Chapter 1, Problem 1C.4P

(a)

Interpretation Introduction

Interpretation: The volume occupied by $1\text{mol}$ of ${\text{N}}_2$, using the van der Waals equation expanded into the form of a virial expansion at its critical temperature has to be calculated.

Concept introduction: The empirical extension of the perfect gas equation which summarises the behaviour of real gases for the range of conditions is called virial equation of state.  The virial equation of state is,

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

Answer to Problem 1C.4P

The volume occupied by $1\text{mol}$ of ${\text{N}}_2$, using the van der Waals equation expanded into the form of a virial expansion at its critical temperature is $\underset{\_}{0.94d{m}^{3}mo{l}^{-1}}$.

Explanation of Solution

The virial equation of state is,

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

Where,

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

Rearrange the equation (1).

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

The value of $B$ can be derived from the van der Waals equation.  The van der Waals equation is,

    $p=\frac{RT}{V_{\text{m}}-b}-\frac{a}{{\left(V_{\text{m}}\right)}^2}$ (3)

Where,

• Ø  $p$ is the pressure.
• Ø  $R$ is the gas constant.
• Ø  $T$ is the temperature.
• Ø  $a$, $b$ are the van der waal parameters.
• Ø  $V_{\text{m}}$ is the molar volume.

Rearrange the equation (3).

  $\begin{array}{l}p=\frac{RT}{V_{\text{m}}-b}-\frac{a}{{\left(V_{\text{m}}\right)}^2}\\ p=\frac{RT}{V_{\text{m}}}\left[\frac{1}{1-b/V_{\text{m}}}-\frac{a}{RT{V}_{\text{m}}}\right]\end{array}$ (4)

Therefore, the bionomial expansion is,

    ${\left(1-x\right)}^{-1}=1+x+x^2+\mathrm{....}$

Since

  $\frac{b}{V_{\text{m}}}<1$

Therefore, apply the binomial expansion to the first term of equation (4).

    $p=\frac{RT}{V_{\text{m}}}\left[1+\frac{b}{V_{\text{m}}}-\frac{a}{RT{V}_{\text{m}}}+\mathrm{...}\right]$

  $p=\frac{RT}{V_{\text{m}}}\left[1+\left(b-\frac{a}{RT}\right)\frac{1}{V_{\text{m}}}+\mathrm{...}\right]$ (5)

Compare equation (2) and (5).

    $B=b-\frac{a}{RT}$

Hence, equation (2) can be rewritten.

    $p=\frac{RT}{V_{\text{m}}}\left(1+\left(b-\frac{a}{RT}\right)\frac{1}{V_{\text{m}}}+\mathrm{.....}\right)$

    $\frac{p{V}_{\text{m}}}{RT}=\left(1+\left(b-\frac{a}{RT}\right)\frac{1}{V_{\text{m}}}+\mathrm{.....}\right)$ (6)

Since,

  $\frac{p{V}_{\text{m}}}{RT}=Z$

Hence, the equation (6) can be rewritten as,

    $\begin{array}{l}\frac{p{V}_{\text{m}}}{RT}=\left(1+\left(b-\frac{a}{RT}\right)\frac{1}{V_{\text{m}}}+\mathrm{.....}\right)\\ Z=1+\left(b-\frac{a}{RT}\right)\frac{p}{RT}+\mathrm{.....}\\ Z=1+\frac{1}{RT}\times \left(b-\frac{a}{RT}\right)\left(p\right)+\mathrm{.....}\end{array}$

Since,

    $\begin{array}{c}Z=\frac{p{V}_{\text{m}}}{RT}\\ V_{\text{m}}=\frac{RT}{p}\left(Z\right)\\ =\frac{RT}{p}\left[1+\left(b-\frac{a}{RT}\right)\left(\frac{p}{RT}\right)+\mathrm{.....}\right]\\ =\frac{RT}{p}+\left(b-\frac{a}{RT}\right)+\mathrm{.....}\end{array}$

The volume occupied by $1\text{mol}$ of ${\text{N}}_2$ has to be calculated at its critical temperature, therefore, temperature, $T$ is equal to $T_{\text{c}}$.

