Chapter 3.8, Problem 94P

A 3.27-m³ tank contains 100 kg of nitrogen at 175 K. Determine the pressure in the tank using (a) the idealgas equation, (b) the van der Waals equation, and (c) the Beattie-Bridgeman equation. Compare your results with the actual value of 1505 kPa.

To determine

The pressure in the tank using the ideal gas equation.

Answer to Problem 94P

The pressure in the tank using the ideal gas equation is $\underset{\_}{1588\text{kPa}}$.

Explanation of Solution

Determine the specific volume in the tank.

$v=\frac{\nu }{m}$ (I)

Here, the volume of the tank is $\nu$ and the mass of the tank is $m$.

Determine the pressure of the tank using the ideal gas equation.

$P=\frac{RT}{v}$ (II)

Here, the universal gas constant is $R$ and the temperature of the tank is $T$.

Conclusion:

Refer to Table A-1 to find the gas constant, molar mass, the critical pressure, and the critical temperature of nitrogen as $0.2968\text{kPa}\cdot {\text{m}}^{\text{3}}/\text{kg}\cdot \text{K}$, $28.013\text{kg}/\text{kmol}$, $126.2\text{K}$, and $3.39\text{MPa}$.

Substitute $3.27{\text{m}}^{\text{3}}$ for $\nu$ and $100\text{kg}$ for $m$ in Equation (I).

$\begin{array}{c}v=\frac{\left(3.27{\text{m}}^{\text{3}}\right)}{\left(100\text{kg}\right)}\\ =0.0327{\text{m}}^{\text{3}}/\text{kg}\end{array}$

Substitue $0.2968\text{kPa}\cdot {\text{m}}^{\text{3}}/\text{kg}\cdot \text{K}$ for $R$, $175\text{K}$ for $T$, and $0.0327{\text{m}}^{\text{3}}/\text{kg}$ for $v$ in Equation (II).

$\begin{array}{c}P=\frac{\left(0.2968\text{kPa}\cdot {\text{m}}^{\text{3}}/\text{kg}\cdot \text{K}\right)\left(175\text{K}\right)}{\left(0.0327{\text{m}}^{\text{3}}/\text{kg}\right)}\\ =\frac{\left(51.94\text{kPa}\cdot {\text{m}}^{\text{3}}/\text{kg}\right)}{\left(0.0327{\text{m}}^{\text{3}}/\text{kg}\right)}\\ =1588\text{kPa}\end{array}$

Therefore, the error compression with the actual value is 5.5%.

Thus, the pressure in the tank using the ideal gas equation is $\underset{\_}{1588\text{kPa}}$.

To determine

The pressure in the tank using the van der Waals.

Answer to Problem 94P

The pressure in the tank using the van der Waals is $\underset{\_}{1495\text{kPa}}$.

Explanation of Solution

Determine the pressure in the tank using the van der Waals.

$\begin{array}{c}P=\frac{RT}{v-b}-\frac{a}{v^2}\\ =\frac{RT}{v-\left(\frac{R{T}_{\text{cr}}}{8P_{\text{cr}}}\right)}-\frac{\left(\frac{27R^2{T}_{\text{cr}}^2}{64P_{\text{cr}}}\right)}{v^2}\end{array}$ (III)

Here, the critical temperature is $T_{\text{cr}}$, the critical pressure is $P_{\text{cr}}$.

Conclusion:

Substitute $0.2968\text{kPa}\cdot {\text{m}}^{\text{3}}/\text{kg}\cdot \text{K}$ for $R$, $175\text{K}$ for $T$, and $0.0327{\text{m}}^{\text{3}}/\text{kg}$ for $v$, $126.2\text{K}$ for $T_{\text{cr}}$, and $3.39\text{MPa}$ for $P_{\text{cr}}$ in Equation (III).

