\underset{\_}{2.25d{m}^{3}mo{l}^{-1}}$ respectively.

Explanation of Solution

The Virial equation of state in terms of $B\text{'}$ is given by the formula as,

    $p{V}_{\text{m}}=RT\left(1+B\text{'}p+C\text{'}{p}^2+\mathrm{...}\right)$ (1)

Where,

• • $p$ is the pressure.
• • $V_{\text{m}}$ is the molar volume.
• • $R$ is the gas constant.
• • $T$ is the temperature.
• • $B\text{'},C\text{'}$ are the Virial coefficients.

The molar volume is given as,

    $V_{\text{m}}=\frac{V}{n}$

Substitute the above equation in equation (1) as,

    $p\left(\frac{V}{n}\right)=RT\left(1+B\text{'}p+C\text{'}{p}^2+\mathrm{...}\right)$ (2)

The number of moles can be calculated using the formula given below as,

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

Rearrange the equation (2) as follows.

    $\begin{array}{c}p\left(\frac{V}{n}\right)=RT\left(1+B\text{'}p+C\text{'}{p}^2+\mathrm{...}\right)\\ p=\frac{nRT}{V}\left(1+B\text{'}p+C\text{'}{p}^2+\mathrm{...}\right)\\ p=\frac{m}{V}\times \frac{RT}{M}\left(1+B\text{'}p+C\text{'}{p}^2+\mathrm{...}\right)\\ p=\rho \left(\frac{RT}{M}\right)\left(1+B\text{'}p+C\text{'}{p}^2+\mathrm{...}\right)\end{array}$

The above equation can also be written as given below.

    $\begin{array}{c}p=\rho \left(\frac{RT}{M}\right)\left(1+B\text{'}p+C\text{'}{p}^2+\mathrm{...}\right)\\ \frac{p}{\rho }=\frac{RT}{M}+B\text{'}p\left(\frac{RT}{M}\right)+\mathrm{...}\end{array}$                                                                 (3)

On neglecting the higher terms in equation (3), the equation obtained is given below as,

    $\frac{p}{\rho }=\frac{RT}{M}+\frac{B\text{'}RTp}{M}$

According the above equation, the plot of $\left(p/\rho \right)$ versus $p$  gives a straight line graph with intercept $\frac{RT}{M}$ and slope $\frac{B\text{'}RT}{M}$

The value of $\left(p/\rho \right)$ for the given data is calculated as given below.

$p/\text{kPa}$	$\rho /\text{kg}{\text{m}}^{-3}$	$\left(p/\rho \right)/{10}^4{\text{m}}^{\text{2}}{\text{s}}^{-2}$
$12.223$	0.225	$5.432$
$25.20$	$0.456$	$5.526$
$36.97$	$0.664$	$5.568$
$60.37$	$1.062$	$5.685$
$85.23$	$1.468$	$5.806$
$101.3$	$1.734$	$5.842$

The graph of $\left(p/\rho \right)$ versus $p$ is shown below as,

The slope of the graph is calculated as,

    $\text{Slope}=\frac{y_2-y_1}{x_2-x_1}$

Where,

• • $x_2,x_1$ are the points on x-axis.
• • $y_2,y_1$ are the points on y-axis.

Substitute the two points $\left(85.23\times {10}^3\text{Pa},5.8\times {10}^4{\text{m}}^{\text{2}}{\text{s}}^{-2}\right)$ and $\left(25.20\times {10}^3\text{Pa},5.5\times {10}^4{\text{m}}^{\text{2}}{\text{s}}^{-2}\right)$ in the above equation as,

    $\begin{array}{c}\text{Slope}=\frac{y_2-y_1}{x_2-x_1}\\ =\frac{5.8\times {10}^4{\text{m}}^{\text{2}}{\text{s}}^{-2}-5.5\times {10}^4{\text{m}}^{\text{2}}{\text{s}}^{-2}}{85.23\times {10}^3\text{Pa}-25.20\times {10}^3\text{Pa}}\\ =0.049{\text{kg}}^{-1}{\text{m}}^3\end{array}$

Also, $\text{Slope}=\frac{B\text{'}RT}{M}$.  Thus, the value of $B\text{'}$ can be calculated as,

    $\begin{array}{c}\frac{B\text{'}RT}{M}=0.049{\text{kg}}^{-1}{\text{m}}^3\\ B\text{'}=\frac{\left(0.049{\text{kg}}^{-1}{\text{m}}^3\right)M}{RT}\\ =\frac{\left(0.049{\text{kg}}^{-1}{\text{m}}^3\right)}{\text{Intercept}}\end{array}$

The intercept for the graph is equal to $5.4\times {10}^4{\text{m}}^{\text{2}}{\text{s}}^{-2}$.

Substitute the corresponding value in the above equation as,

    $\begin{array}{c}B\text{'}=\frac{\left(0.049{\text{kg}}^{-1}{\text{m}}^3\right)}{\text{Intercept}}\\ =\frac{\left(0.049{\text{kg}}^{-1}{\text{m}}^3\right)}{\left(5.4\times {10}^4{\text{m}}^{\text{2}}{\text{s}}^{-2}\right)}\\ =9.07\times {10}^{-7}{\text{kg}}^{-1}\text{m}{\text{s}}^{-2}\\ =\underset{\_}{9.07\times 10^{-7}P{a}^{-1}}\end{array}$

The relation between $B\text{'}$ and $B$ is given as,

    $B\text{'}=\frac{B}{RT}$

Substitute the corresponding values in the above equation as,

    $\begin{array}{c}B\text{'}=\frac{B}{RT}\\ 9.07\times {10}^{-7}\times 1.0133\times {10}^5{\text{atm}}^{-1}=\frac{B}{\left(0.0821{\text{dm}}^3\text{atm}{\text{mol}}^{-1}{\text{K}}^{-1}\right)\left(25+273.15\right)\text{K}}\\ B=\left(9.07\times {10}^{-7}\times 1.0133\times {10}^5{\text{atm}}^{-1}\right)\left(0.0821{\text{dm}}^3\text{atm}{\text{mol}}^{-1}{\text{K}}^{-1}\right)\left(298.15\right)\text{K}\\ =\underset{\_}{2.25d{m}^{3}mo{l}^{-1}}\end{array}$

Thus, the values of $B\text{'}$ and $B$ are $\underset{\_}{9.07\times 10^{-7}P{a}^{-1}}$ and $\underset{\_}{2.25d{m}^{3}mo{l}^{-1}}$ respectively.