It is given that $T_{\text{c}}=126.3\text{K}$ and $p=10\text{atm}$

The values of the van der Waal parameters for ${\text{N}}_2$ gas are: $a=1.390\text{atm}{\text{dm}}^{\text{6}}{\text{mol}}^{-2}$ and $b=0.0391{\text{dm}}^{\text{3}}{\text{mol}}^{-1}$

Substitute the values of $R,T_{\text{c}},p,b,a$ in the above equation.

  $\begin{array}{c}{V}_{\text{m}}=\frac{0.0821\text{atm}{\text{dm}}^{\text{3}}{\text{K}}^{-1}{\text{mol}}^{-1}\times 126.3\text{K}}{10\text{atm}}+\left(\begin{array}{l}0.0391{\text{dm}}^{\text{3}}{\text{mol}}^{-1}-\\ \frac{1.390\text{atm}{\text{dm}}^{\text{6}}{\text{mol}}^{-2}}{0.0821\text{atm}{\text{dm}}^{\text{3}}{\text{K}}^{-1}{\text{mol}}^{-1}\times 126.3\text{K}}\end{array}\right)\\ =1.037{\text{dm}}^{\text{3}}{\text{mol}}^{-1}+\left(0.0391{\text{dm}}^{\text{3}}{\text{mol}}^{-1}-0.13{\text{dm}}^{\text{3}}{\text{mol}}^{-1}\right)\\ =1.037{\text{dm}}^{\text{3}}{\text{mol}}^{-1}-0.095{\text{dm}}^{\text{3}}{\text{mol}}^{-1}\\ =\underset{\_}{0.94d{m}^{3}mo{l}^{-1}}\end{array}$

Hence, the volume occupied by $1\text{mol}$ of ${\text{N}}_2$, using the van der Waals equation expanded into the form of a virial expansion at its critical temperature is $\underset{\_}{0.94d{m}^{3}mo{l}^{-1}}$.

(b)

Interpretation Introduction

Interpretation: The volume occupied by $1\text{mol}$ of ${\text{N}}_2$, using the van der Waals equation expanded into the form of a virial expansion at its Boyle’s temperature has to be calculated.

Concept introduction: The empirical extension of the perfect gas equation which summarises the behaviour of real gases for the range of conditions is called virial equation of state.  The virial equation of state is,

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

Answer to Problem 1C.4P

The volume occupied by $1\text{mol}$ of ${\text{N}}_2$, using the van der Waals equation expanded into the form of a virial expansion at its Boyle’s temperature is $\underset{\_}{2.67d{m}^{3}mo{l}^{-1}}$.

Explanation of Solution

The virial equation of state is,

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

Where,

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

Rearrange the equation (1).

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

The value of $B$ can be derived from the van der Waals equation.  The van der Waals equation is.

    $p=\frac{RT}{V_{\text{m}}-b}-\frac{a}{{\left(V_{\text{m}}\right)}^2}$ (3)

Where,

• Ø  $p$ is the pressure.
• Ø  $R$ is the gas constant.
• Ø  $T$ is the temperature.
• Ø  $a$, $b$ are the van der waal parameters.
• Ø  $V_{\text{m}}$ is the molar volume.

Rearrange the equation (3).

  $\begin{array}{l}p=\frac{RT}{V_{\text{m}}-b}-\frac{a}{{\left(V_{\text{m}}\right)}^2}\\ p=\frac{RT}{V_{\text{m}}}\left[\frac{1}{1-b/V_{\text{m}}}-\frac{a}{RT{V}_{\text{m}}}\right]\end{array}$ (4)

The bionomial expansion is,

    ${\left(1-x\right)}^{-1}=1+x+x^2+\mathrm{....}$

Since

  $\frac{b}{V_{\text{m}}}<1$

Therefore, apply binomial expansion to the first term of equation (4).