$\begin{array}{c}P=\left[\begin{array}{l}\frac{\left(0.2968\text{kPa}\cdot {\text{m}}^{\text{3}}/\text{kg}\cdot \text{K}\right)\left(175\text{K}\right)}{\left(0.0327{\text{m}}^{\text{3}}/\text{kg}\right)-\left(\frac{\left(0.2968\text{kPa}\cdot {\text{m}}^{\text{3}}/\text{kg}\cdot \text{K}\right)\left(126.2\text{K}\right)}{8\times \left(3.39\text{MPa}\right)}\right)}\\ -\frac{\frac{\left(0.2968\text{kPa}\cdot {\text{m}}^{\text{3}}/\text{kg}\cdot \text{K}\right){\left(126.2\text{K}\right)}^2}{64\times \left(3.39\text{MPa}\right)}}{\left(0.0327{\text{m}}^{\text{3}}/\text{kg}\right)}\end{array}\right]\\ =\left[\begin{array}{l}\frac{\left(0.2968\text{kPa}\cdot {\text{m}}^{\text{3}}/\text{kg}\cdot \text{K}\right)\left(175\text{K}\right)}{\left(0.0327{\text{m}}^{\text{3}}/\text{kg}\right)-\left(\frac{\left(0.2968\text{kPa}\cdot {\text{m}}^{\text{3}}/\text{kg}\cdot \text{K}\right)\left(126.2\text{K}\right)}{8\times \left(3.39\text{MPa}\times \left(\frac{1000\text{kPa}}{1\text{MPa}}\right)\right)}\right)}\\ -\frac{\frac{\left(0.2968\text{kPa}\cdot {\text{m}}^{\text{3}}/\text{kg}\cdot \text{K}\right){\left(126.2\text{K}\right)}^2}{64\times \left(3.39\text{MPa}\times \left(\frac{1000\text{kPa}}{1\text{MPa}}\right)\right)}}{\left(0.0327{\text{m}}^{\text{3}}/\text{kg}\right)}\end{array}\right]\\ =\left[\begin{array}{l}\frac{\left(0.2968\text{kPa}\cdot {\text{m}}^{\text{3}}/\text{kg}\cdot \text{K}\right)\left(175\text{K}\right)}{\left(0.0327{\text{m}}^{\text{3}}/\text{kg}\right)-\left(0.00138{\text{m}}^{\text{3}}/\text{kg}\right)}\\ -\frac{\left(0.75{{\text{m}}^{\text{6}}\cdot \text{kPa}/\text{kg}}^2\right)}{\left(0.0327{\text{m}}^{\text{3}}/\text{kg}\right)}\end{array}\right]\\ =1495\text{kPa}\end{array}$

Therefore, the error compression with the actual value is 0.7%.

Thus, the pressure in the tank using the van der Waals is $\underset{\_}{1495\text{kPa}}$.

To determine

The pressure in the tank using the Beattie-Bridgeman equation.

Answer to Problem 94P

The pressure in the tank using the Beattie-Bridgeman equation is $\underset{\_}{1504\text{kPa}}$.

Explanation of Solution

Determine the pressure in the tank using the Beattie-Bridgeman equation.

$\begin{array}{c}P=\frac{R_{u}T}{{\overline{v}}^2}\left(1-\frac{c}{\overline{v}{T}^3}\right)\left(\overline{v}+\left(B_{\text{0}}\left(1-\frac{b}{\overline{v}}\right)\right)\right)-\frac{\left(A_0\left(1-\frac{a}{\overline{v}}\right)\right)}{{\overline{v}}^2}\\ =\frac{R_{u}T}{{\left(Mv\right)}^2}\left(1-\frac{c}{\left(Mv\right)T^3}\right)\left(\left(Mv\right)+\left(B_{\text{0}}\left(1-\frac{b}{\left(Mv\right)}\right)\right)\right)-\frac{\left(A_0\left(1-\frac{a}{\left(Mv\right)}\right)\right)}{{\left(Mv\right)}^2}\end{array}$ (IV)

Here, the five constant are $A_0,a,B_0,b,\text{and}c$.

Conclusion:

From the Table 3-4, “Constant that appear in the Beattie-Bridgeman and the Benedict-Webb-Rubin equation of state” to obtain of value of the five constant are $A_0,a,B_0,b,\text{and}c$ as 136.2315, 0.02617, 0.05046, -0.00691, and $4.20\times {10}^4$.

Substitute 136.2315 for $A_0$, 0.02617 for $a$, 0.05046 for $B_0$, -0.00691 for $b$, $4.20\times {10}^4$ for $c$, $0.0327{\text{m}}^{\text{3}}/\text{kg}$ for $v$, $28.013\text{kg}/\text{kmol}$ for $M$ $175\text{K}$ for $T$, and $8.314\text{kPa}\cdot {\text{m}}^{\text{3}}/\text{kmol}\cdot \text{K}$ for $R$ in Eqaution (IV).

$\begin{array}{l}P=\left[\begin{array}{l}\left(\frac{\left(8.314\text{kPa}\cdot {\text{m}}^{\text{3}}/\text{kmol}\cdot \text{K}\right)\left(175\text{K}\right)}{{\left(\left(28.013\text{kg}/\text{kmol}\right)\times \left(0.0327{\text{m}}^{\text{3}}/\text{kg}\right)\right)}^2}\right)\\ \times \left(1-\frac{\left(4.20\times {10}^4{\text{m}}^{\text{3}}\cdot {\text{K}}^{\text{3}}/\text{kmol}\right)}{\left(\left(28.013\text{kg}/\text{kmol}\right)\times \left(0.0327{\text{m}}^{\text{3}}/\text{kg}\right)\right){\left(175\text{K}\right)}^3}\right)\times \\ \left(\begin{array}{l}\left(\left(28.013\text{kg}/\text{kmol}\right)\times \left(0.0327{\text{m}}^{\text{3}}/\text{kg}\right)\right)\\ +\left(\left(0.05046\right)\left(1-\frac{\left(-0.00691\right)}{\left(\left(28.013\text{kg}/\text{kmol}\right)\times \left(0.0327{\text{m}}^{\text{3}}/\text{kg}\right)\right)}\right)\right)\end{array}\right)\\ -\frac{\left(\left(136.2315\right)\left(1-\frac{\left(0.02617\right)}{\left(\left(28.013\text{kg}/\text{kmol}\right)\times \left(0.0327{\text{m}}^{\text{3}}/\text{kg}\right)\right)}\right)\right)}{{\left(\left(28.013\text{kg}/\text{kmol}\right)\times \left(0.0327{\text{m}}^{\text{3}}/\text{kg}\right)\right)}^2}\end{array}\right]\\ =\left[\begin{array}{l}\left(\frac{\left(1454.95\text{kPa}\cdot {\text{m}}^{\text{3}}/\text{kmol}\right)}{{\left(0.9160{\text{m}}^{\text{3}}/\text{kmol}\right)}^2}\right)\times \left(1-\frac{\left(4.20\times {10}^4{\text{m}}^{\text{3}}\cdot {\text{K}}^{\text{3}}/\text{kmol}\right)}{\left(0.9160{\text{m}}^{\text{3}}/\text{kmol}\right)\left(5359375{\text{K}}^3\right)}\right)\\ \times \left(\begin{array}{l}\left(0.9160{\text{m}}^{\text{3}}/\text{kmol}\right)\\ +\left(0.05084{\text{m}}^{\text{3}}/\text{kmol}\right)\end{array}\right)-\frac{\left(132.339{\text{m}}^{\text{3}}/\text{kmol}\right)}{{\left(0.9160{\text{m}}^{\text{3}}/\text{kmol}\right)}^2}\end{array}\right]\\ =\left[\begin{array}{l}\left(\frac{\left(1454.95\text{kPa}\cdot {\text{m}}^{\text{3}}/\text{kmol}\right)}{{\left(0.9160{\text{m}}^{\text{3}}/\text{kmol}\right)}^2}\right)\times \left(1-\frac{\left(4.20\times {10}^4{\text{m}}^{\text{3}}\cdot {\text{K}}^{\text{3}}/\text{kmol}\right)}{\left(0.9160{\text{m}}^{\text{3}}/\text{kmol}\right)\left(5359375{\text{K}}^3\right)}\right)\\ \times \left(\begin{array}{l}\left(0.9160{\text{m}}^{\text{3}}/\text{kmol}\right)\\ +\left(0.05084{\text{m}}^{\text{3}}/\text{kmol}\right)\end{array}\right)-\frac{\left(132.339{\text{m}}^{\text{3}}/\text{kmol}\right)}{\left(0.839056{{\text{m}}^{\text{6}}/\text{kmol}}^2\right)}\end{array}\right]\\ =1504\text{kPa}\end{array}$

Therefore, the error compression with the actual value is 0.07%.

Thus, the pressure in the tank using the Beattie-Bridgeman equation is $\underset{\_}{1504\text{kPa}}$.