    $p=\frac{RT}{V_{\text{m}}}\left[1+\frac{b}{V_{\text{m}}}-\frac{a}{RT{V}_{\text{m}}}+\mathrm{...}\right]$

  $p=\frac{RT}{V_{\text{m}}}\left[1+\left(b-\frac{a}{RT}\right)\frac{1}{V_{\text{m}}}+\mathrm{...}\right]$ (5)

From equation (2) and (5).

    $B=b-\frac{a}{RT}$

Hence, equation (2) can be rewritten.

    $p=\frac{RT}{V_{\text{m}}}\left(1+\left(b-\frac{a}{RT}\right)\frac{1}{V_{\text{m}}}+\mathrm{.....}\right)$

    $\frac{p{V}_{\text{m}}}{RT}=\left(1+\left(b-\frac{a}{RT}\right)\frac{1}{V_{\text{m}}}+\mathrm{.....}\right)$ (6)

Since,

  $\frac{p{V}_{\text{m}}}{RT}=Z$

Hence, the equation (6) can be rewritten as,

    $\begin{array}{l}\frac{p{V}_{\text{m}}}{RT}=\left(1+\left(b-\frac{a}{RT}\right)\frac{1}{V_{\text{m}}}+\mathrm{.....}\right)\\ Z=1+\left(b-\frac{a}{RT}\right)\frac{p}{RT}+\mathrm{.....}\\ Z=1+\frac{1}{RT}\times \left(b-\frac{a}{RT}\right)\left(p\right)+\mathrm{.....}\end{array}$

Since,

    $\begin{array}{l}Z=\frac{p{V}_{\text{m}}}{RT}\\ V_{\text{m}}=\frac{RT}{p}\left(Z\right)\\ V_{\text{m}}=\frac{RT}{p}\left[1+\left(b-\frac{a}{RT}\right)\left(\frac{p}{RT}\right)+\mathrm{.....}\right]\\ V_{\text{m}}=\frac{RT}{p}+\left(b-\frac{a}{RT}\right)+\mathrm{.....}\end{array}$

The volume occupied by $1\text{mol}$ of ${\text{N}}_2$ has to be calculated at its critical temperature, therefore, temperature, $T$ is equal to $T_B$.

It is given that $T_B=327.2\text{K}$ and $p=10\text{atm}$

The values of the van der Waal parameters for ${\text{N}}_2$ gas are: $a=1.390\text{atm}{\text{dm}}^{\text{6}}{\text{mol}}^{-2}$ and $b=0.0391{\text{dm}}^{\text{3}}{\text{mol}}^{-1}$

Substitute the values of $R,T_B,p,b,a$ in the above equation.

  $\begin{array}{c}{V}_{\text{m}}=\frac{0.0821\text{atm}{\text{dm}}^{\text{3}}{\text{K}}^{-1}{\text{mol}}^{-1}\times 327.2\text{K}}{10\text{atm}}+\left(\begin{array}{l}0.0391{\text{dm}}^{\text{3}}{\text{mol}}^{-1}-\\ \frac{1.390\text{atm}{\text{dm}}^{\text{6}}{\text{mol}}^{-2}}{0.0821\text{atm}{\text{dm}}^{\text{3}}{\text{K}}^{-1}{\text{mol}}^{-1}\times 327.2\text{K}}\end{array}\right)\\ =2.6863{\text{dm}}^{\text{3}}{\text{mol}}^{-1}+\left(0.0391{\text{dm}}^{\text{3}}{\text{mol}}^{-1}-0.052{\text{dm}}^{\text{3}}{\text{mol}}^{-1}\right)\\ =2.6863{\text{dm}}^{\text{3}}{\text{mol}}^{-1}-0.0126{\text{dm}}^{\text{3}}{\text{mol}}^{-1}\\ =\underset{\_}{2.67d{m}^{3}mo{l}^{-1}}\end{array}$

Hence, the volume occupied by $1\text{mol}$ of ${\text{N}}_2$, using the van der Waals equation expanded into the form of a virial expansion at its Boyle’s temperature is $\underset{\_}{2.67d{m}^{3}mo{l}^{-1}}$